LINEAR ALGEBRA A Pathway to Abstract Mathematics 3rd Edition
G. Viglino
Ramapo College of New Jersey January 2018
Contents
CONTENTS
CHAPTER 1 MATRICES AND SYSTEMS OF LINEAR EQUATION 1.1 Systems of Linear Equations 1 1.2 Consistent and Inconsistent Systems of Equations 14 Chapter Summary 27
CHAPTER 2 VECTOR SPACES 2.1 Vectors in the plane, and Beyond 31 2.2 Abstract Vector Spaces 40 2.3 Properties of Vector Spaces 51 2.4 Subspaces 59 2.5 Lines and Planes 68 Chapter Summary 75
CHAPTER 3 BASES AND DIMENSION 3.1 Spanning Sets 77 3.2 Linear Independence 86 3.3 Bases 94 Chapter Summary 109
CHAPTER4 LINEARITY 4.1 Linear Transformations 111 4.2 Image and Kernel 124 4.3 Isomorphisms 135 Chapter Summary 149
CHAPTER 5 MATRICES AND LINEAR MAPS 5.1 Matrix Multiplication 151 5.2 Invertible Matrices 165 5.3 Matrix Representation of Linear Maps 178 5.4 Change of Basis 191 Chapter Summary 202 Contents
CHAPTER 6 DETERMINANTS AND EIGENVECTORS 6.1 Determinants 205 6.2 Eigenspaces 218 6.3 Diagonalization 233 6.4 Applications 246 6.5 Markov Chains 259 Chapter Summary 275
CHAPTER 7 INNER PRODUCT SPACES 7.1 Dot Product 279 7.2 Inner Product 292 7.3 Orthogonality 301 7.4. The Spectral Theorem 315 Chapter Summary 307 Appendix A Principle of Mathematical Induction Appendix B Solutions to Check Your Understanding Boxes
Appendix C Answers to Selected Odd-Numbered Exercises Preface
PREFACE There is no mathematical ramp that will enable you to continuously inch your way higher and higher in mathematics. The climb calls for a ladder consisting of discrete steps designed to take you from one mathematical level to another. You are about to take an important step on that lad- der, one that will take you to a plateau where mathematical abstraction abounds. Linear algebra rests on a small number of axioms (accepted rules, or “laws”), upon which a beautiful and practi- cal theory emerges. Technology can be used to reduce the time needed to perform essential but routine tasks. We fea- ture the TI-84+ calculator, but any graphing utility or Computer Algebraic System will do. The real value of whatever technological tool you use is that it will free you to spend more time on the development and comprehension of the theory and its applications. In any event, if you haven’t already discovered in other courses:
MATHEMATICS DOES NOT RUN ON BATTERIES
Systems of linear equations are introduced and analyzed in Chapter 1. Graphing utilities can be used to solve such systems, but understanding what those solutions represent plays a dominant role throughout the text. We begin Chapter 2 by sowing the seeds for vector spaces in the fertile real number field, where they soon blossom into the concept of an abstract vector. The remainder of Chapter 2 and all of Chapter 3 are dedicated to a study of vector spaces in isolation. Functions from one vector space to another which, in a sense, respect the algebraic structure of those spaces are investigated in Chapters 4 and 5. The sixth chapter focuses on Eigenvalues and Eigenvectors, along with some of their important applications. The first six chapters may provide a full plate for most one-semester courses. If not, then Chap- ter 7 (on inner product spaces) is offered for dessert.
We have made every effort to provide a leg-up for the step you are about to take. Our primary goal was to write a readable book, without compromising mathematical integrity. Along the way, you will encounter numerous Check Your Understanding boxes designed to challenge your under- standing of each newly introduced concept. Complete solutions to the problems in those boxes appear in Appendix B, but please don’t be in too much of a hurry to look at our solutions. You should TRY to solve the problems on your own, for it is only through ATTEMPTING to solve a problem that one grows mathematically. In the words of Descartes: We never understand a thing so well, and make it our own, when we learn it from another, as when we have discovered it for ourselves.
1.1 Systems of Linear Equations 1
1 CHAPTER 1 MATRICES AND SYSTEMS OF LINEAR EQUATIONS Much of the development of linear algebra calls for the solution and interpretation of systems of linear equations. While the “solution part” can be relegated to a calculator, the “interpretation part” cannot. We focus on the solution-part of the process in Section 1, and on the more important interpretation-part in Sections 2. §1. SYSTEMS OF LINEAR EQUATIONS To solve the system of equations: 2x +64y – 4z = 2x ++6y 4z = 0 xy++2z = – 2 is to determine values of the variables (or unknowns) x, y, and z for which each of the three equations is satisfied. You certainly solved such systems in earlier courses, and if you take the time to solve the above Brackets are used to denote sets. In particular, system, you will find that it has but one solution: –11– 1 x ===–11 y z –1 . We can also say that the three-tuple denotes the set containing but one element—the ele- –11– 1 is a solution of the given system of equation, and that ment –11– 1 . –11– 1 is its solution set. In general: An (ordered) n-tuple is an expression of the form c1c2 cn , where each ci is a real number (written ci ), for 1 in .
We say that the n-tuple c1c2 cn is a solution of the system of m equations in n unknowns:
The x s denote variables a11x1 ++a12x2 a1nxn = b1 i (or unknowns), while the a x ++a x a x = b 21 1 22 2 2n n 2 aij ’s and bi ’s are con- . ... stants (or scalars). . ... . ... am1x1 ++am2x2 amnxn = bm
if each equation in the system is satisfied when ci is substi- tuted for xi , for 1 in . The set of all solutions of a system of equations is said to be the solution set of that system. 2 Chapter 1 Matrices and Systems of Linear Equations
EQUIVALENT SYSTEMS OF EQUATIONS Consider the system of equations: – 3x +2y = x 1 ---2+ y = --- 2 2 As you know, you can perform certain operations on that system which will not alter its solution set. For example, you can: (1) Interchange the order of the equations: – 3x +2y = x 1 ---2+ y = --- x 1 2 2 ---2+ y = --- 2 2 – 3x +2y = (2) Multiply both sides of the resulting top equation by 2: x 1 ---2+ y = --- x +14y = 2 2 – 3x +2y = – 3x +2y = (3) Multiply the resulting top equation by 3 and add it to the bottom You used this third equation: maneuver a lot when eliminating a variable x +14y = x +14y = from a given system of equations For example: – 3x +2y = 13y = 5 (i) x +13yz– = multiply by 3 3x +312y = (2) 2x –35y +3z = – 3x +2y = (3) – 3x ++y 2z = 2 add: 13y = 5 multiply (1) by -2 and add it to (2) The above three operations, are said to be elementary equation –511y +1z = operations: 10yz–5= multiply (1) by 3 and add it ELEMENTARY OPERATIONS ON to (3) SYSTEMS OF LINEAR EQUATIONS Interchange the order of any two equations in the system. Multiply both sides of an equation by a nonzero number. Add a multiple of one equation to another equation.
EQUIVALENT SYSTEM Two systems of equations sharing a common solution set are said to OF EQUATIONS be equivalent. As you may recall: THEOREM 1.1 Performing any sequence of elementary opera- tions on a system of linear equations will result in an equivalent system of equations. 1.1 Systems of Linear Equations 3
AUGMENTED MATRICES Matrices are arrays of numbers (or expressions representing numbers) arranged in rows and columns:
47 17 0 10 2134 10 6 36 5 12 7 4 1 –739 –38 11 2– 12 9 (i) (ii) (iii) (iv) (v) Matrix (i) contains 2 rows and 3 columns and it is said to be a 23 (two-by-three) matrix. Similarly, (ii) is a 32 matrix, and (iii) is a 33 matrix (a square matrix). In general, an mn matrix is a matrix consisting of m rows and n columns. In particular, (iv) is a 13 matrix (a row matrix), and (v) is a 31 matrix (a column matrix). It is often convenient to represent a system of equations in a more com- pact matrix form. The rows of the matrix in Figure 1.1(b), for example, concisely represents the equations in Figure 1.1(a). Note that the vari- ables x, y, and z are suppressed in the matrix form, and that the vertical line recalls the equal sign in the equations. Such a matrix is said to be the augmented matrix associated with the given system of equations. 2x +64y – 4z = 24–6 4 2x ++6y 4z = 0 26 4 0 AUGMENTED MATRIX xy++2z = – 2 11 2– 2 System of Equations Augmented Matrix (a) (b) Figure 1.1 Switching two equations in a system of equations results in the switching of the corresponding rows in the associated augmented matrix. Indeed, each of the three previously introduced elementary equation operations corresponds with one of the following elementary matrix row operations:
ELEMENTARY MATRIX ROW OPERATIONS Interchange the order of any two rows in the matrix. Multiply each element in a row of the matrix by a nonzero number. Add a multiple of one row of the matrix to another row of the matrix. The following terminology is motivated by Theorem 1.1: DEFINITION 1.1 Two matrices are equivalent if one can be EQUIVALENT derived from the other by performing elemen- MATRICES tary row operations. 4 Chapter 1 Matrices and Systems of Linear Equations
HERE IS WHERE WE ARE AT THIS POINT: A system of linear equations can be represented by an aug- mented matrix, and every augmented matrix represents a sys- tem of linear equations. Moreover:
SYSTEMS OF EQUATIONS ASSOCIATED WITH EQUIVA- LENT AUGMENTED MATRICES ARE THEMSELVES EQUIVALENT (SAME SOLUTION SET).
AND HERE IS WHERE WE ARE GOING: Suppose you want to solve the system of equations [1] in Figure 1.2. Assume, for the time being, that you can go from its augmented matrix ([2]) to matrix [3], via elementary row operations. System [4], which is associated with the augmented matrix [3], is easily seen to have the solution: x ===–1 y 1 z –1 . But this must also be the solution of system [1], since the two systems of equations are also equivalent!
2x +64y – 4z = 24–6 4 augmented matrix [1] 2x ++6y 4z = 0 2640 [2] xy++2z = – 2 112– 2
Same Solution Set Elementary Row Operations
x ++0y 0z = – 1 100– 1 [4] system of equations 0xy++0z = 1 0101 [3] 0x ++0yz= – 1 001– 1 Figure 1.2 The remainder of this section is designed to illustrate a method which can be used to go from matrix [2] of Figure 1.2 to matrix [3], via ele- mentary row operations.
PIVOTING ABOUT A PIVOT POINT Capital letters are used to represent matrices, and double subscripted lower case letters for their entries; as in:
a11 a12 a13 a14
A = a21 a22 a23 a24
a31 a32 a33 a34 1.1 Systems of Linear Equations 5
Note that the first subscript of the element aij denotes its row: i, and the second subscript, its column: j. For example, if: 2762 Aa==ij 3631 –2231
then a12 ===7 a21 3 a34 3 , and so on. In the next example, we specify a location in a given matrix (called PIVOT POINT the pivot-point), which contains a non-zero entry. We then illustrate a PIVOTING process (called pivoting) designed to turn the given matrix into an equivalent matrix with a 1 in the pivot-point, and with each entry above or below the pivot-point equal to 0. It is a routine process that plays a dominant role in a number of matrix applications, so please make sure that you understand it fully. The following notation will be used to represent elementary row operations:
ELEMENTARY ROW OPERATION NOTATION Switch row i with row j: Ri Rj
Multiply each entry in row i by a nonzero cR R number c: i i
Multiply each entry in row i by a number cR + R R c, and add the resulting row to row j: i j j
EXAMPLE 1.1 Pivot the matrix: 2 64–2 Aa==ij 36315 –2231
about the pivot point a11 with pivot entry 2, and then again about the pivot point a22 of the resulting equivalent matrix.
SOLUTION: Step 1. Get a 1 in the pivot-point position by multiplying each entry
in row 1 by --1- : 2 1 ---R1 R1 2 64–22 1 32–1 36315 36315 –2231 –2231 6 Chapter 1 Matrices and Systems of Linear Equations
Step 2.Get 0’s below (there is no above) the pivot point position. Multiply row 1 by –3 and add it to row 2 (see margin), and then multiply row 1 by 1 and add it to row 3: –3 1 3 – 2 1
–3R 1 32–1 1 32–1 1 32–1 1 –93 –63– – 3R + R R 1 2 2 1R1 + R3 R3 R 2 36315 36315 0 –9123 0 –9123 – 3R +:R 1 2 0 –9123 –2231 –2231 0 504
13–1 2 –2231 0 504 Repeating the above two-step procedure, we now pivot about
a22 = –3 :
1 32–11 –---R R 1 32–1 1 0 713 1 0 713 3 2 2 – 3R2 + R1 R1 – 5R + R R 0 –9123 0 1 –43 – 0 1 –43 – 2 3 3 0 1 –43 – 0 504 0 50 4 0 50 4 0015 24
03–912 05–1520 13–1 2 0504 1 0 713 0015 24
GRAPHING CALCULATOR GLIMPSE 1.1 We utilize a graphing calculator to perform the first of the two pivot- ing processes in the above example, and invite you to use your calcu- lator to address the other pivoting process. The TI-84+ calculator is fea- tured throughout the text. 1.1 Systems of Linear Equations 7
CHECK YOUR UNDERSTANDING 1.1
10 7 13 ??0 ? Pivot about a33 = 15 to go from: 01–4 3– to ??0 ? Answer: See page B-1. 0015 24 ??1 ?
ROW-REDUCED-ECHELON FORM A matrix may have many different equivalent forms. Here is the nic- est of them all: DEFINITION 1.2 A matrix is in row-reduced-echelon form A matrix satisfying (i), (ii) when it satisfies the following three conditions: and a slightly weaker form ROW-REDUCED of (i): ECHELON FORM (i) The first non-zero entry in any row is 1 The first non-zero entry (called its leading-one), and all of the entries in any row is 1, and the above or below that leading-one are 0. entries below (only) that leading-one are 0 (ii) In any two successive rows, not consisting is said to be in row-echelon entirely of zeros, the leading-one in the form. lower row appears further to the right than the leading-one in the row above it. (iii) All of the rows that consist entirely of zeros are at the bottom of the matrix. These three matrices are in row-reduced-echelon form: The matrices 10012 1000 10012 19012 103 and A = 010 5 B = 0011 C = 010 5 014 5 010 001 0 001 0 001 0 0000 are in row-echelon form
CHECK YOUR UNDERSTANDING 1.2 Determine if the given matrix is in row-reduced-echelon form. If not, list the condition(s) of Definition 1.2 which are not satisfied.
0010 10–0 3 01–05 3 100 a 0000 b 0001 c d 00013 Answer: Yes: (a), (c), and (d). 014 No: (b) [fails (ii)] 0000 0100 00000 Though a bit tedious, reducing a matrix to its row-reduced-echelon form is a routine task. Just focus on getting those all-important leading- ones (which are to be positioned further to the right as you move down), with zeros above and below them. Consider the following example: 8 Chapter 1 Matrices and Systems of Linear Equations
EXAMPLE 1.2 Perform elementary row operations to obtain the row-reduced-echelon form for the matrix: 24–6 4 26 4 0 11 2– 2
SOLUTION: Leading-one Step. We could divide the first row by 2 to get a leading-one in that row, but choose to switch the first row and third row instead:
24–6 4 1 12– 2 R R 26 4 0 1 3 26 4 0 11 2– 2 24–6 4 Zeros-above-and-below Step: –22 –4–4 –22 –4–4 2640 24–6 4 0 404 0 28–10
1 12– 2 – 2R + R R 1 12– 2 1 12– 2 1 2 2 – 2R1 + R3 R3 26 4 0 0 40 4 0 40 4 24–6 4 24–6 4 0 28–10 Next leading-one Step:
1 12– 2 1 1 12– 2 ---R2 R2 0 40 4 4 0 1 01 0 28–10 0 28–10
Zeros-above-and-below Step: 01–01– 02–02– 112– 2 02–10 8 1 0 23– 00–88
1 12– 2 – 1R + R R 1 0 23– 1 0 23– 2 1 1 – 2R2 + R3 R3 0 1 01 0 1 01 0 1 01 0 28–10 0 28–10 00–88 1.1 Systems of Linear Equations 9
Next leading-one Step:
1 0 23– 1 1 0 23– –---R3 R3 0 1 01 8 0 1 01 00–88 001 –1
Zeros-above-and-below Step:
00–2 2 102– 3 1 00–1
1 0 23– 1 00–1 – 2R3 + R1 R1 0 1 01 0 1 0 1 001 –1 001 –1 We are now at a row-reduced-echelon form, and so we stop.
While not difficult, the above example illustrates that obtaining the row-reduced-echelon form of a matrix can be a bit tedious. It’s a dirty job, but someone has to do it:
GRAPHING CALCULATOR GLIMPSE 1.2
Row-reduced-echelon form
In harmony with graphing calculators, we will adopt the notation rrefA to denote the row-reduced-echelon form of a matrix A.
EXAMPLE 1.3 Solve the system: 2x +64y – 4z = 2x ++6y 4z = 0 xy++2z = – 2 10 Chapter 1 Matrices and Systems of Linear Equations
SOLUTION: All the work has been done: Example 1.2 augmented matrix x y z x y z 2x +64y – 4z = x = –1 24–6 4 100– 1 2x ++6y 4z = 0 26 4 0 010 1 y = 1 xy++2z = – 2 11 2– 2 001– 1 z = –1
These two systems of equations are equivalent (same solution sets) From the above we can easily spot the solution of the given system:
x ===–1 y 1 z –1
CHECK YOUR UNDERSTANDING 1.3 Proceed as in Example 1.3 to solve the given system of equations. xyz++= 6 3x +42yz– = Answer: x ===1 y 2 z 3 3xy++2z = 11
A bit of human-intervention was used in the pivoting process of Example 1.2. If it is “freedom from choice” that you want, then you can use the following algorithm to reduce a given matrix to its row- reduced-echelon form: Gauss, Karl Friedrich (1777 -1855), the great German mathematician Gauss-Jordan Elimination Method and astronomer. Wilhelm Jordan (1842- Step 1. Locate the left-most column that does not consist 1899) German professor entirely of zeros, and pick a nonzero element in that of geodesy. column. Let the position of that chosen element be the pivot-point. Step 2. Pivot about the pivot-point of Step 1. Step 3. If necessary, switch the pivot-row with the furthest row above it (nearest the top) that does not already contain a leading-one, to the left of the pivot col- umn. Step 4. If the matrix is in row-reduced-echelon form, then you are done. If not, return to Step 1, but ignore all rows with established leading-ones for that step of the process. 1.1 Systems of Linear Equations 11
EXERCISES
Exercises 1-2. Write down the augmented matrix associated with the given system of equations.
3x – 3y +2z = 2x +53y – 4w = 1. 5x +15y – 9z = – x – 4z +1w = – 2. –43x – y +0z = x –04y = – x – y ++z 4w = 9
Exercises 3-4. Write down the system of equations associated with the given augmented matrix.
514 3 2410 9 3. –32 –1 4 4. 0552 2 1 --- –01 0 21–8 3 11 2
Exercises 5-8. Perform elementary row operations to obtain the row-reduced echelon form for the given matrix.
024 1002 025 2301 5. 6. 102 2121 7.442 8. 1012 245 0224 103 1001 235 2201
Exercises 9-11. Solve the system of equations corresponding to the given row-reduced-echelon form matrix. x y z w x y z x1 x2 x3 x4 x5 10002 10 01 100001 9. 0100– 3 010002 01 00 10. 11. 00 12 00101 001002 00010 000102 00001– 1
Exercises 12-15. Proceed as in Example 1.2 to solve the given system of equations.
x – 2y +1z = xy– –2z = 12.–53x +5y – 2z = – 13. 4x –52y –2z = – 4x –38y +6z = –3x ++y 6z = 0 12 Chapter 1 Matrices and Systems of Linear Equations
2xy–2= zw++1 2x +15y – 2z = – wx– = y 14. – 4x – y +4z = – 15. 1 4y + 3z = –--- x +42yz– = 2 x + 3y = wz–
16. Construct a system of three equations in three unknowns, x, y, and z such that x ===1 y 2 z 3 is a solution of the system. 17. Construct a system of four equations in four unknowns, x, y, z, and w with solution set x ===1,2 y ,3 z ,4 w =. Exercises 18-20. (Row-Echelon Form) A matrix is said to be in row-echelon form if it satisfies all of the conditions of Definition 1.2, except that elements above a leading 1 need not be zero (the entries below leading ones must still be zero). Determine if the matrix is in row-echelon form. If not, indicate why not.
1231 1032 0012 18.0123 19.0120 20. 0100 0001 0002 0001
Exercises 21-22. Perform elementary row operations to transform the given matrix to the given row-echelon form (see Exercise 18-20).
24–6 4 12–3 2 13–1 2 26–2 4 21. 2640 01 4– 3 01–4 3– 22. 36315 112– 2 00 1– 1 8 –2231 001--- 5
Exercises 23-25. Determine the solution set of the system of equations corresponding to the given row-echelon form matrix (see Exercise 18-20). Note: If you are using a graphing calculator, then you might as well use the row-reduced- echelon command, for that is the most revealing form. If you are doing things by hand, however, you may be able to save some time by going with the row-echelon form. x y z x y z x y z w 12 01 11 02 12012 23.01 10 24. 01 12 25. 0123– 3 00 12 00 10 00111 00010 26. Offer an argument to justify the following claim: If the jth column of a matrix A consists entirely of zeros, and if the matrix B is equiva- lent to A, then the jth column of B also consists entirely of zeros. 1.1 Systems of Linear Equations 13
In the remaining exercises you are to decide whether the given statement is True or False. If True, then you are to present a general argument to establish the validity of the statement in its most general setting. If False, then you are to exhibit a concrete specific example, called a counterexample, showing that the given statement does not hold in general. To illustrate: Prove or Give a Counterexample: (a) The sum of any two even integers is again an even integer. (b) Every odd number is prime. (a) Yes, each time you add two even integers, out pops another even integer, suggesting that statement (a) is true. But you certainly can’t check to see if (a) holds for all even integers—case by case—as there are infinitely many such cases. A general argument is needed: If a and b are even integers, then a = 2n and b = 2m for some integers n and m. We then have: ab+2==n +22m nm+ . Since ab+ is itself a multiple of 2, it is even. (b) Surely (b) is false. Why? Because 9 is odd, but 9 is not prime, that’s why. To be sure, we could offer a different counterexample, say 15, or 55, but we did have to come up with a specific concrete counterexample to shoot down the claim.
PROVE OR GIVE A COUNTEREXAMPLE
ax+0 by = 27. The system of equations has a solution for all abcd . cx+0 dy =
ax+1 by = 28. The system of equations always has a solution for all abcd . cx+1 dy =
ax+0 by = 29. The system of equations can never have more than one solution. cx+0 dy = 30. The systems of equations associated with the two augmented matrices:
ab 0 and a b 0 cd 0 c d 0 will have the same solution set only if aa===', b b', c c' , and dd= ' .
31. If the matrix A has n rows, and if rrefA contains less than n leading ones, then the last row of rrefA must consist entirely of zeros. 14 Chapter 1 Matrices and Systems of Linear Equations
1
§2. CONSISTENT AND INCONSISTENT SYSTEMS OF EQUATIONS A system of equations may have a unique solution, infinitely many solutions, or no solution whatsoever. If it has no solution, then the sys- CONSISTENT tem is said to be inconsistent, otherwise it is said to be consistent. As INCONSISTENT is illustrated in the following examples, the solution set of any system of equations can be spotted from the row-reduced-echelon form of its augmented matrix. EXAMPLE 1.4 Determine if the following system of equations is consistent. If so, find its solution set. 4x –72y –5z = –56x ++y 10z = – 11 –23x ++y 5z = – 5
SOLUTION: Proceeding as in the previous section, we have: x y z x y z 4x –72y –5z = x = 6 42–7–5 1 006 –56x ++y 10z = – 11 –6 5 10– 11 rref 0 1 01– y = –1 –23x ++y 5z = – 5 –2553 – 001 3 z = 3 We see that the given system is consistent, and that it has but one solution: x ==6 y –1 z =3 .
EXAMPLE 1.5 Determine if the following system of equations is consistent. If so, find its solution set. 3x –72y –5z = –56x ++y 10z = – 11 –32x ++y 4z = – 3 –23x ++y 5z = – 5 SOLUTION: Proceeding as in the previous section, we have: x y z x y z 3x –72y –5z = 32–7–5 1 0 0 0 –56x ++y 10z = – 11 –6 5 10– 11 0 1 0 0 rref –32x ++y 4z = – 3 –3432 – 00 1 0 –23x ++y 5z = – 5 –2553 – 00 0 1 0x ++0y 0z = 1 Since the equation represented by the last row in the above rref-matrix cannot be satisfied, the given system of equations is inconsistent. 1.2 Consistent and Inconsistent Systems of Equations 15
EXAMPLE 1.6 Determine the solution set of the system: 3x –36y +9w = –42x ++y 2zw– = – 11 3x –68y ++z 7w = – 5 1 SOLUTION: x y z w x y z w 1 0002 3x –36y +9w = 36–03 9 1 1 2 rref 0 1 0 –--- –--- -1/2 –42x ++y 2zw– = – 11 –4212 – –11 2 2 -5/2 3x –68y ++z 7w = – 5 1 5 38–67–5 001 --- –--- 2 2 Figure 1.3 We know that the solution set of the above system of equations coin- cides with that of the one stemming from the row-reduced-echelon form of its augmented matrix; namely: x +++0y 0z 0w = 2 x = 2 1 1 1 1 0xy++0z – ---w = –--- y = – --- + ---w 2 2 or: 2 2 1 5 1 0x +++0yz---w = –--5- z = – --- – ---w 2 2 2 2 Any variable that is not associated with a leading As you can see, that variable w, which we moved to the right of one in the row-reduced the equations, can be assigned any value whatsoever, after which echelon form of an aug- the values of x, y, and z (the variables associated with leading- mented matrix is said to be a free variable. In the cur- ones in Figure 1.3) are determined. For example, setting w = 0 rent setting, the variable w leads to the particular solution: is a free variable (see rref 1 5 in Figure 1.3). x ==2 y –--- z =–--- w =0 2 2 We can generate another solution by letting w = 1 : x ===2 y 0 z –3 w =1 Indeed, the solutions set of the system of equation is obtained by letting wc= , where c can be any real number whatsoever:
1 c 5 c x ==2 y – --- + --- z =–--- – --- w =c c 2 2 2 2 Read: such that We can arrive at a nicer representation of the solution set by replacing each c with 2c: 1 5 x ==2 y – --- + cz =–2--- – c w =c 2c 2 2 and then observing that as “2c runs through all of the numbers,” so does c: 1 5 x ==2 y – --- + cz =–2--- – c w =c c 2 2 16 Chapter 1 Matrices and Systems of Linear Equations
The system of equations of Example 1.4 has a unique solution, that of Example 1.5 has no solutions, and the one in Example 1.6 has infinitely many solutions. These examples cover all of the bases, for if a system of equations has more than one solution then it must have infinitely many solutions (Exercise 22).
CHECK YOUR UNDERSTANDING 1.4 Determine if the system associated with the given row-reduced-eche- lon augmented matrix is consistent. If it is, find its solution set.
Answer: (a) Inconsistent 10– 2 1 10– 2 1 1 2 01 1 (b) 12+ r 45– rr r (a) 01 5 4 (b) 01 5 4 (c) –2 (c): 0 014 12– r – sr –42 – ssrs R 00 0 2 00 0 0 0000 0
Our concern thus far has been with systems of equations with fixed con- stants on the right side of the equations. We now turn to the question of whether or not a system of equations has a solution for all such constants: EXAMPLE 1.7 Determine if the following system of equations is consistent for all a, b, and c. 2xz+ = a 3xy+ = b –5x – y – z = c
SOLUTION: x y z 1 a 2xz+ = a 201a 1 1 0 ------ ---R1 R1 2 2 3xy+ = b 310b 2 310b –5x – y – z = c –51 –1– c –51 –1– c
–3R1 + R2 R2 1R1 + R3 R3
1 a 1 0 ------1 a 2 2 1 0 ------2 2 3 3a 5R2 + R3 R3 Unlike the TI-84+, the TI-89 0 1 –--- –------+ b 3 3a and above have symbolic capa- 2 2 0 1 –--- – ------+ b bilities. In particular: 2 2 1 a 0 – 5–------+ c 00 –8 – 7a ++5bc 2 2 1 –---R R 8 3 3 x y z 1 a a ++5bc 1 0 ------1 0 0 ------2 2 16 1 3 3a –---R + R R – 3a + b – 3c 0 1 –--- – ------+ b 2 3 1 1 0 1 0 ------2 2 3 16 ---R + R R 7a – 5b – c 2 3 2 2 7a – 5b – c 00 1 ------001 ------8 8 1.2 Consistent and Inconsistent Systems of Equations 17
We see that the given system of equations has a solution for all a, b, a ++5bc – 3a + b – 3c 7a – 5b – c and c; namely: x = ------ y = ------ z = ------16 16 8 EXAMPLE 1.8 Determine if the following system of equations is consistent for all possible values of a, b, c, and d. 2xy++2zw + = a x –32y ++z 2w = b 2yz++4w = c xz–4– w = d SOLUTION: If you go through the Gauss-Jordon elimination method without making a mistakes you will find that: x y z w 2xy++2zw + = a 2121a x –32y ++z 2w = b 12–32b 0214c 0x +++2yz4w = c 10–4 1– d x + 0yz–4– w = d x y z w 8a –73b – c 1 0 0 – 2 ------13 Here, unlike with the a –42b + c 0 1 0 1 ------smaller system of equa- 13 tions in Example 1.8, the rref –42a ++b 5c TI-89 (or higher) is of lit- 0 0 1 2 ------tle help: 13 –710a ++b 12c +13d 0 0 0 0 ------13 Figure 1.4 The last row of the above rref-matrix represents the equation: 0x +++0y 0z 0w = ------–710a ++b 12c +13d 13 As such, that matrix tells us that the given system of equation will have a solution if and only if: –710a +b +12c +13d = 0 (*) The last row of the above In particular, if you choose random numbers for a, b, c, and d, then it rref matrix tec ++3ab= 0 lls us that is very unlikely that the system will have a solution, for what are the there is no solution to the odds that those four numbers will satisfy (*)? system, but it “lies,” for solutions do exist for cer- tain values of a, b, c, and d CHECK YOUR UNDERSTANDING 1.5 [see (*)]. Determine if the given system of equations has a solution for all a, b, and c. If not, find some specific values of a, b, and c for which a solu- tion does not exist. 4x – 2y + z = a x –44y – z = a Answer: (a) It is consistent for all a, b, and c. (a) –42x ++y 2z = b (b) 2x + 8y – 12z = b (b) Consistent if and only if c ++3ab= 0 5xy–4+ z = c –12x ++y 2z = c 18 Chapter 1 Matrices and Systems of Linear Equations
COEFFICIENT MATRIX The coefficient matrix of an mn system of equation is the mn matrix obtained by eliminating the last column of the augmented matrix of the system. For example, referring to system of equations of Examples 1.8, we have: augmented 2121a matrix 12–32b 2xy++2zw + = a 0214c x –32y ++z 2w = b 10–4 1– d 2yz++4w = c 21 2 1 xz–4– w = d coefficient matrix 12–32 0214 10–4 1– At this point, it behooves us to introduce a bit of notation. To begin with, we will use S to represent a general system of linear equations. We will then let aug(S) and coef (S) denote the augmented and coeffi- cient matrices of S, respectively. To illustrate: 2x +13yz– = 23–1 1 23– 1 For S: 3xy–2+5z = augS ==31–25 and coefS 31–2 –2x +2y – 3z = –231 –2 –231 – The following theorem will enable us to invoke a graphing calculator to resolve the issues of Examples 1.7 and 1.8: THEOREM 1.2 The system of equations: Let P and Q be two proposi- SPANNING a11x1 ++a12x2 a1nxn = b1 tions (a proposition is a math- THEOREM ematical statement that is a x ++a x a x = b S: 21 1 22 2 2n n 2 either true or false). To say “P . ... if and only if Q,” (also writ- ...... ten in the form PQ ) is to a x ++a x a x = b say that if P is true then so is m1 1 m2 2 mn n m Q (also written PQ ), and has a solution for all values of b1b2 bm if if Q is true then so is P (also written QP ). and only if rref coefS does not contain a row consisting entirely of zeros.
PROOF: If rref coefS does not contain a row consisting entirely of free: set to 0 zeros, then each row of rref coefS will have a leading one, as will x x x x x x 1 2 3 4 5 6 every row of rref augS . For any given values of b1b2 bm , a 120003– 9 solution for S can then be obtained by setting each of the nm– free 0010012 variables in rref aug S to zero, and letting the variable associated 0001001 th 0000155 with a leading one in the i row of rref augS equal the last entry in that row (see margin for an illustration). a solution:–902150 1.2 Consistent and Inconsistent Systems of Equations 19
For the converse, assume that the last row of rref coefS consists entirely of zeros. The only difference between rref coefS and rref augS is that the latter has an additional column, the last entry of which (as was the case in Figure 1.6) must be a linear expression involving b1b2 bm , say Fb1b2 bm :
x1 x2 x3 . . . xm x1 x2 x3 . . . xm same
0 0 0 . . . 0 0 0 0 . . . 0 Fb1b2 bm rref coefS rref augS (see Example 1.8)
It follows that for any values of b1b2 bm for which Fb1b2 bm 0 , the resulting system of equations will not have a solution, for here is its last equation:
0x1 +++0x2 0xn = Fb1b2 bm
EXAMPLE 1.9 Use the spanning theorem to determine if the given system of equations has a solution for all values of the constants on the right side of the equations. 2xy++2zw + = a 2xz+ = a x –32y ++z 2w = b (a) 3xy+ = b (b) 2yz++4w = c –5x – y – z = c xz–4– w = d See Example 1.7 See Example 1.8
SOLUTION: 2xz+ = a 201 1 00 (a) S: 3xy+ = b coef(S) 310 rref coef S 0 1 0 (see margin) –5x – y – z = c –51 –1– 001
does not contain a row of zeros: system has a solution for all values of a, b, and c
2xy++2zw + = a 2121 1 00– 2 x –32y ++z 2w = b coef(S) 12–32rref coefS 0 1 01 (b) S: 2yz++4w = c 0214(see margin) 001 2 xz–4– w = d 10–4 1– 000 0
contain a row of zeros: system does not have a solution for all values of a, b, c, and d 20 Chapter 1 Matrices and Systems of Linear Equations
Note that while the matrix rref coefS in (b) shows that the sys- tem S is not consistent for all values of a, b, c, and d, it does not reveal the specific values of a, b, c, and d for which a solution does exist. That information can be derived from the matrix rref augS (see Exam- ple 1.8).
CHECK YOUR UNDERSTANDING 1.6 Use the spanning theorem to determine if the given system of equa- tions has a solution for all values of the constants on the right side of the equations. x – 3y + w = a 3x + 7yz– = a 3xy–2+ z – 3w = b (a) 13x –24y + z = b (b) xz+ – 5w = c 2x –24y + z = c Answer: (a) Yes (b) No 2xy– + z – 2w = d
HOMOGENEOUS SYSTEMS OF EQUATIONS A system of linear equations is said to be homogeneous if all of the A system with fewer equa- constants on the right side of the equations are zero: tions than unknowns (“wide”) is said to be a11x1 + a12x2 + + a1nxn =0 underdetermined. H: a x + a x + + a x =0 A system with more equa- 21 1 22 . 2 2n. n . tions than unknowns . . . . (“tall”) is said to be overde- . . . . termined. am1x1 + am2x2 + + amnxn =0 A square system is a system which contains as many (A homogeneous system of m equations in n unknowns) equations as unknowns. It is easy to see that every homogeneous system is consistent, with trivial solution: x1 ==0 x2 0 xn =0 . In the event that the homogeneous system is “wide”, then it has more than one solution:
THEOREM 1.3 Any homogeneous system S of m linear equa- FUNDAMENTAL THEO- tions in n unknowns with nm has nontrivial REM OF HOMOGENEOUS SYSTEMS OF EQUATIONS solutions.
PROOF: Having more columns than rows, rref augS must have free variables, and therefore the system has infinitely many solutions. EXAMPLE 1.10 Determine the solution set of: 2x ++03y –54z w = – 3x ++y 4zw += 0 x ++07y –114z w = 1.2 Consistent and Inconsistent Systems of Equations 21
SOLUTION: Theorem 1.3 tells us that the system has nontrivial solu- tions. Let’s find them: x y z w 2x ++03y –54z w = 23–50 4 aug S S: – 3x ++y 4zw += 0 –14103 x ++07y –114z w = 17–110 4
x y z w free variables 16 2 16 2 211 1 0 –------0 x = ------z – ------w 17 11 11 11 11 11 rref 4 17 4 17 0 1 –------0 y = ------z – ------w 11 11 11 11 00 0 0 0 Figure 1.5 Assigning arbitrary values to the two free variables z and w we arrive at the solution set of the system: x y z w
} 16 2 4 17} ------a – ------b ------a – ------bab ab R = 16a – 2b 4a – 17b11a 11b ab R 11 11 11 11 If S is a homogeneous system of equations, then the last column of rref augS will always consist entirely of zeros (Exercise 20). Con- sequently, when solving a homogeneous system of equations, one might as well start with coefS rather than with augS (one less col- umn to carry along in the rref-process, that’s all). In particular, the solu- tion set of the homogeneous system of the last example can easily be read from rref coefS (just mentally add a column of zeros to the right of the matrix):
x y z w x y z w 2 1 0 –-----16------0 2x ++03y –54z w = 23–5 4 11 11 coef S rref – 3x ++y 4zw += 0 4 17 S : –1413 0 1 –------0 11 11 x ++07y –114z w = 17–11 4 0 0 0 0 0
CHECK YOUR UNDERSTANDING 1.7 Determine the solution set of: 2x +++3y 4z 5w = 0 3xy++4zw += 0 Answer: x +++7y 4z 11w = 0 4r –2r–23r r rR 22 Chapter 1 Matrices and Systems of Linear Equations
While underdetermined (“wide”) homogeneous systems of equations are guaranteed to always have non-trivial solu- tions, this is not the case with overdetermined (“tall”) sys- tems of equations [see Exercises 27-28], or with square systems of equations [see Exercises 29-30].
We end this section with a rather obvious result, but one that will play an important role in future developments; so much so, that we label it accordingly: THEOREM 1.4 A homogeneous system S of m linear equa- LINEAR INDEPEN- tions in n unknowns has only the trivial solu- DENCE THEOREM tion if and only if rref coefS has n leading ones.
PROOF: Since there are n unknowns, to say that rref coefS has n leading ones, is to say that it has no free variables. 1.2 Consistent and Inconsistent Systems of Equations 23
EXERCISES
Exercises 1-6. Determine if the system S with given rref augS is consistent. If it is, find its solution set.
01 2 0101 3 100 –2 1. 00 0 2. 0010 –2 3. 010 –2 00 0 0000 0 001 –2
10 3 5 1000130 100005–0 2 5. 4. 01– 4 2 0010021 6. 010002–3 1 00 0 1 0001201 001000 1 2 000010 0 0
Exercises 7-12. Determine if the system of equations is consistent. If it is, find its solution set.
2x ++3yz= 4 x –44y –1z = 4x – 2y +4z = 7. 8. xy++2z = 5 2xy+8– 2z = 9. –42x ++y 2z = 10 5xy–4+2z =
x –44y –1z = 2x ++3yz–2w = 4 – x +++wyz= 3 10. 2x +88y – 12z = xy–2+3zw– = 6x ++44z –32y w = 11. 12. –12x ++y 2z = 1 2yz+1– 2w = 5y –63x – w –1z = – 6x –63y +15z = –720x –10w –8z +18y = –
Exercises 13-14. Does the system of equations have a solution for all a, and b? If not, find some specific values of a and b for which a solution does not exist, and some specific values of a and b, not both zero, for which a solution does exist.
2x ++3yz= a x –44y – z = a 13. 14. xy++2z = b 2xy+ – 2z = b
Exercises 15-16. Does the system have a solution for all a, b, and c? If not, find some specific val- ues of a, b, and c for which a solution does not exist, and some specific values of a, b, and c, not all zero, for which a solution does exist.
x –44y – z = a 4x – 2y + z = a 15. 2x + 8y – 12z = b 16. –42x ++y 2z = b –12x ++y 2z = c 5xy–4+ z = c 24 Chapter 1 Matrices and Systems of Linear Equations
Exercises 17-19. Use the Spanning Theorem to determine if the system of equations has a solu- tion for all values of a, b, c, and d.
2xy– = a 4x – 2y + z = a z – 3w = b –42x ++y 2z = b 4x – 2y + z = a 17. 18. 19. 2x + 2z = c 5xy–4+ z = c –42x ++y 2z = b y + 2z = d 2xyz++ = d
20. Let S is a homogeneous system of equations. Prove that the last column of rref augS contains only zeros. 21. Prove that if mn , then the system of equations:
a11x1 ++a12x2 a1nxn = b1 a x ++a x a x = b 21 1 22 2 2n n 2 . ... . ... . ... am1x1 ++am2x2 amnxn = bm
cannot have a solution for all values of b1b2 bm .
22. (a) Show that if xx==0 y y0 and xx==1 y y1 are solutions of the system: ax+ by = c , then, xx==0 + kx1 – x0 y y0 + ky1 – y0 is also a solution for any dx+ ey = f given k . Suggestion: Substitute the above expressions for x and y into the given system. (b) Generalize the above argument to show that if a system of equations has more than one solution, then it must have infinitely many solutions Exercises 23-26. Determine the solution set of the given underdetermined (“wide”) homogeneous system of equations.
2x +03yz– = 2x +03yz– = 23. 24. 4x ++6y 2z = 0 4x +03y – 2z =
2x ++3yz–4w = 0 2x ++03y –42z w = 25. –53x –2y +0z – 3w = 26. –53x –2y +0z – 3w = –3x –2y ++z 7w = 0 –3x –2y ++z 7w = 0 1.2 Consistent and Inconsistent Systems of Equations 25
Exercises 27-28. Determine if the given overdetermined (“tall”) homogeneous system of equa- tions has a unique solution. 2x +++3y 4z 6w = 0 2x +03y – 4z = x +++3y 5z 2w = 0 3x ++2yz= 0 27. 2xy++6z +7w = 0 x +04y – 9z = 28. 5x +++3y 2zw= 0 – 4x –6y –0z = 2x +++4y 6z 2w = 0 3xy++4zw += 0 Exercises 29-30. Determine if the given square homogeneous system of equations has a unique solution.
29. 5x ++03y –54z w = 30. 2x +++5yz4w = 0 – x –2y +0z – 9w = –23x –4y ++z 6w = 0 3x –23y +0zw– = 4xy++–62z w = 0 11xy+0–92z – w = 9x –23y –0z +0w = Exercises 31-33. For what values of a will the given homogeneous system of equations have a unique solution?
xay+0= xyz++= 0 xyz++= 0 31. ax+0 y = 32. x +02yz– = 33. xayz+0– = – x ++yaz= 0 – x ++yaz= 0
Exercises 34-36. For what values of a and b will the given homogeneous system of equations have a unique solution?
xay+0= xay+0= xyz++= 0 34. 35. 2xby+0= bx+0 y = 36. xayz+0– = – x ++ybz= 0
ax+0 by = 37. For what values of a, b, c, and d will the homogeneous system of equations cx+0 dy = have a unique solution:
38. Show that if x0 y0 is a solution of a given two by two homogeneous system of equations,
then kx0 ky0 is also a solution for any k .
39. Show that if x0 y0 and x1 y1 are solutions of a given two by two homogeneous system
of equations, then x0 + x1 y0 + y1 is also a solution. 26 Chapter 1 Matrices and Systems of Linear Equations
a11xa+ 12y = b1 40. Let M be the solution set of S: and let T be the solution set of the corre- a21xa+ 22y = b2
a11xa+012y = sponding homogeneous system H: . Show that: a21xa+022y =
(a) If x0 y0 M and x1 y1 M , then x0 – x1 y0 – y1 T .
(b) If x0 y0 M and x1 y1 T , then x0 + x1 y0 + y1 M
a11xa+ 12y = b1 41. Let M be the solution set of S: and let T be the solution set of the corre- a21xa+ 22y = b2
a11xa+012y = sponding homogeneous system H: . Show that for any x0 y0 M , a21xa+022y =
Mx= 0 + x1 y0 + y1 x1 y1 T .
PROVE OR GIVE A COUNTEREXAMPLE
42. The system of equations associated with the augmented matrix abc 0 is consistent, def a independent of the values of the entries a through f.
43. The system of equations associated with the augmented matrix abc 1 is consistent, def 0 independent of the values of the entries a through f.
44. The system of equations associated with the augmented matrix 16 3 is consistent if 0 d 0 and only if d = 0 . 45. If a homogeneous system of equations has a nontrivial solution, then it has infinitely many solutions.
a11xa+012y = 46. If the homogeneous system has only the trivial solution, then the system a21xa+022y = a11xa+ 12y = b1 has a unique solution for all b1 b2 . a21xa+ 22y = b2
47. Any system S of m linear equations in n unknowns with nm has nontrivial solutions.
48. A system of n linear equations in m unknowns S is consistent if and only if rref coefS has m leading ones. Chapter Summary 27
CHAPTER SUMMARY
N-TUPLE An (ordered) n-tuple is an expression of the form c1c2 cn , where each ci is a real number, for 1 in .
SOLUTION SET OF A An n-tuple c1c2 cn is a solution of the system of m SYSTEM OF EQUATIONS equations in n unknowns:
a11x1 ++a12x2 a1nxn = b1 a21x1 ++a22x2 a2nxn = b2 S: . ... . ... . ... am1x1 ++am2x2 amnxn = bm
if each of the m equations is satisfied when ci is substituted for
xi , for 1 in . The solution set of S is the set of all solutions of S.
CONSISTENT AND A system of equations is said to be consistent if it has non- INCONSISTENT SYS- empty solution set. A system of equations that has no solution is said to be inconsistent. TEMS OF EQUATIONS
EQUIVALENT SYSTEMS Two systems of equations are said to be equivalent if they have OF EQUATIONS equal solution sets.
OVERDETERMINED, A system of m equations in n unknowns is said to be: UNDERDETERMINED, Overdetermined if nm (more equations than unknowns). AND SQUARE SYSTEMS Underdetermined if nm (fewer equations than unknowns). OF EQUATIONS Square if nm= .
ELEMENTARY The following three operations on a system of linear equations EQUATION OPERATIONS are said to be elementary equation operations: Interchange the order of any two equations in the system. Multiply both sides of an equation by a nonzero number. Add a multiple of one equation to another equation. Elementary row operations Performing any number of elementary equation operations on a do not alter the solution sets system of linear equations will result in an equivalent system of of systems of equations. equations (same solution set). 28 Chapter 1 Matrices and Systems of Linear Equations
MATRICES Matrices are arrays of numbers arranged in rows and columns, such as the matrix A below:
a11 a12 a13 a14
A ===a21 a22 a23 a24 also: A34 aij or A aij
a31 a32 a33 a34 Since A has 3 rows and 4 columns, it is said to be a three-by- four matrix. When the number of rows of a matrix equals the number of columns, the matrix is said to be a square matrix.
ELEMENTARY ROW The following three operations on any given matrix are said to OPERATIONS be elementary row operations: Interchange the order of any two rows in the matrix. Multiply each element in a row of the matrix by a nonzero num- ber. Add a multiple of one row of the matrix to another row of the matrix.
EQUIVALENT MATRICES Two matrices are equivalent if one can be derived from the other by means of a sequence of elementary row operations.
AUGMENTED MATRIX The augmented matrix of a system of equations S is that matrix aug(S) composed of the coefficients of the equations in, along with the constants to the right of the equations in S. For example:
2xyz+2– = – 21– 1 –2 aug(S) S: x ++3y 2z = 9 132 9 – x –2y +1z = –11 –2 1
Equivalent systems of equa- Two systems of equations, S1 and S2 , are equivalent if and tions corresponding to equiv- only if their corresponding augmented matrices, aug S and alent augmented matrices. 1 aug S2 , are equivalent.
ROW-REDUCED-ECHELON All rows consisting entirely of zeros are at the bottom of the FORM OF A MATRIX matrix. All of the other rows contain a leading-1 (with zeros in all entries above or below it), and those leading-ones “move” to the right, as you “move” down the matrix.
Gauss-Jordan Elimination The Gauss-Jordan Elimination Method of page 10 can be Method. used to obtain the row-reduced-echelon form rrefA of a given matrix A. Chapter Summary 29
COEFFICIENT MATRIX The coefficient matrix of a system of equations S is that matrix coef (S) composed of the coefficients of S. For example:
2xyz+2– = – 21– 1 coef (S) S: x ++3y 2z = 9 132 – x –2y +1z = –11 –2 forget about the constants on the right of the equal signs
Spanning Theorem The system of equations:
a11x1 ++a12x2 a1nxn = b1 a x ++a x a x = b S: 21 1 22 2 2n n 2 . ... . ... . ... am1x1 ++am2x2 amnxn = bm
has a solution for all values of b1b2 bm if and only if rref coefS does not contain a row consisting entirely of zeros.
HOMOGENEOUS SYSTEM OF A system of equations of the form: EQUATIONS a x + a x + + a x =0 11 1 12 2 1n n a x + a x + + a x =0 21 1 22 . 2 2n. n . S: ...... am1x1 + am2x2 + + amnxn =0 with zeros to the right of the equal sign, is said to be homoge- neous.
TRIVIAL SOLUTION x1 ==0 x2 0 xn =0 is a solution of the above homoge- neous system of equation. It is said to be the trivial solution of the system.
Fundamental Theorem of Any homogeneous system S of m linear equations in n Homogeneous Systems unknowns with nm has nontrivial solutions.
You can use rref coefS to Let S be a homogeneous system of equations. The only differ- solve a homogeneous system ence between rref augS and rref coefS is that the of equations S former contains an additional column of zeros. Being aware of this, you might as well focus on rref coefS to derive the solution set of S. Linear Independence A homogeneous system S of m linear equations in n unknowns Theorem has only the trivial solution if and only if rref coefS has n leading ones. 30 Chapter 1 Matrices and Systems of Linear Equations 2.1 Vectors in the Plane and Beyond 31
2 CHAPTER 2 VECTOR SPACES We begin this chapter with a geometrical consideration of vectors as directed line segments in the plane and in three dimensional space, and then extend the vector concept to higher dimensional Euclidean spaces. Abstraction is the nature of mathematics, and we let the “essence” of Euclidean vector spaces guide us, in Section 2, to the definition of an abstract vector space. In Section 3 we begin to uncover some of the beautiful (and very practical) theory of abstract vector spaces, an exca- vation that will keep us well-occupied for the remainder of the text. Subsets of vector spaces which are themselves vector spaces are con- sidered in Section 4. In Section 5, we return to the two and three dimen- sional Euclidean spaces of the first section and derive a vector representation for lines and planes in those spaces.
§1. VECTORS IN THE PLANE AND BEYOND
We begin by considering vectors in the plane, such as those in Figure 2.1, which are depicted as directed line segments (“arrows”). In that geometrical setting, the arrow is pointing in the direction of the vector, with the length of the arrow representing its magnitude. B B v = AB D . v = AB . A A v = CD C. (a) (b) Figure 2.1 Vectors will be denoted by boldface lowercase letters. The vector v = AB in Figure 2.1 is said to have initial point A, and terminal point B. One defines two vectors to be equal if they have the same mag- nitude and direction. If you pick up the vector v in Figure 2.1(a) and move it in a parallel fashion as we did in Figure 2.1(b) to the vector with initial point C and terminal point D, then you will still have the same vector: v ==AB CD
In particular, the vector v in Figure 2.2 with initial point Ax= 0 y0 and terminal point Bx= 1 y1 can be moved in a parallel fashion so that its initial point coincides with the origin. When so placed, the vector is said to be in standard position. 32 Chapter 2 Vector Spaces
y Bx= 1 y1 . v Ax= 0 y0
v x1 – x0 y1 – y0 tor in standard position vec . x Figure 2.2
EXAMPLE 2.1 Sketch the vector with initial point –23 and terminal point 41 – . Position that vector in standard position, and identify its terminal point.
SOLUTION: The figure below tells the whole story: y 6 –23 . Pick up the top vector and move it 2 units down and 3 4 units to the right to the right v so that its initial point x –23 . In the process, the original terminal point v 41 – 41 – is also moved 2 4 units to the right at 3 units – 1 – 3 down, coming to rest at terminal point 64 – . 6 64 – 42– – Figure 2.3 A standard position vector in the plane is completely determined by the coordinates of its terminal point. This observation enables us to identify the vector in Figure 2.3 as an ordered pair of numbers or 2- Note that the two-tuple in tuple; namely: the expression v = 64 – v = 64 – appears in bold-face, so as to distinguish it from the In a similar fashion we may refer to the vectors v and w in Figure 2.4(a) form 64 – which rep- resents a point in the plane. as v = 23 and w = –32 – . z y 23 4 v 134 u x -3 -2 -1 1 2 3 y w 3 1 –32 – x (a) (b) Figure 2.4 2.1 Vectors in the Plane and Beyond 33
Likewise, the vector u in the 3 dimensional space of Figure 2.4(b) can be described by the bold-faced 3-tuple u = 134 The beauty of all of this is that while we cannot geometrically repre- sent a vector in 4 dimensional space, we can certainly consider 4- tuples, and beyond. With this in mind, let us agree to denote the set of n all (ordered) n-tuples a1a2 an by the symbol : n = a1a2 an ai
The real numbers ai in the n-tuple a1a2 an are said to be the components of the n-tuple, and we define two n-tuples to be equal if their corresponding components are identical:
DEFINITION 2.1 a1a2 an = b1b2 bn if n-TUPLE EQUALITY ai = bi , for 1 in .
SCALAR PRODUCT AND SUMS OF VECTORS Vectors evolved from the need to adequately represent quantities which are characterized by both magnitude and direction. In a way, these quantities themselves tell us how we should go about defining cer- tain algebraic operations on vectors. Suppose, for example, that the vec- tor v = 32 of Figure 2.5(a) represents a force. Doubling the magnitude of that force without changing its direction would result in the vector force labeled 2v in that figure, as that vector is in the same direction as v = 32 , with length twice that of v.
2v = 64 v = 32
v = 32 1 –---v = –1 –--2- 3 3 (a) (b) Figure 2.5 Length of v = 32 : Similarly, if a force that is one-third that of v = 32 is applied in the 32 +1322 = opposite direction of v, then the vector representing that new force is 1 the vector –---v in Figure 2.5(b); for that vector is in the opposite direc- 3 Length of –1 –--2- : 3 tion of v = 32 , with length one-third that of v.
2 2 4 –1 2 +1–--- = + --- This stretching or shrinking of a vector, in one direction or the other, 3 9 is an important operation which we now formalize and extend to the set ==-----13------13- 9 3 of n-tuple-vectors for any positive integer n: 34 Chapter 2 Vector Spaces
DEFINITION 2.2 n To any vector vv= 1v2 vn in , SCALAR and any r , we let: PRODUCT rvrv= 1rv2 rvn The vector rv is said to be a scalar multiple of v. For example: 315 = 315 –510– 4 = –5020 213– 45 = 23 2– 4 25 2
VECTOR ADDITION If two people pull on an object positioned at the origin with forces v and w, then the observed combined effect is the same as that of one individual pulling with force z, where z is the vector coinciding with the diagonal in the parallelogram formed by the vectors v and w [Fig- ure 2.6(a)]. y y v1 + w1 v2 + w2
w2 w v1 v2 z z v v v2 w
x x v1 w1 (a) (b) Figure 2.6 The above vector z is said to be the sum of the vectors v and w, and is denoted by vw+ . Figure 2.6(b) reveals that:
zvw==+ v1 v2 + w1 w2 =v1 + w1 v2 + w2 Generalizing, we have: While identical in shape, the “+” in vw+ differs in DEFINITION 2.3 The sum of the vectors vv= 1v2 vn spirit from that in vi + wi : ECTOR UM n the latter represents the V S and ww= 1w2 wn in , is denoted familiar sum of two num- bers, as in 37+ , while the by vw+ and is given by: former represents the vw+ = v + w v + w v + w newly defined sum of two 1 1 2 2 n n n-tuples, as in: 32 – + 711 2.1 Vectors in the Plane and Beyond 35
EXAMPLE 2.2 For v ==–2315 w 150– 2, and r = 2 , determine the vector rvw+ .
SOLUTION: rvw+2= –2315 + 150– 2 Definition 2.2: = –46210 + 150– 2 Definition 2.3: = – 4 + 16 + 52 + 0102 – = –31128
CHECK YOUR UNDERSTANDING 2.1 For v ====32– 2 w –310 r 2 s –3 , determine the Answer: 15 1– 4 vector rv + sw
EUCLIDEAN VECTOR SPACES
The set of n-tuples n , together with the above defined operations of vector addition: v1v2 vn + w1w2 wn = v1 + w1 v2 + w2 vn + wn and scalar multiplication: rv1v2 vn = rv1rv2 rvn is called the Euclidean n-space. Every Euclidean space contains a most distinguished vector: DEFINITION 2.4 The zero vector in n , denoted by 0, is that ZERO VECTOR vector with each component the number 0: 000= 0 For example 0= 000 is the zero vector in 3 , and 0= 0000 is the zero vector in 4 . No direction is associated with the zero vector. A zero force, for example, is no force at all, and its “direction” would be a moot point.
Every real number r has an additive inverse –r , namely that number which when added to r yields 0. Similarly:
DEFINITION 2.5 n For given vv= 1v2 vn the ADDITIVE INVERSE additive inverse of v is denoted by –v and is OF A VECTOR given by: –v = –v1–v2 –vn
n Note that for every vv= 1v2 vn : vv+ – ==v1v2 vn + –v1–v2 –vn 0 36 Chapter 2 Vector Spaces
We are now in a position to list the properties of Euclidean spaces which will morph into the definition of an abstract vector space in the next section: THEOREM 2.1 Let u, v, and w be vectors in the Euclidean n- space n , and let r and s be scalars (real num- bers). Then: (i)uv+ = vu+ (Commutative Property)
(ii)uv+ + w = uvw+ + (Vector Associative Property) Addition: (iii) v + 0 = v (Zero Property)
{ (iv)vv+ – = 0 (Inverse Property)
(v)ruv+ = ru + rv (Vector Distributive Property) Scalar and Addition: { (vi)rs+ u = ru + su (Scalar Distributive Property) (vii)rsu = rs u (Scalar Associative Property) Scalar: { (viii)1uu= (Identity Property)
PROOF: We establish (ii) in 2 , (v) in 3 , and (vii) in 3 , and invite you to verify the rest on your own.
To emphasize the important role played by definitions, the symbol instead of = will tem- porarily be used to indicate a step in the proof which follows directly from a definition. In addi- tion, the abbreviation “PofR” will be used to denote that a step follows directly from a Property of the Real numbers, all of which will be assumed to hold; for example, the additive associative property of the real numbers: ab+ + c = abc+ + .
(ii): uv+ + w = uvw+ + (in 2 ): This associative property If uu== u v v v and w =w w , then: eliminates the need for 1 2 1 2 1 2 including parenthesis when summing more than two uv+ + w u1 u2 + v1 v2 + w1 w2 vectors. In particular, Definition 2.3: uvw++ u1 + v1 u2 + v2 + w1 w2 is perfectly well defined. Definition 2.3: u1 + v1 + w1 u2 + v2 + w2
POFR: = u1 + v 1 + w1 u2 + v2 + w2 Definition 2.3: u1 u2 + v1 v2 + w1 w2 uvw+ + 2.1 Vectors in the Plane and Beyond 37
(v): ruv+ = ru + rv (in 3 ):
If uu==1u2 u3 and vv1v2 v3 , then:
ruv+ ru1u2 u3 + v1v2 v3 Definition 2.3: ru1 + v1 u2 + v2 u3 + v3 Definition 2.2: ru1 + v1 ru2 + v2 ru3 + v3 PofR: = ru1 + rv1ru2 + rv2 ru3 + rv3 Defintion 2.3: ru1 ru2 ru3 + rv1 rv2 rv3 Definition 2.2: ru1u2 u3 + rv1v2 v3 ru + rv (vii): rsu = rs u (in Rn ):
rsu rsu1u2 un
Definition 2.2: rsu1su2 sun
Definition 2.2: rsu1 rsu2 rsun PofR: = rs u1rs u2 rs un Defintion 2.2: rs u1u2 un rs u
In this, and any other abstract math course: DEFINITIONS RULE! Just look at the above proof. It contains but one “logical step,” the step labeled PofR; all other steps hinge on DEFINITIONS.
CHECK YOUR UNDERSTANDING 2.2
Answer: See page B-3. Establish Theorem 2.1(iv), vv+ – = 0 , in 3 and in n . 38 Chapter 2 Vector Spaces
EXERCISES
Exercises 1-6. Sketch the vector with given initial point A and terminal point B. Sketch the same vector in standard position in the plane, identifying its terminal point.
1.A ==–12 B 01 2.A ==33 B 01 – 3. A ==11 B –22
4.A ==10 B 01 – 5.A ==–12 – B 11 – 6. A ==22 B 12 –
Exercises 7-10. Express, as a 3-tuple, the vector with given initial point A and terminal point B.
7.A ==123 B 321 8. A ==–450 B 251 –
9.A ==01– 9 B –902 10. A ==–35– 3 B 353 –
Exercises 11-14. Perform the indicated vector operations.
11. 532 – ++01 –24 – 12. 25 ++13 ––23
13.– 231– 5 + –1 – 200 14. – ––1234 + 312– 2
Exercises 15-18. Find the vector v such that:
15.v + 24 – = –42 16. v +204135 = –
17.42 +35–v = + –12 – 18. 43– 1 +2132–v = –
19. For u = 13 , v = 24 , and w = 62 – , find scalars r and s such that: (a) ru + sv = w (b) –ru + sw = v (c) rv + –sw = u
20. Find scalars r, s, and t, such that: r130 ++s216 t146 = 756
21. Find scalars r, s, and t, such that: –r130 ++s216 –t146 = 756 22. Show that there do not exist scalars r, s, and t, such that r235 ++s325 t123 = 124
23. Find the vector ab 2 of length 5 that has the same direction as the vector with initial point 13 and terminal point 31 .
24. Find the vector ab 2 of length 5 that is in the opposite direction of the vector with ini- tial point 13 and terminal point 31 .
25. Prove Theorem 2.1(ii) for: (a) 3 (b) n . 2.1 Vectors in the Plane, and Beyond 39
26. Prove Theorem 2.1(v) for: (a) 2 (b) n .
27. Prove Theorem 2.1(i) for: (a) 2 (b) 3 (c) n
28. Prove Theorem 2.1(iii) for: (a) 2 (b) 3 (c) n
29. Prove Theorem 2.1(vi) for: (a) 2 (b) 3 (c) n
30. Prove Theorem 2.1(viii) for: (a) 2 (b) 3 (c) n
31. Prove that if v, w, and z, are vectors in 3 such that vw+ = vz+ , then wz= .
PROVE OR GIVE A COUNTEREXAMPLE
32. For v n , if rv = sv then rs= .
33. For v 0 n , if rv = sv then rs= .
n 34. For v1 v2 and r 0 , if rv1 = rv2 then v1 = v2 .
35. For v n , rv = 0 if and only if r = 0 or v = 0 . 40 Chapter 2 Vector Spaces
2
§2. ABSTRACT VECTOR SPACES
One of the main objectives of abstract mathematics is to isolate and analyze a particular structure of the real number system, so as to better focus on its “essence.” The essence of the vector structure in n tabu- lated in Theorem 2.1, page 36, leads us to the definition of an abstract vector space: DEFINITION 2.6 A (real) vector space is a nonempty set V along with two operations, called vector addi- The elements of a vector VECTOR SPACE space V are called vec- tion and scalar multiplication. The operation tors, and will be denoted of addition assigns to any two element u and by bold-faced letters (like v). Scalars will continue v in V, another element uv+ in V. The oper- to be denoted by non- ation of scalar multiplication assigns to any bold-faced letters (like r). real number r (also called a scalar), and any element v in V, another element rv in V. These operations must satisfy the following eight axioms for all uvw V , and all rs :
(i)uv+ = vu+ (Commutative Axiom) (ii)uv+ + w = uvw+ + (Vector Associative Axiom) (iii) There is a vector in V, denoted Addition: by 0 such that v + 0 = v for every vector v in V. (Zero Axiom) (iv) For every vector v in V, there is a vector in V, denoted by –v such that vv+ – = 0 . (Additive Inverse Axiom) (v)ruv+ = ru + rv (Vector Distributive Axiom) Scalar and Addition: { (vi)rs+ v = rv + sv (Scalar Distributive Axiom)
Scalar: (vii)rsv = rs v (Scalar Associative Axiom) { (viii)1vv= (Identity Axiom) While eight axioms are specifically listed in the above definition, two A set is said to be closed, more are lurking within the above so-called closure statements: with respect to an operation, if elements of that set sub- V is closed under addition: For every v and w in V, vw+ V jected to that operation remain in the set. For exam- V is closed under scalar ple, the set of positive inte- For every v V and r , rv V gers is closed under addition multiplication: (the sum of two positive integers is again a positive integer), but that set is not closed under subtraction. 2.2 Abstract Vector Spaces 41
It is important to note that while the two addition signs in rs+ v = rv + sv are identical in appearance, they do not represent a common operator: The “+” in rs+ v denotes the sum of two real numbers, as in 25+ v = 7v , while the “+” in rv + sv denotes the sum of two vectors, as in 2v + 5v . By the same token, the two “products” in the expression rs v also denote We also point out that, by distinct operators: convention, no meaning is The rs in rs v denotes the product of two real numbers, attributed to an expression of the form vr , wherein a resulting in another number, as in 25 = 10 , while the scalar vector v appears to the left product 10v represents a vector. of a scalar r. We already have infinitely many vector spaces at our disposal, namely the Euclidean n-spaces. We now turn our attention to several others.
MATRIX SPACES
EXAMPLE 2.3 The set of two-by-two matrices: ab M22 = abcd R cd with addition and scalar multiplication given by:
ab + a b = aa+ bb+ cd c d cc+ dd+
r ab = ra rb cd rc rd is a vector space.
SOLUTION: We content ourselves with verifying Axiom (iii) (the zero axiom), and Axiom (iv) (the additive inverse axiom).
00 ab Axiom (iii):Let 0 = . Then, for any v = we have: 00 cd
We are again using to PofR indicate that equality fol- lows from a definition, v + 0 ab + 00 a + 0 b + 0 = ab v and “PofR” for “Property cd 00 c + 0 d + 0 cd of the Real numbers.”
Axiom (iv): For any given v = ab , we show there exists a vector –v , cd
namely –v = –a –b , such that vv+ – = 0 : –c –d 42 Chapter 2 Vector Spaces
PofR
vv+ – ab + –a –b aa– bb– = 00 0 cd –c –d cc– dd– 00
Generalizing Example 2.3 to accommodate matrices of all dimen- sions, we have:
THEOREM 2.2 Let Mmn denote the set of all mn matrices. MATRIX SPACE For A = aij and B = bij in Mmn , let:
AB+ ==aij + bij aij + bij (The ijth entry of the sum matrix is the sum of the ijth entry in matrix A with the ijth entry in the matrix B.)
For r and A = aij Mmn , let:
rA ==raij raij (The ijth entry in the matrix rA is r times the ijth entry in the matrix A.)
The set Mmn with the above operations is a vector space.
PROOF: We content ourselves with verifying Axioms (i) and (vi).
Axiom (i): For every A = aij and B = bij : PofR
AB+ aij + bij aij + bij = bij + aij bij + aij BA+
Axiom (vi): For scalars r and s, and A = aij : PofR
rs+ A rs+ aij rs+ aij = raij + saij
raij + saij
raij + saij rA + sA
CHECK YOUR UNDERSTANDING 2.3 Verify the associative axiom rsv = rs v for the vector space Answer: See page B-3. Mmn of Theorem 2.2. 2.2 Abstract Vector Spaces 43
POLYNOMIAL SPACES
n n – 1 A function of the form px= anx +++an – 1x a0 , with an 0 is said to be a polynomial function of degree n. For any given In particular: integer n 0 , Pn will represent the set of polynomials of degree less P0x ==a0 a0 than or equal to n We note that the polynomial: n n – 1 anx +++an – 1x + a1xa0 can be written the other way around: 2 n a0 ++a1xa2x ++ anx and can also be expressed in Sigma-notation form: n The Greek letter i 2 n (Sigma) is used to denote aix = a0 ++a1xa2x ++ anx a sum. i = 0
THEOREM 2.3 The set of polynomials Pn of degree less than or equal to n, with operations: POLYNOMIAL SPACES n n n a xi + b xi = a + b xi i i i i i = 0 i = 0 i = 0
n n raxi = ra xi i i i = 0 i = 0 is a vector space.
PROOF: We establish the two distributive axioms, and relegate the remaining six to the exercises. Axiom (v) ruv+ = ru + rv :
n n n n raxi + b xi ra+ b xi ra+ b xi i i i i i i i = 0 i = 0 i = 0 i = 0 n n n = ra + rb xi ra xi + rb xi i i i i PofR i = 0 i = 0 i = 0 n n raxi + rbxi i i i = 0 i = 0 44 Chapter 2 Vector Spaces
Axiom (vi) rs+ v = rv + sv : n n PofR n i i i i rs+ aix rs+ aix = raix + saix i = 0 i = 0 i = 0 n n n n ra xi + sa xi raxi + saxi i i i i i = 0 i = 0 i = 0 i = 0
CHECK YOUR UNDERSTANDING 2.4 Referring to Theorem 2.3, verify the commutative axiom: n n n n i i i i aix + bix = bix + aix Answer: See page B-3. i = 0 i = 0 i = 0 i = 0
FUNCTION SPACES You are probably accustomed of thinking that a function as some sort of dynamic creature that “takes numbers into numbers.” At this point, however, you want to think of a function as being an object, in the same All “objects” in mathemat- way that you see the number 5 as an object. Indeed, the set of all such ics are sets, and functions are no exceptions. The function functions f : X from a set X (the domain of the function) to the 2 f given by fx = x , for set of real numbers, can be turned into a vector space: example, is that subset of the plane, typically called the graph of f: THEOREM 2.4 Let FX denote the set of all real-valued 2 f = xx xR FUNCTION SPACE functions defined on a non-empty set X: Pictorially: FX= f f : X R For f and g in FX , and r R , let fg+ , and rf be given by: fg+ x = fx+ gx (*) and rf x = rf x With respect to these operations, FX is a vector space.
A function f : X is defined to be equal to a function g : X , if fx= gx for every xX . 2.2 Abstract Vector Spaces 45
PROOF: We verifying Axioms (i), (iii), (iv) and (v): The fact that FX is Axiom (i) (Commutative Axiom). For fg FX and xX : closed under addition and scalar multiplication is PofR self-evident. fg+ x fx+ gx= gx+ fx gf+ x Since fg+ x = gf+ x for every xX , fg+ = gf+ .
Axiom (iii) (Zero Axiom). Let Z : X be the function given by As you can see, we Zx = 0 for every xX . For any fF and xX : elected to use the letter Z, rather than the symbol 0, PofR for our zero vector. It’s just that an expression fZ+ x fx+ Zx fx+ 0 = fx fx like 0x would strongly suggest that a multiplica- Since fZ+ x = fx for every xX , fZ+ = f . tion by zero is being per- formed, which is not the case. Axiom (iv) (Additive Inverse Axiom). For given f FX let –f be the function given by –f x = –fx . Then, for any xX : PofR ff+ – x fx+ –fx = 0 Zx
Since ff+ – x = Zx for every xX , ff+ – = Z .
Axiom (v) [Distributive Axiom (vector)]: For any fg FX , xX and r R : PofR rfg+ x rfg+ x rfx+ gx = rf x + rg x rf x + rg x Since rfg+ x = rf x + rg x for every xX , rfg+ = rf + rg .
CHECK YOUR UNDERSTANDING 2.5
Verify the distributive axiom rs+ v = rv + sv for the func- Answer: See page B-3. tion space of Theorem 2.4.
ADDITIONAL EXAMPLES
As you will see in the next two examples, addition and scalar multi- plication in a vector space can be somewhat “counter-intuitive.” In addition, both the zero vector 0 and the additive inverse vector –v , may appear somewhat strange in a vector space 46 Chapter 2 Vector Spaces
EXAMPLE 2.4 Show that the set + = aa 0 of positive real numbers with addition and scalar multipli- cation given by: ab+ ==ab and raar is a vector space.
SOLUTION: + is certainly closed under both of the above opera- tions. Moreover: PofR Axiom (i): ab+ ab = ba ba+
Axiom (ii): PofR ab+ + c ab c = abc abc+ +
Axiom (iii):This axiom professes the existence of a vector 0 such that a + 0 = a for every a + . It certainly cannot be the number 0, for that number is not even in the set + . Looking for a clue, we ask ourself: 5 +5what = ? which is to say: 5w hat = 5? since rs+ = rs Yes, 1 is the zero vector in this space: a + 1 a 1 = a
Axiom (iv): Looking for a clue, we ask ourselves “What is the additive inverse of 7:” 7 + what = 1? (remember that 1 is the zero vector) which is to say: 7w hat = 1?
Yes, in this space, –a = --1- : a 1 aa+ – a--- = 1 (the zero vector) a
Axiom (v): rab+ = ra + rb : PofR rab+ ab r = arbr ar + b r ra + rb You are invited to establish the remaining three axioms of Definition 2.6 in the exercises, thereby establishing the fact that + , with given operations, is a vector space. 2.2 Abstract Vector Spaces 47
EXAMPLE 2.5 Show that the set Vxy= xy under addition: xy + x y = xx+ – 1 yy++ 1 and scalar multiplication rxy = rx–1 r + ry+ r – 1 is a vector space:
SOLUTION: V is certainly closed under both of the above operations. We content ourselves by establishing the zero and inverse axioms, and leave it for you to verify the remaining six axioms in the exercises. Zero Axiom: Does there exist a vector 0 such that v + 0 = v for every v V ? Don’t be to quick to say “no,” basing you answer on the observation that xy + 00 ==x + 01– y ++01 x – 1 y + 1 xy But you have no right to assume that if a zero vector exists, then it must be the one you would like it to be! Putting partiality aside, let’s see if we can find a vector 0 = ab such that xy + 0 = xy , for every xy V : xy + 0 = xy xy + a b = xy xa+ – 1 ==x a 1 xa+ – 1 yb++1 = xy yb++1 ==y b –1 That’s right, in this vector space, 0 = 11 – , for: xy + 1 –1 ==x + 1 – 1 y ++–1 1 xy Additive Inverse Axiom: Now that we have a zero vector, 0 = 11 – , we can ask if, for any given v = xy , there exists a vector –v = ab such that xy +11ab = – . There does: xy +11a b = – xa+1– 1 == a – x + 2 xa+ – 1 yb++1 = 11 – yb++1 ==– 1 b – y – 2 So, the additive inverse of xy turns out to be –2x + –2y – , for: xy + – x + 2 – y – 2 ==x – x + 2 – 1 y – y – 2 + 1 11 –
the zero vector
CHECK YOUR UNDERSTANDING 2.6 Verify that Axioms (iii) and (iv) of Definition 2.6 are satisfied for the set V= xyz xyz with imposed addition: Answer: Zero vector: 12– 3 xyz + x y z = xx+ – 1 yy++ 2 zz+ – 3 Inverse of xyz : and scalar multiplication: – x + 2– y + 4 – z + 6 rxyz = rx– r + 1 ry – 2r – 2 rz – 3z + 3 48 Chapter 2 Vector Spaces
EXAMPLE 2.6 Let V denote the upper-half plane: y Vxy= xy y 0 V with standard addition and scalar multiplication: xy + x y = xx+ yy+ x rxy = rx ry Is V a vector space?
SOLUTION: No, since V is not closed under scalar multiplication. But we can’t just say this, for our claim has to be established. We have to demonstrate that the scalar product involving some specific element of V and some specific real number ends up being outside of V, and so we shall: 52 V but – 7 5 2 = –35 – 14 V
EXAMPLE 2.7 Let X be the set of two-tuples with operations: xy + x y = xx+ yy+ rxy = rx y Is X a vector space?
SOLUTION: X is certainly closed under both operations. We need not challenge the first four axioms of Definition 2.6, as they involve only the addition operator which coincides with that of Euclidean two- space. The scalar operator, however, is a bit different from that of 2 , and we must therefore determine whether or not Axioms (v) through (viii) are satisfied. What you may want to do is to quickly check to see if they hold for some specific vectors and scalars: Let’s Challenge Axiom (vi), rs+ v = ru + sv , with r = 7 , s = 3 , and u = 42 – : 73+ 42 – ==10 4 – 2 40 – 2 and 7 4 – 2 +28234 – 2 == – +40412 – 2 – Oops! We need go no further, for the above shows that Axiom (vi) does not hold, and can therefore conclude that under the given operations, X is not a vector space.
CHECK YOUR UNDERSTANDING 2.7
THE TRIVIAL VECTOR SPACE: Define addition and scalar multiplication on the set V = 0 and verify Answer: See page B-4. that V is a vector space with respect to those operations. 2.2 Abstract Vector Spaces 49
EXERCISES
Exercises 1-13. Verify that the set S, with given operations, fails to be a vector space. 1.S = , xy+ = xy– , and rx = rx.
2.Sxy= xy , xy + x y = xx+ 0 , and rxy = rx ry .
3.Sxy= xy , xy + x y = xx yy , and rxy = rx ry .
4.Sxy= xy , xy +00x y = , and rxy = rx ry .
5.Sxy= xy , xy +2x y = x + 2x yy+ , and rxy = rx ry .
6.Sxy= xy , xy + x y = xx+ yy+ , and rxy = 00 .
7.Sxy= xy , xy + x y = xy+ yx+ , and rxy = rx ry .
8.Sxy= 0 xy , xy0 + xy 0 = xx+ yy+ 0 , and rxy0 = rx ry 0 .
ab a b aa+0 ab ra rb 9.SM= 22 , + = , and r = . cd c d 0 dd+ cd rc rd
ab a b ac+ bd+ ab ra rb 10.SM= 22 , + = , and r = . cd c d ca+ db+ cd rc rd
11.Sax= 2 ++bx c , ax2 ++bx c + ax2 ++bxc = aa+ x2 + bb+ x and rax2 ++bx c = ra x2 ++rb xrc.
12.S = 01 ; 00+0= , 11+0= , 01+10==+1 ; and r00= , r11= .
13.S = a a 0 , ab+ = lnab+ ln , and ra = lnar .
Exercises 14-16. Verify that the set V, with given operations, is a vector space. 14.V = 1 , 11+1= , and r11= .
15.Vxy= 0 xy , xy0 + xy 0 = xx+ yy+ 0 , and rxy0 = rx ry 0 .
16.Vxy= xy , xy + x y = xx++ 2 yy+ , rxy = rx + 2r – 2 ry . 50 Chapter 2 Vector Spaces
17. Complete the proof of Theorem 2.2. 18. Complete the proof of Theorem 2.3. 19. Complete the proof of Theorem 2.4. 20. Establish the remaining three axioms for the space of Example 2.4. 21. Establish the remaining six axioms for the space of Example 2.5.
i 22. A polynomial is an expression of the form px= aix for which there exists an m such i = 0
that ai = 0 for im . Show that, with respect to the following operations, the set P of all polynomials is a vector space: i i i i i aix + bix = ai + bi x and r aix = rai x i = 0 i = 0 i = 0 i = 0 i = 0
PROVE OR GIVE A COUNTEREXAMPLE
23. Let V be a vector space, and let v V . If rv = 0 , then r = 0 .
24. Let V be a vector space, and let v V . If rv = 0 and v 0 , then r = 0 .
25. Let V be a vector space, and let v V . If rv = 0 and r 0 , then v = 0 .
26. Let V be a vector space, and let v V . If rv = sv , then rs= .
27. Let V be a vector space, and let v V . If rv = sv and v 0 , then rs= . 2.3 Properties of Vector Spaces 51
2
§3. PROPERTIES OF VECTOR SPACES
For the sake of convenience, we again list the vector space axioms: (i)uv+ = vu+ (Commutative Axiom) (ii)uv+ + w = uvw+ + (Vector Associative Axiom) (iii) There is a vector in V, denoted Addition: by 0 such that v + 0 = v for every vector v in V. (Zero Axiom) (iv) For every vector v in V, there is a vector in V, denoted by –v such that vv+ – = 0 . (Additive Inverse Axiom) (v)ruv+ = ru + rv (Vector Distributive Axiom) Scalar and Addition: { (vi)rs+ v = rv + sv (Scalar Distributive Axiom)
Scalar: (vii)rsv = rs v (Scalar Associative Axiom) { (viii)1vv= (Identity Axiom)
For aesthetic reasons, a set of axioms should be independent, in that no part of an axiom is a consequence of the rest. One should not, for example, replace Axiom (iii) with: There is a vector in V, denoted by 0 such that v + 0 = v and 0 + v = v for every vector v in V. Reason: Axiom (i) already implies that the 0 of Axiom (iii) can be on either side of the v: THEOREM 2.5 Let V be a vector space. (a) For every vector v in V: v +00 ==+ v v (b) For every vector v in V, there exists a vector –v such that: vv+ – ==– v +0v
PROOF: (a): Since v + 0 = v [Axiom (iii)], and vw+ = wv+ [Axiom (1)]: v ==v + 0 0 + v (b): Since vv+ – = 0 [Axiom (v)], and v + 0 = v [Axiom (iii)], and vw+ = wv+ [Axiom (1)]: 0 ===vv+ – –v + v – v + v 52 Chapter 2 Vector Spaces
Our next theorem tells us that there is but one 0 vector in any vector Axiom (iii) asserts the space, and that every vector v has a unique additive inverse –v . While existence of a zero vec- tor, but makes no claim you might have taken these two facts for granted, neither is given to as to its uniqueness, and you free of charge: Axiom (iv) only asserts that every vector has an Let V be any vector space, then: additive inverse (could THEOREM 2.6 it have several?). (a) There is but one vector 0 which satisfies the property that v + 0 = v for every v in V. (b) For any given vector v in V, there is but one vector –v in V such that vv+ – = 0 .
PROOF:
Strategy for (a): Assume that 0 and 0 are any two zeros, and then go on to show 00= .
Let 0 and 0 be such that, for every vector v: v + 0 ==v and v + 0 v (*) v + 0 ==v and (**) v + 0 v
Substituting 0 for v in (*), we have (i): 00+ = 0 .
Substituting 0 for v in (**), we also have (ii): 00+ = 0 . Then: (i) (ii)
000===+ 00+ 0 00= commutativity
Strategy for (b): Assume that a vector v has two additive inverses, –v and v , and then go on to show that –v = v .
Let –v and v be such that:
vv+ – = 0 and v + v (*) vv+ – ==0 and (**) v + v 0 Adding v to both sides of (*) we have: v + vv+ – = v + 0 Axioms (ii) and (iii): vv+ + –v = v Axiom (i) and (**): 0 + –v = v Theorem 2.5(a): –v = v –v = v The above proof illustrates the important fact that a mathematical the- ory is based on a set of rules or axioms, on which sit logically derived 2.3 Properties of Vector Spaces 53
results, or theorems. Once established, a1.5 theorem can be used to prove other theorems. At some point, the axioms and theorems kind of blend into each other—they are just facts, with some of them being dic- tated (the axioms), while others are established (the theorems).
CHECK YOUR UNDERSTANDING 2.8 Show that in any vector space: If: vz+ = wz+ Answer: See page B-4. Then: vw=
Two different zeros come into play in the following theorem: The real number 0 that is involved in the scalar prod- uct at the left of the equality, and the vector 0 that appears to the right of the equality.
THEOREM 2.7 For any vector v in a vector space V: 0v = 0
PROOF: At times, as is the case here, a proof almost writes itself, once an appropriate initial step is taken (in this case, to write 0 as 00+ ): PofR Axiom (ii) (Distributive Axiom) 0v ==00+ v 0v + 0v The end is now in sight: just add the additive inverse of the vector 0v to both sides of the equation: 0v = 0v + 0v – 0v + 0v = – 0v + 0v + 0v Axiom (iv) and (ii): 0 = – 0v + 0v + 0v Axiom (iv): 00= + 0v Theorem 2.5: 0 = 0v In words, the above theorem tells us that:
Multiplying any vector by the scalar 0 results in the vector 0.
CHECK YOUR UNDERSTANDING 2.9 Answer: See page B-4. Prove that in every vector space V, r00= for every r . The above CYU together with Theorem 2.7 tells us that if either r = 0 or v = 0 , then the scalar product rv = 0 . The converse also hods: 54 Chapter 2 Vector Spaces
THEOREM 2.8 In any vector space V: If rv = 0 then r = 0 or v = 0
PROOF: If r = 0 , then surely the statement r = 0 or v = 0 holds, and we are done. If r 0 , then: rv = 0 1 1 ---rv = ---0 r r 1 Axiom (vii) and CYU 2.8: --- r v = 0 r 1v = 0 Axiom (viii): v = 0
CHECK YOUR UNDERSTANDING 2.10 Establish the following Cancellation Properties: (a) If r 0 and rv = rw , then vw= . Answer: See page B-4. (b) If v 0 and rv = sv , then rs= . Here is what the next theorem is saying: Multiplying any vector by the scalar –1 results in the additive inverse of that vector.
THEOREM 2.9 For any vector v in a vector space V: –1v = –v PROOF: Strategy: Show that if you add –1v to v you end up with the vector 0.
Axiom (viii) – 1v +1v ==–1v +1v –1+ v ==0v 0
Axiom (vi) PofR Theorem 2.7
CHECK YOUR UNDERSTANDING 2.11 Establish the following results, for v in a vector space V, and r .
Answer: See page B-5. (a) ––v = v (b) –r v = –rv (c) r–v = –rv 2.3 Properties of Vector Spaces 55
SUBTRACTION
We all want to replace the expression vw+ – with vw– . Let’s do it, but officially: DEFINITION 2.7 For vectors v and w in a vector space V, we “SUBTRACTION” define v minus w, denoted by vw– , to be the vector given by: vw– = vw+ –
A definition is the introduction of a new word in the language of mathematics. As such, one must understand all of the words used in its description. This is so in Definition 2.7, where the “new word “vw– ” on the left of the equal sign is described by previ- ously defined words “vw+ – ” on the right of the equal sign. Here are a few results featuring the operation of subtraction. They are very reminiscent of the familiar subtraction operation of real numbers. This should come as no surprise since the real number system is itself a vector space.
THEOREM 2.10 For any vectors v, w, and z in a vector space V, and scalars r, and s in : (a) vv– = 0 (b) vw+ – z = vwz+ – (c) vw+ – w = v
PROOF: (a) vv– ==vv+ – 0 Definition 2.7 (b) vw+ – z ===vw+ + –z vw+ + – z vwz+ –
Definition 2.7 (c) vw+ – w ===vww+ – v + 0 v
(b) (a)
CHECK YOUR UNDERSTANDING 2.12 (a) Show that for any two vectors v and w: –vw+ = – v – w (b) Use the Principle of Mathematical Induction (see Appendix A) to show that for any n vectors v1v2 vn : Answer: See page B-5. –v1 ++v2 +vn = – v1 – v2 – – vn 56 Chapter 2 Vector Spaces
We complete this section with a list of results; some of which we proved, some of which appeared in Check Your Understanding boxes, and others which you are invited to establish in the exercises.
THEOREM 2.11 For every v , w , and z in a vector space V, and every rst : (i) There exists a unique vector 0 V v +00 ==+ v v .
(ii) There exists a unique vector –v V such that vv+ – ==– v +0v . Cancellation (iii) If vz+ = wz+ , then vw= Properties { (iv) If r 0 and rv = rw , then vw= . (v) If v 0 and rv = sv , then rs= . (vi) 0v = 0 (vii) r00= (viii)rv = 0 if and only if r = 0 or v = 0
(ix)rsv + tw = rs v + rt w (x) –1v = –v
(xi)––v = v (xii) r–v ==–r v –rv (xiii) rvw– = rv – rw (xiv) rs– v = rv – sv (xv) –vw+ = – v – w (xvi) vw– = – w + v
(xvii) vwz– + = vw– – z (xviii) vwz– – = vw– + z 2.3 Properties of Vector Spaces 57
EXERCISES
Exercises 1-8. Prove: 1. Theorem 2.11 (iv): If r 0 and rv = rw , then vw= .
2. Theorem 2.11 (v): If v 0 and rv = sv , then rs= .
3. Theorem 2.11 (ix): rsv + tw = rs v + rt w .
4. Theorem 2.11 (xiii): rvw– = rv – rw .
5. Theorem 2.11 (xiv): rs– v = rv – sv .
6. Theorem 2.11 (xvi): vw– = – w + v .
7. Theorem 2.11 (xvii): vwz– + = vw– – z .
8. Theorem 2.11 (xviii): vwz– – = vw– + z .
9. Show that for any vector v in a vector space V, and any r : rv = ––rv . 10. Show that for any vector v in a vector space V and any integer n 1 : nv = n – 1 vv+ . 11. Let v, w, and z be any vectors in a vector space V, and let abc , with a 0 . Show that c b if av + bw = cz , then v = ---z – ---w . a a 12. Let v and w be vectors in a vector space V, with v 0 . Show that if rvw+ = svw+ , then rs= . 13. Let v and w be vectors in a vector space V. Show that if r 1 and rvw+ = v + rw , then vw= . 14. Show that for any v and w in a vector space V, and for any ab : ab+ vw+ = av ++bv aw +bw 15. Let v and w be non-zero vectors in a vector space V. Show that if rv + sw = 0 , with not both r and s equal to 0, then there exist unique numbers a and b such that v = aw and w = bv . Hint: Show that the condition that not both r and s equal 0 implies that neither is 0. 58 Chapter 2 Vector Spaces
PROVE OR GIVE A COUNTEREXAMPLE
16. All vector spaces contain infinitely many vectors. 17. Any vector space that contains more than one vector must contain an infinite number of vec- tors. 18. For any vector v in a vector space V and any r : rv = r – 1 vv+ 19. Let V and W be vector spaces. Let VW = vw v V w W with operations given by: v1 w1 + v2 w2 ==v1 + v2 w1 + w2 and rvw rv rw Then VW is a vector space.
20. Let V and W be vector spaces. Let VW = vw v V w W with operations given by:
v1 w1 + v2 w2 ==v1 – v2 w1 – w2 and rvw rv rw Then VW is a vector space. 2.4 Subspaces 59
2
§4. SUBSPACES
DEFINITION 2.8 A subspace of a vector space V is a non- SUBSPACE empty subset S of V which, together with the imposed operations of addition and scalar multiplication of V, is itself a vector space.
If S is to be a subspace of V, then it is itself a vector space and must therefore be closed under addition and scalar multiplication:
If s1 and s2 are in S, then s1 + s2 S . If s S and r , then rs S . In addition, the eight axioms listed in Definition 2.6, page 40, must also hold for S. Actually, the eight axioms come “free of charge,” once clo- sure is established; for: THEOREM 2.12 A subset S of a vector space V is a subspace of V if and only if: 1. S is nonempty. 2. S is closed under addition. 3. S is closed under scalar multiplication.
PROOF: If S is a subspace of the vector space V, then it is itself a vec- tor space and must therefore satisfy the three stated conditions. We now show that if those three conditions hold, then S is a subspace of V. Of the eight axioms of Definition 2.6, we need not worry about Axi- oms (i), (ii), (v), (vi), (vii), and (viii): (i)vu+ = uv+ (ii) uv+ + w = uvw+ + (v)ruv+ = ru + rv (vi) rs+ u = ru + su (vii)rsu = rs u (viii)1uu= Why not? Because, since they hold for all u, v and w in the given vec- tor space V, then they will surely hold for all u, v and w in the subset S of V. Why do we have to worry about the zero axiom? Because though we know that there is a vector 0 in V such that v + 0 = v for every v V (and therefore for every s S ), we have no assurance that 0 sits in S. To see that it does, take any vector v in S (we are given that S is nonempty), and then scalar multiply that vector by 0: 60 Chapter 2 Vector Spaces
Since v S and S is closed under scalar mutliplication 0v = 0 S Theorem 2.7, page 53 We now complete the proof by showing that if v S , then its addi- tive inverse –v is also in S: Since v S and S is closed under scalar mutliplication –1vv= – S Theorem 2.9, page 54
When challenging the “S is nonempty” condition of Theorem 2.12, one typically looks to see if the zero vector is contained in S. For: If 0 S , then S is certainly nonempty. If 0 S , then S is not a subspace, period.
EXAMPLE 2.8 Verify that S = abc cab= + is a subspace of 3 .
SOLUTION: We show that S satisfies the three conditions of Theorem 2.12. 1. Since the sum of the first two components of 0= 000 is equal to its third component, 0 S (and therefore S is not empty). 2. To show that S is closed under addition, we take two arbitrary ele- The “ticket” to be in S is ments of S: that the third component is equal to the sum of its s1 ==abab+ s2 cdcd+ first two components. and consider their sum:
s1 + s2 ==abab+ + cdcd+ acbd+ + abcd+++
Since s1 + s2 has the Since the third component of s1 + s2 , abcd+++ , equals the “ticket,” it is in S. sum of its first two components, s1 + s2 S . 3. We now show S is closed under scalar multiplication. For aba + b S and r : rabab+ ==ra rb r a+ b ra rb ra+ rb S
third component is sum of first two 2.4 Subspaces 61
CHECK YOUR UNDERSTANDING 2.13 Show that: a 2a S = a –0a Answer: See page B-5. is a subspace of the vector space M22 of Example 2.3, page 41.
The following theorem merges two of the properties of Theorem 2.12 into one: THEOREM 2.13 A nonempty subset S of a vector space V is a subspace of V if and only if:
For every s1 s2 S and r , rs1 + s2 S .
PROOF: For S a nonempty subspace of V, let s1 s2 S and r . Since S is closed under scalar multiplication: rs1 S . Since S is closed under addition: rs1 + s2 S .
Conversely, assume that for every s1 s2 S and r :
rs1 + s2 S (*) We show that S is closed under addition and scalar multiplication:
For any given s1 s2 S , simply choose r = 1 , and apply (*):
1s1 + s2 = s1 + s2 S To show that S is also closed under scalar multiplication, we first observe that (*) implies that 0 S : Since S is nonempty, we can choose an s S . Letting
s1 ==s2 s , and r = –1 in (*) brings us to: – 1s + s S 0 S Now consider any s S and r . Appealing to (*)
with s1 = s and s2 = 0 S , we find that: rs + 0 = rs S
EXAMPLE 2.9 Let u and v be any two vectors in a vector space V. Show that the set: Sa= u + bv ab is a subspace of V.
SOLUTION: Since 0 = 0u + 0v S , 0 S : S is not empty. 62 Chapter 2 Vector Spaces
For au + bv S and cu + dv S , and for r :
s1 s2 S and rR rau+ bv + cu+ dv = rau + cu + rbv + dv = ra+ c u + rb+ d v S
It is of the form Au+ Bv rs1 + s2 S (has the “ticket”)
CHECK YOUR UNDERSTANDING 2.14 Answer: See Page B-5. Show that Sxyz= xyz++= 0 is a subspace of 3 .
EXAMPLE 2.10 Let F denote the function space FX of Theorem 2.4, page 44, with domain X = . Show that: S = f F f9 = 0 is a sub- The “ticket” needed for a function f to get into S is space of F . that it maps 9 to 0. SOLUTION: As you recall, the zero in F turned out to be the function Z given by: Zx= 0 for every number x. In particular, since Z9 = 0 , Z S . Hence, S . If fg S and r , then: fg S and rR since f and g are in S
rfg+ 9 ===rf9 + g9 r00+0 rfg+ S rfg+ S (has the “ticket”)
CHECK YOUR UNDERSTANDING 2.15 Let F denote the function space of Theorem 2.4, page 44 (with domain X = ). Determine whether or not S = f F f0 = 9 Answer: Not a subspace. is a subspace of FR .
EXAMPLE 2.11 Show that for any ab : Sxy= ax+ by = c is a subspace of 2 if and only if c = 0 .
SOLUTION: We have to establish two results: The “if -condition:” If c = 0 , then S is a subspace of 2 . The “only if-condition:” If c 0 , then S is not a subspace of 2 . 2.4 Subspaces 63
The “only if-condition” is easily dispensed with: If c 0 , then 000= S , for: a 0 +0b 0 = c . Since S does not contain 0, S is not a subspace of 2 . To establish the “if-condition,” we assume the c = 0 , and first observe that S is not empty: Since a 0 +0b 0 ==c , 000= S .
We complete the proof by showing that if x1 y1 x2 y2 S and r , then:
rx1 y1 + x2 y2 = rx1 + x2 ry1 + y2 S which is to say, that:
arx1 + x2 +0bry1 + y2 = Here goes:
arx1 + x2 + bry1 + y2 = arx1 ++ax2 bry1 +by2
= rax1 + by1 + ax2 + by2
Since x1 y1 x2 y2 S: ==r00+ 0
CHECK YOUR UNDERSTANDING 2.16 In Example 1.10, page 20, we showed that: S = 16a – 2b 4a – 17b11a 11b ab is the solution set of the homogeneous system of equations: 2x ++03y –54z w = – 3x ++y 4zw += 0 x ++07y –114z w = Answer: See page B-5. Show that S is a subspace of 4 .
INTERSECTION AND UNION OF SUBSPACES
Let S and T be subspaces of a vector space V. Is their intersection ST = vv S and v T [Figure 2.7(a)] necessarily a subspace of V?Is their union ST = vv S or v T [Figure 2.7(b)] neces- sarily a subspace of V? The answer to the first question is “Yes” (Theo- rem 2.14 below), while the answer to the second question is “No” (Example 2.12 below). 64 Chapter 2 Vector Spaces
V V ST ST
ST = vv S and v T ST = vv S or v T S intersect T S union T (a) (b) Figure 2.7
In the exercises you are THEOREM 2.14 If S and T are subspaces of a space V, then so asked to show that the is their intersection: intersection of any num- ber of subspaces of V is ST = vv S and v T again a subspace of V. PROOF: ST is not empty: Since 0 S and 0 T (why?), 0 ST . uv ST and rR Let uv ST and rR . Being in the intersection of S and T, u and v are both in S and in T. Since S and T are subspaces, ruv+ S ruv+ ST and ruv+ T . Being in both S and T, ruv+ ST .
EXAMPLE 2.12 Show that the union of two subspaces S and T of a vector space V: ST = vv S or v T need not be a subspace of V. SOLUTION: While one can easily show that the set ST is non- empty, and that it is closed under scalar multiplication, one cannot show that it is closed under addition, for it need not be! And how do we show that this is the case? By exhibiting a specific vector space V, along with two specific subspaces S and T, such that their union fails T to be closed under addition. Let’s do it: Let V = 2 , S = x 0 x , and T = 0 y y . We 1 leave it for you to verify that both S and T are subspaces of 2 . To see S 1 that ST is not closed under addition, simply note that while 10 ST and 01 ST , 10 + 01 = 11 ST (for 11 is neither in S nor in T).
CHECK YOUR UNDERSTANDING 2.17
PROVE OR GIVE A COUNTEREXAMPLE: If S and T are subsets of a vector space V, and if ST is a subspace Answer: See page B-6. of V, then either S is a subspace of V or T is a subspace of V. 2.4 Subspaces 65
EXERCISES
Exercises 1-6. Determine if the given subset S of the vector space 2 is a subspace of 2 . Jus- tify your answer.
1.S = xy y = 2x 2. S = xy y = 2x + 1
3.S = xy xy+0= 4. S = xy xy
5. S = xy yx= 2 6. S = xy xy = 0
Exercises 7-12. Determine if the given subset S of the vector space 3 is a subspace of 3 . Jus- tify your answer.
7.S = xyz z = 2xy– 8. S = xyz z = 2xy+
9.S = xyz z =12xy– + 10. S = xyz z = 2xy++1
11. S = xyz xyz++= 0 12. S = xyz y = 0
Exercises 13-18. Determine if the given subset S of the matrix space M22 of Example 2.3, page
41 is a subspace of M22 . Justify your answer.
ab ab 13.S = d = 0 14. S = dab= + cd cd
ab ab 15. S = ad==0 c =2b 16. S = ab+2= c – 3d cd cd
ab ab 17.S = abcd+++= 1 18. S = abcd+++= 0 cd cd
Exercises 19-22. Determine if the given subset S of the polynomial space P2x of Theorem 2.3, page 43, is a subspace of P2x . Justify your answer.
19.Sax= 2 ++bx c a = 2 20. Sax= 2 ++bx c b = 0
21.Sax= 2 ++bx c a ++0 b c = 22. Sax= 2 ++bx c b = 2a 66 Chapter 2 Vector Spaces
Exercises 23-26. Determine if the given subset S of the polynomial space P3x of Theorem 2.3, page 43, is a subspace of P3x . Justify your answer.
23.Sax= 2 ++bx c b = 0 24. Sax= 2 ++bx c b = –2a + c 25.Sax= 2 ++bx c a + b = 0 26. Sax= 2 ++bx c b = 2ac = a + 1
Exercises 27-38. Determine if the given subset S of the function space F of Theorem 2.4 (with X = ), page 44, is a subspace of F . Justify your answer. 27. Sff= 0 = 0 28. Sff= 1 = 0 29. Sff= 1 = 1 30. Sff= 2x = 2fx 31. The subset of even functions: Sf= fx– = fx 32. The subset of odd functions: Sf= fx– = –fx 33. The subset of increasing functions: Sf= a bfa fb 34. The subset of decreasing functions: Sf= a bfa fb 35. The subset of bounded functions: Sf= fx M for every x , for some M 0 36. (Calculus dependent) Sf= f is continuous 37. (Calculus dependent) Sf= f is differentiable 38. (Calculus dependent) Sf= f is integrable 39. Let V be a vector space. Show that: (a)0 is a subspace of V. (b) V is a subspace of V.
40. (PMI) Establish the following generalization of Theorem 2.14.
(a) If S1S2 Sn are subspaces of a vector space V, then so is their intersection: n Si = S1 S2 Sn i = 1
(b) If S1S2 Sn are subspaces of a vector space V, then so is their intersection: Si = S1 S2 Sn i = 1
(c) Let A be a nonempty set. If S is a subspace of a vector space V for every A , then the set S is also a subspace of V. A 2.4 Subspaces 67
41. (PMI) Let v1v2 vn be vectors in a vector space V. Show that Sa= 1v1 +++a2v2 anvn ai 1 in is a subspace of V.
42. Let S and T be subspaces of a vector space V. Show that ST+ = st+ s S and t T is a subspace of V.
43. Let S and T be subspaces of a vector space V, with ST = 0 . Show that every vector in the subspace ST+ of the previous exercise can be uniquely expressed as a sum of a vector in S with a vector in T.
Suggestion: Show that if st+ = s1 + t1 , then ss= 1 and tt= 1 .
PROVE OR GIVE A COUNTEREXAMPLE
44. If S and T are both subsets of a vector space V, and if neither S nor W is a subspace of V, then ST cannot be a subspace of V. 45. If S and T are both subsets of a vector space V, and if neither S nor W is a subspace of V, then ST cannot be a subspace of V. 46. If S and T are subspaces of a vector space V, then ST+ = ST (see Exercise 43). 47. If S and T are subspaces of a vector space V, then STST + (see Exercise 43). 48. If S, T, and W are subspaces of a vector space V, then ST + TW is also a subspace of V (see Exercise 43). 49. If S, T, and W are subspaces of a vector space V, then STW + = ST + W (see Exercise 42). 50. If S and T are subspaces of a vector space V with ST = 0 , then ST is a subspace of V. 51. If S is a subspace of a vector space V, and if T is a subspace of S, then T is a subspace of V. 52. If a vector space has two distinct subspaces, then it has infinitely many distinct subspaces. 68 Chapter 2 Vector Spaces
2
§5. LINES AND PLANES
This chapter began with a consideration of vectors in the plane and in three dimensional space, both from a geometrical point of view (directed line segments, or “arrows”), and from an analytical perspec- tive (2-tuples and 3-tuples). The main focus of this section is to deter- mine and classify all of the subspaces of those Euclidean spaces. Additional insight for the material of this section will surface in the fol- lowing chapter on Dimension Theory. This section, in turn, is a nice lead-in to the following chapter, for Euclidean spaces have a dimension component built right into their terminology. It should come as no sur- prise, for example, to find that the Euclidean spaces 2 and 3 will turn out to have dimensions 2 and 3, respectively.
SUBSPACES OF 2
A subset of a vector space V that is neither V or 0 is said to be a proper subset of V. The following theorem serves to characterize the proper subspaces of 2 :
THEOREM 2.15 S is a proper subspace of 2 if and only if St= ab t for ab 00
PROOF: Let St= ab t for ab 00 . Appealing to Theorem 2.13, page 61, we first observe that S is not empty [it clearly contains ab ]. Moreover, for t1ab t2ab S and r :
rt1ab + t2ab = rt1 + t2 ab S .ab At this point we know that S is a non-empty subspace of 2 . Can it be all of 2 ? No, for the set ta tb t is the line in the plane passing through the origin and the point ab (see margin). 2 tab ab Conversely, assume that T is a proper subspace of . Since it is not empty, it contains a nonzero vector ab . Since T is a subspace, every scalar multiple of ab must be in T, which is to say: St= ab t T . We now show that, in fact, ST= , by demonstrating that if T were to contain any vector cd S then T For any vectorvR 2 : must be all of 2 (see margin)”: Can we find scalars A, B, such that: Aa+ Bc = x (a,b) .v Aab + Bcd = xy ? . Ab+ Bd = y
Yes, providing the last row of rref ac does not consist entirely (c,d) bd . of zeros (see the Spanning Theorem, page 18) — and it doesn’t (Exercise 61). 2.5 Lines and Planes 69
While no line in the plane that does not pass through the origin can represent a subspace of 2 (why not?), every line in the plane is paral- lel to one that does pass through the origin—a translation of a subspace of 2 : THEOREM 2.16 Let L be the line passing through two distinct The vector v is said to be a points Px= y and Qx= y in the direction vector for the line, 1 1 2 2 and the vector u is said to be plane [Figure 2.9(a)]. Then, in terms of vectors: a translation vector. Lu= + rv r (*)
where v ==PQ x2 – x1 y2 – y1 , and
ux= 0 y0 , for x0 y0 any chosen point on L [see Figure 2.8(b)]. [(*) is said to be a vector form repre- sentation for the line L.]
PROOF: Figure 2.8(b) may serve as a geometrical “proof” of the theo- rem; for if you take the vector v which is in the same direction as the line L, and stretch it every which way, then you will get a line that is parallel to L passing through the origin. Adding the vector u to every rv “moves” that line up to L, in a parallel fashion. Those not satisfied with y y u + rv L this geometrical proof L are invited to consider . x0 y0 Exercise 62. P Q r . to c u . e v n o ti a sl n v a r ctor rv x t ion ve direct x (a) (b) Figure 2.8 EXAMPLE 2.13 Find a vector form representation for the line L passing through the points 15 and 23 Note that the set: L = 15 + r12 – r SOLUTION: We take v ==2135– – 12 – to be our direction = 1 + r 52– r r vector, and u = 15 as our translation vector, leading us to the vec- This brings us to the so- tor form: L ==u + rv r 15 + r12 – r . called parametric repre- sentation of L: x ==1 +52ry – r CHECK YOUR UNDERSTANDING 2.18
(a) Referring to Example 2.13, find the vector form representation for the line L, when v is the vector from 23 to 15 , and u = 23 . (b) Your vector form representation in (a) will look different from Answer: that of Example 2.12: L = 15 + r12 – r . Appear- (a) 2 +32r – r ances aside, show that your set of two-tuples in (a) and ours of (b) 1 +52r – r r Example 2.13 are one and the same. 70 Chapter 2 Vector Spaces
SUBSPACES OF 3
We now move up a notch and focus our attention on the Euclidean space 3 . The first order of business is to arrive at vector form repre- sentations for lines and planes in 3 . As it was in 2 , a line in three- space is determined by any two points, P and Q, in 3 . In the exer- cises, you are invited to establish the following result (compare with Theorem 2.16): THEOREM 2.17 Let L be the line passing through two distinct points Px= y z and Qx= y z in One cannot envision a line in 1 1 1 2 2 2 n for n 3 . We can, three-space. Then, in vector form: however define, in vector L = u + rv r form, the line passing through: where v is the direction vector: Pa= 1a2 an v ==PQ x2 – x1 y2 – y1 z2 – z1 and Qb= 1b2 bn in n to be the set: and ux= 0y0 z0 is a translation vector, L = u + rv r with x0y0 z0 any chosen point on L. where: v = b – a b – a z . L 1 1 n n rv and: Q x y z . 0 0 0 u = a a a 1 2 n P . . u translation vector v = PQ direction vector y
x
EXAMPLE 2.14 Find a vector form representation for the line L passing through the points 20– 3 and 142 .
SOLUTION: We take the vector from 20– 3 to 142 The line can also be v ==12402– – – – 3 –145 as our direction vector. expressed in parametric form (see margin note Selecting the translation vector u = 20– 3 , we have: of Example 2.13): x ===2 –4ry rz 35+ r L ==u + rv r 20– 3 + r–145 r
CHECK YOUR UNDERSTANDING 2.19
Consider the line L = 135 + r21– 1 r . Find two points on L, both different from 135 , and proceed as in Example 2.14 to obtain another representation for the set L. Verify that “your Answer: See page B-6 set” is equal to the set 135 + r21– 1 r . 2.5 Lines and Planes 71
In Theorem 2.16, we noted that the line L in 2 which passes through the origin and the point ab (distinct from the origin) can be expressed in vector form: Lr= v r , where vab= . A sim- ilar result, which you are invited to establish in the exercises, holds for a plane in 3 :
THEOREM 2.18 Let P be a plane in 3 passing through the ori-
gin [Figure 2.9(a)]. Let x1y1 z1 and x2y2 z2 be any two points on P which do not both lie on a common line passing through the origin. Then, in vector form: Pr= u + sv rs
where ux= 1y1 z1 and vx= 2y2 z2 .
z
.B P . A x y z . u 0 0 0 . .C v . w u . y y 0 v x x (a) (b) Figure 2.9 In this general setting, we have (compare with Theorem 2.17): THEOREM 2.19 Let P be the plane passing through three non- colinear (not lying on a common line) points
Ax= 1y1 z1 , Bx= 2y2 z2 and
Cx= 3y3 z3 [Figure 2.9(b)]. Then, in vec- u and v are said to be direction vectors, and w tor form: is said to be a translation P = w ++ru sv rRs vector. where u ==AB x2 – x1y2 – y1 z2 – z1
v ==AC x3 – x1y3 – y1 z3 – z1 ,
and wx= 0y0 z0 for x0y0 z0 any cho- sen point on P. 72 Chapter 2 Vector Spaces
EXAMPLE 2.15 Find a vector form representation for the Plane P passing through the points 32–2 , 25– 3 and 41– 3 .
SOLUTION: We elect 32–2 to play the role of both w and of
x1y1 z1 in Theorem 2.19; with: u ==25– 3 – 32– 2 –175– and P consists of all points v ==41– 3 – 32– 2 13– 5 xyz such that: Then: x = 3 – r + s P = w ++ru sv rs y = –72 ++r 3s z = 25–5r – s = 32–2++r–751– s13– 5 rs The above is said to be a parametric representa- = 3 – r + s –72 ++r 3s 25–5r – s rs tion of the plane (with parameters r and s). CHECK YOUR UNDERSTANDING 2.20
(a) Repeat our solution to Example 2.15, but this time letting 41– 3 play the role of w, instead of 32–2 ; all of the rest remaining as before. (b) Your set representation for P in (a) will look different from that of Example 2.15. Appearances aside, show that your set in (a) Answer: See page B-6 and ours of Example 2.15 are one and the same.
We previously showed that the only proper subspaces of 2 are the lines passing through the origin (as sets). The following theorem, a proof of which is relegated to the exercises, settles the subspace issue in 3 :
THEOREM 2.20 S is a proper subspace of 3 if and only if S is the set of points on a line that passes through the origin, or the set of points on a plane that passes through the origin. 2.5 Lines and Planes 73
EXERCISES
Exercises 1-4. Determine a vector form representation for the line in 2 passing through the ori- gin and the given point. 1.15 2.–24 3.51 – 4. –22 –
Exercises 5-8. Determine a vector form representation for the line in 2 passing through the two given points. 5.13 24 – 6.35 55 7.35 37 8. 24 –52 –
Exercises 9-12. Determine two different vector form representations for the line in 2 passing through the two given points, and then proceed to show that the set of points associated with the two vector forms are one and the same. 9.13 24 – 10.35 55 11.35 37 12. 24 –52 –
Exercises 13-20. Determine a vector form representation for the line in 2 that passes through the point 37 and is parallel1 to the line of: 13. Exercise 1 14. Exercise 2 15. Exercise 3 16. Exercise 4 17. Exercise 5 18. Exercise 6 19. Exercise 7 20. Exercise 8
Exercises 21-28. Determine a vector form representation for the line in 2 that passes through the point 37 and is perpendicular2 to the line of: 21. Exercise 1 22. Exercise 2 23. Exercise 3 24. Exercise 4 25. Exercise 5 26. Exercise 6 27. Exercise 7 28. Exercise 8
Exercises 29-32. Determine a vector form representation for the line in 3 passing through the origin and the given point. 29.245 30.100 31.–240 32. –441 – –
Exercises 33-36. Determine a vector form representation for the line in 3 passing through the two given points. 33.245 311 34. 012 102
35.34– 1 210 36. 511 – – 222
1. Two lines are parallel they have equal slopes, or if both lines are vertical. 2. Two lines are perpendicular if and only if the slope of one is the negative reciprocal of the slope of the other, or if one line is horizontal and the other is vertical. 74 Chapter 2 Vector Spaces
Exercises 37-40. Determine two different vector form representations for the line in 3 passing through the two given points, and then proceed to show that the set of points associated with the two vector forms are one and the same. 37.245 311 38. 012 102
39.34– 1 210 40. 511 – – 222
Exercises 41-48. Determine a vector form representation for the line in 3 that passes through the point 12– 1 and is parallel to the line of: 41. Exercise 29 42. Exercise 30 43. Exercise 31 44. Exercise 32 45. Exercise 33 46. Exercise 34 47. Exercise 35 48. Exercise 36 Exercises 49-52. Determine a vector form representation for the plane passing through the origin and the two given points. 49.132 214 50. 314 200
51.200 020 52. –321 – – 241 Exercises 53-56. Determine a vector form representation for the plane passing through the three given points. 53.341 215 11– 1 54. 210 – 211 – 123
55.24– 3515 41– 1 56. –112 – 231 – 121 Exercises 57-60. Determine two different vector form representations for the plane passing through the two given points, and then proceed to show that the set of points associated with the two vector forms are one and the same. 57.341 215 11– 1 58. 210 – 211 – 123
59.24– 3515 41– 1 60. –112 – 231 – 121 61. Complete the proof of Theorem 2.15. Incidentally:
You can also show directly that if v1 = ab and v2 = cd are such that v2 rv1 for 2 any r , then for any v there exist r1 r2 such that v = r1v1 + r2v2 . 62. Prove Theorem 2.16. 63. Prove Theorem 2.17. 64. Prove Theorem 2.18. 65. Prove Theorem 2.19. 66. Prove Theorem 2.20. Chapter Summary 75
CHAPTER SUMMARY
EUCLIDEAN VECTOR The set of n-tuples, with addition and scalar multiplication given by: SPACE n v1v2 vn + w1w2 wn = v1 + w1 v2 + w2 vn + wn
rv1v2 vn = rv1rv2 rvn ABSTRACT VECTOR A nonempty set V, closed under addition and scalar multiplication, satisfying the following eight properties: SPACE (i)uv+ = vu+ (ii)uv+ + w = uvw+ + (iii) There is a vector in V, denoted by 0 such that v + 0 = v for every vector v in V. (iv) For every vector v in V, there is a vector –v in V, such that vv+ – = 0 . (v) ruv+ = ru + rv (vi) rs+ u = ru + su (vii) rsu = rs u (viii) 1uu= SUBTRACTION vw– = vw+ – Uniqueness of 0 and –v There is but one vector 0 which satisfies the property that v + 0 = v for every v in V. For any given vector v in V, there is but one vector –v in V such that vv+ – = 0 . Cancellation Properties If vz+ = wz+ , then vw= If r 0 and rv = rw , then vw= . If v 0 and rv = sv , then rs= . Zero Properties 0v = 0 r00= rv = 0 if and only if r = 0 or v = 0 Inverse Properties –1v = –v ––v = v r–v ==–r v –rv 76 Chapter 2 Vector Spaces
SUBSPACE A nonempty subset S of V which is itself a vector space under the vector addition and scalar multiplication operations of the space V. Closure says it all A nonempty subset S of a vector space V is a subspace of V if and only if it is closed under addition and under scalar multipli- cation. A one liner A nonempty subset S of a vector space V is a subspace of V if
and only if for every s1 s2 S and r , rs1 + s2 S . Intersection of subspaces The intersection of any collection of subspaces in a vector space V is again a subspace of V. PROPER SUBSPACES A subspace of a vector space V distinct from the trivial sub- space 0 and V itself is said to be a proper subspace of V.
Vector form of lines Let L be the line in 2 passing through the origin, and let ab be any point on L distinct from the origin, then, in vector form: Lr= v r , where v = ab
Let L be the line in the plane passing through Px= 1 y1 and Qx= 2 y2 . Then, in vector form, L = u + rv r
where v = PQ , and ux= 0 y0 , with x0 y0 any chosen point on L. Let L be the line in 3 passing through two distinct points Px= 1y1 z1 and Qx= 2y2 z2 . Then L = u + rv rR where v is the direction vector
v ==PQ x2 – x1 y2 – y1 z2 – z1 and u = x0y0 z0 is a translation vector, with x0y0 z0 any chosen point on L.
Vector form of planes Let P be a plane in 3 passing through the origin. Let
x1y1 z1 and x2y2 z2 be any two points on P which do not both lie on a common line passing through the origin. Then, in
vector form: Pr= u + sv rs where ux= 1y1 z1
and vx= 2y2 z2 . Let P be the plane passing through three non-colinear points
P1 = x1y1 z1 , P2 = x2y2 z2 and P3 = x3y3 z3 .
Then P = w ++ru sv rRs where u = P1P2 ,
v = P1P3 , and w = x0y0 z0 is any chosen point on P 3.1 Spanning Set 77
3 CHAPTER 3 BASES AND DIMENSION It is easy to see that every vector xyz in 3 can uniquely be expressed in terms of the three vectors 100 , 010 , and 001 . For example: 52– 9 = 5100 ++2010 –9 001 In this chapter, we consider an arbitrary vector space V to see if we can
find a set of vectors v1v2 vn in V, called a basis for V, such that every vector v in V can be uniquely expressed in the form:
v = c1v1 +++c2v2 cnvn for some scalars c1c2 cn . As you will see, while many such sets of vectors v1v2 vn may exist, the number of vectors in those sets will always be the same. For example, if you find a basis for a vec- tor space V that contains 5 vectors, v1v2 v3 v4 v5 , and someone else finds another basis, that other basis must also consist of 5 vectors. That being the case, we will then be in a position to say that the vector space V has dimension 5.
§1. SPANNING SETS
Using vector addition and scalar multiplication in 2 , one can build the vector v = 914 from the vectors v1 = 34 and v2 = 12 : 914 = 234 + 312 and we say that v = 914 is a linear combination of v1 = 34 and v2 = 12 . In general:
DEFINITION 3.1 A vector v in a vector space V is said to be a linear combination of vectors v v v in LINEAR 1 2 n COMBINATION V, if there exist scalars c1c2 cn such that: v = c1v1 +++c2v2 cnvn
EXAMPLE 3.1 Determine whether or not the vector 0224 in 3 is a linear combination of the vectors 138 and 254 . 78 Chapter 3 Bases and Dimension
SOLUTION: We are to see if we can find scalars a and b such that: 0224 = a138 + b254
or: a + 2b 3a + 5b 8a + 4b = 0224 Equating coefficients, we come to the following system of three equations in two unknowns: a b a b a +02b = 120 104 augS S: 3a +25b = 352 rref 01– 2 8a +244b = 8424 000
Solution: a ==4 b –2 Conclusion: 0224 is a linear combination of the vectors 138 and 254 : 0224 = 4138 + –2 254
Some Added Insight on Example 3.1
The set of all linear combinations of z 138 138 and 254 is the plane in 3 containing those vectors. As such, were we to pick an arbitrary point in 3 , say 254 243 , then there would be little chance that it would lie in that plane, and would therefore not be a linear combina- tion of the two given vectors: x a b a b a +22b = Note that, except for the last augS 12 2 10 0 3a +45b = rref column, this augmented S: 35 4 01 0 matrix is the same as that of Example 3.1. 8a +34b = 84 3 00 1
a +00b = 0ab+0= 0a +10b = No solution!
CHECK YOUR UNDERSTANDING 3.1
Determine if the given vector is a linear combination of the vectors 138 and 254 .
Answer: (a) No. (b) Yes. (a) –238 – (b) –248 – 3.1 Spanning Set 79
THEOREM 3.1 The set of linear combinations of the set of vec- tors v1v2 vn in a vector space V is a sub- space of V.
PROOF: Let Sc= 1v1 +++c2v2 cnvn . We first observe that since 0 = 0v1 +++0v2 0vn S , S is not empty. Moreover, for u = a1v1 +++a2v2 anvn and w = b1v1 +++b2v2 bnvn in uw S and r S, and r :
ruw+ = ra1v1 +++a2v2 anvn + b1v1 +++b2v2 bnvn
= ra1 + b1 v1 +++ra2 + b2 v2 ran + bn vn S ruw+ S
a linear combination of the vis
DEFINITION 3.2 The set of linear combinations of a set of vec- tors v v v in a vector space V is SPANNING 1 2 n called the subspace of V spanned by v1v2 vn and will be denoted by Spanv1v2 vn .
In the event that V = Spanv1v2 vn , v1v2 vn is said to span V.
EXAMPLE 3.2 Determine if the vectors 2x2 + 3x – 1 , x – 5 , 2 and x – 1 span the space P2x of polynomi- als of degree less than or equal to two.
2 SOLUTION: We consider an arbitrary vector ax ++bx c in P2x to see whether or not we can find scalars rsand t such that: r2x2 + 3x – 1 ++sx– 5 tx2 – 1 = ax2 ++bx c Expanding and combining the left side of the above equation brings us to: 2rt+ x2 ++3rs+ xr–5– s – t = ax2 ++bx c
Equating coefficients we come to the following system of three equa- tions, in the unknowns r, s, and t (the a, b, and c are not variables, they are the fixed coefficients of the polynomial ax2 ++bx c ): System S was solved directly in Example 1.7, page 16. In that example, we labeled the variables x, y, and z, instead of r, s, and t. 80 Chapter 3 Bases and Dimension
2rt+ = a S: 3rs+ = b –5r – s – t = c Can we find values for r, s, and t such that the above three equations hold? The Spanning Theorem (page 18), tells us that the answer is “yes” if and only if rref coefS does not contain a row consisting entirely of zeros. Let’s see:
Since the rref-matrix does not contain a row of zeros, system S has a solution for all values of a, b, and c, and we conclude that the vectors 2 2 2x + 3x – 1 , x – 5 , and x – 1 span the space P2x . While the above does not tell you how to build ax2 ++bx c from 2x2 + 3x – 1 , x – 5 , and x2 – 1 , it does tell you that it can be done, for
each and every polynomial in P2x . If you want to see how to build any particular polynomial, say the polynomial 4x2 + 10x – 6 , then that’s not a problem, for the task reduces to finding scalars r, s, and t, such that: 4x2 + 10x – 6 = r2x2 + 3x – 1 ++sx– 5 tx2 – 1 Or: 4x2 +210x – 6 = rt+ x2 ++3rs+ xr–5– s – t Equating coefficients, we have: r s t r s t 2rt+4= 2014 1003 augS rref S: 3rs+10= 31010 0101 –5r – s –6t = – –51 –1–6– 001– 2 Conclusion: 4x2 +3210x – 6 = x2 + 3x – 1 ++1x – 5 –2 x2 – 1 3.1 Spanning Set 81
CHECK YOUR UNDERSTANDING 3.2 (a) Use the Spanning Theorem (page 18) to show that the vectors 12 10 01 04 span M22 . 34 10 01 20
(b) Express –51 as a linear combination of the above four vectors. Answer. See pare B-7. 113
EXAMPLE 3.3 Do the vectors 2101 , 12– 20 , 231– 1 and 124– 4 span 4 ?
SOLUTION: Let abcd be an arbitrary vector in 4 . Are there scalars x, y, z, and w such that: abcd = x2101 +++y12– 20 z231– 1 w124– 4 which is to say: does the following system of equations have a solu- tion for all values of a, b, c, and d? 2xy++2zw + = a x –32y ++z 2w = b S: 0x +++2yz4w = c x + 0yz–4– w = d The Spanning Theorem tells us that it does not, as rref coefS con- tains a row consisting entirely of zeros:
While the above argument establishes the fact that the vectors 2101 , 12– 20 , 231– 1 and 124– 4 do not span 4 , it does not define for us the subspace of 4 spanned by those four vectors; bringing us to:
EXAMPLE 3.4 Determine the subspace of 4 spanned by 2101 , 12– 20, 231– 1 and 124– 4. 82 Chapter 3 Bases and Dimension
SOLUTION: We are to find the set of all vectors abcd for which there exist scalars x, y, z, and w such that: abcd = x2101 +++y12– 20 z231– 1 w124– 4 which again boils down to a consideration of a system of equations: 2xy++2zw + = a x –32y ++z 2w = b S: 0x +++2yz4w = c x + 0yz–4– w = d for to say that abcd is in the space spanned by the four given vec- tors is to say that system S has a solution for those numbers a, b, c, and d. That system appeared earlier in Example 1.8, page 17 where it was noted that the given system of equation has a solution if and only if: – 10a ++7b 12c +13d = 0 Bringing us to a representation for the space spanned by the four given vectors: Span 2101 12– 20 231– 1 124– 4 = abcd –710a ++b 12c +13d = 0
CHECK YOUR UNDERSTANDING 3.3 Determine the space spanned by the vectors 215 , 12– 2 , Answer: 3 3 abc b = 5c – 12a 051 . If it is not all of , exhibit a vector in that is not in See Page B-8. Span 215 12– 2 051 . The following example differs from the previous two in that it is not confined to a specific concrete vector space, like 4 .
EXAMPLE 3.5 Let the set of vectors v1v2 v3 and w1 w2 be such that wi Spanv1v2 v3 for 1 i 2 . Show that Spanw1 w2 Spanv1v2 v3 .
SOLUTION: In any non-routine problem, it is important that you chart out, in one way or another, what is given and that which is to be established:
We are given that the vectors w1 and w2 are in the space spanned by the three vectors v1v2 and v3 , and are to show that for any given scalars a and b, the vector aw1 + bw2 can be written as a linear combination of the vectors v1v2 and v3 ; which is to say that we can find scalars c1c2 c3 such that:
aw1 + bw2 = c1v1 ++c2v2 c3v3 3.1 Spanning Set 83
From the given information, we know that there exist scalars
a1a2 a3 and b1b2 b3 such that:
w1 ==a1v1 ++a2v2 a3v3 and w2 b1v1 ++b2v2 b3v3 Consequently:
aw1 + bw2 = aa1v1 ++a2v2 a3v3 + bb1v1 ++b2v2 b3v3
= aa1v1 +++++aa2v2 aa3v3 bb1v1 bb2v2 bb3v3
= aa1 + bb1 v1 ++aa2 + bb2 v2 aa3 + bb3 v3
= c1v1 ++c2v2 c3v3
The following theorem generalizes the situation of Example 3.5.
THEOREM 3.2 Let the set of vectors v1v2 vn and w1w2 wm be such that
wi Spanv1v2 vn for 1 im.Then:
Spanw1w2 wm Spanv1v2 vn .
PROOF: If w Spanw1w2 wm then, for some scalars ci :
w = c1w1 +++c2w2 cmwm We also know that for each 1 im , there exist scalars ai1ai2 ain , such that:
wi = ai1v1 +++ai2v2 ainvn Consequently: w = c1w1 ++ cmwm
= c1a11v1 ++ a1nvn ++cmam1v1 ++ amnvn
= c1a11 ++ cmam1 v1 ++ c1a1n ++ cmamn vn
Since w can be expressed as a linear combination of v1v2 vn , w Spanw1w2 wm . Consequently:
Spanw1w2 wm Spanv1v2 vn
CHECK YOUR UNDERSTANDING 3.4
Prove that for any three vectors v1v2 v3 in a vector space V:
Spanv1v2 v3 = Spanv1v1 + v2 v1 ++v2 v3 Answer: See Page B-8. 84 Chapter 3 Bases and Dimension
EXERCISES
Exercises 1-4. Determine whether or not the given vector in 3 is a linear combination of the vectors –121 and 203 . 1.234 2.14– 2 3.–560 4. –110 11
Exercises 5-8. Determine whether or not the given vector in M22 is a linear combination of the
vectors 1201 12. 03 23 30
5.59 6.49 7.11 6 8. 611 73 73 67 76
Exercises 9-12. Determine whether or not the given vector in P3 is a linear combination of the vectors x3 +21x2 –3 and x + 1 .
9. 2x3 ++x 1 10.3x2 –2x – 11.2x3 + 4x2 –6x – 12. 2x3 ++6x2 x – 6
Exercises 13-16. Show that the given vector in the function space Fx of Theorem 2.4, page 44, is in the space spanned by the vectors sinxx cos sin2x cos2x tan2x cot2x .
13. cos2x 14.sin--- – x 15.sin--- – x 16. sin--- – x 2 7 7
Exercises 17-21. Determine if the given vectors span 4 . If not, find a specific vector in 4 which is not contained in that subspace. 17. 1000 0200 0030 0004 18. 1000 1100 1110 1111 19. 1111 1001 0110 6444 20. 2131 1213 3112 1123 21. –1234 – – – –3112 1213 7032– –
Exercises 22-25. Determine if the given vectors span P3 . If not, find a specific vector in P3 which is not contained in that subspace.
22.1 x2x2 3x3 23. 1 +1x2 + xx – + x2 1 ++xx2 +x3 24.x 1 + x 1 ++xx2 1 ++xx2 +x3 25. 1xx2 + 1 x3 – x2 26. For what values of c do the vectors 213 435 00c span 3 ?
2 2 27. For what values of c do the vectors c 2x 2x 2x ++2xc span P2 ? 3.1 Spanning Set 85
28. For what values of a and b do the vectors ab and –ba span 2 ?
29. Show that for any given set of vectors v1v2 vn , vi Spanv1v2 vn for every 1 in.
30. Let the set of vectors v1v2 vn and w1w2 wm be such that wi Spanv1v2 vn for 1 im and vi Spanw1w2 wm for 1 in . Prove that Spanw1w2 wm = Spanv1v2 vn .
31. Show that if v1v2 v3 span a vector space V, then for any vector v4 the vectors v1v2 v3 v4 also span V. 32. Show that a nonempty subset S of as vector space V is a subspace of V if and only if Spanv1v2 vn S for every v1v2 vn S . 33. Let P denote the vector space of all polynomials of Exercise 22, page 50. Show that no finite set of vectors in P spans P . 34. Let S be a subset of a vector space V. Prove that SpanS is the intersections of all subspaces of V which contain the set S.
PROVE OR GIVE A COUNTEREXAMPLE
35. If the vectors u and v span V, then so do the vectors u and uv+ . 36. If the vectors u and v span V, then so do the vectors u and uv– . 37. If the vectors u and v are contained in the space spanned by the vectors w and z, then Spanuv = Spanwz .
38. If Spanv1v2 v3 = V , and if vi Spanw1 w2 for 1 i 3 , then Spanw1 w2 = V .
39. If S1 and S2 are finite sets of vectors in a vector space V, then: SpanS1 S2 = SpanS1 SpanS2 .
40. If S1 and S2 are finite sets of vectors in a vector space V, then: SpanS1 S2 SpanS1 SpanS2
41. If S1 and S2 are finite sets of vectors in a vector space V, then: SpanS1 S2 = SpanS1 SpanS2 .
42. If S1 and S2 are subspaces of a vector space V, then:
SpanS1 S2 = SpanS1 SpanS2 . 86 Chapter 3 Bases and Dimension
3
§2. LINEAR INDEPENDENCE
The subspace of 3 spanned by the vectors 100 , 010 , 250 is built from those vectors. But there is a kind of inefficiency with the three building blocks 100 010 250 , in that whatever can be built from those three vectors can be built with just two of them; as one of them, say the vector 250 , can itself be con- structed from the other two: 250 = 2100 + 5010 The following concept, as you will soon see, addresses the above “inefficiency issue:” Note that if each ci = 0 , then surely DEFINITION 3.3 A set of vectors v1v2 vn are linearly c v +++c v c v 1 1 2 2 n n Linearly independent if: will equal zero. Independent c v +c v + +c v == 0 each c 0 To say that v1v2 vn is 1 1 2 2 n n i linearly independent, is to A collection of vectors that are not linearly say that no other linear combination of the vectors independent is said to be linearly dependent. equals 0. EXAMPLE 3.6 Is x2 x2 + xx 2 ++x 1 a linearly indepen- dent set in the vector space P2x ? SOLUTION: To resolve the issue we consider the equation: ax2 ++bx2 + x cx2 ++x 1 = 0x2 ++0x 0 Equating coefficients, brings us to the following system of equations: abc++= 0 bc+0= c = 0 Working from the bottom up, we see that c ===0 b 0 a 0 is the only solution for the above system of equations, and therefore con- clude that the given set of vectors is linearly independent. EXAMPLE 3.7 212 120 10– 2 13–0 Is 340 –011 –213 26– 1
a linearly independent set in M23 ?
SOLUTION: If:
a 212 ++b 120 c 10– 2 +d 13–0= 000 340 –011 –213 26– 1 000 3.2 Linear Independence 87
Then: Most graphing calculators do not have the capability of 2abcd+++ a +22b – 3d a – 2c = 000 “rref-ing” a “tall matrix.” 3ab–3–2c +4d a ++2c 6d bcd+ – 000 But you can always add enough zero columns to Leading us to the homogeneous system: arrive at a square matrix: a b c d a b c d 2abcd+++= 0 2111 1 001 a +02b – 3d = 120– 3 0 1 02– 2a –02c = 20–0 2 S: coef(S) rref 001 1 3ab–3–2c +0d = 31–3–2 0000 4a ++2c 6d = 0 4026 0000 bcd+0– = 011– 1 0000 Since rref[coef(S)] has a free variable, the system has nontrivial solu- tions, and we therefore conclude that the given set of vectors is lin- early dependent.
CHECK YOUR UNDERSTANDING 3.5
2 2 Answer: Yes. Is x 2x + x x – 3 a linearly independent set in P2 ?
EXAMPLE 3.8 Let v1v2 v3 be a linearly independent set of vectors in a vector space V. Show that v1v2 + v1 v3 – v2 is also linearly independent.
SOLUTION: We start with: av1 ++bv2 + v1 cv3 – v2 = 0 and go on to show that abc===0 :
av1 ++bv2 + v1 cv3 – v2 = 0
regroup: ab+ v1 ++bc– v2 cv3 ==ab+0
Since v1v2 v3 is linearly independent: ab+0= bc–0= a ===0 b 0 c 0 c = 0
CHECK YOUR UNDERSTANDING 3.6
Let v1v2 v3 v4 be a linearly independent set of vectors in a vec-
tor space V. Show that v1v1 + v2 v1 ++v2 v3 v1 +++v2 v3 v4 Answer: See page B-8. is also a linearly independent set. 88 Chapter 3 Bases and Dimension
Here is a useful consequence of the Linear Independence Theorem of page 22:
THEOREM 3.3 n A set of vectors v1v2 vm in is lin- LINEAR INDEPENDENCE early independent if and only if the row-reduced- THEOREM FOR n echelon form of the nm matrix with ith col-
umn the (vertical) n-tuple vi has m leading ones.
PROOF: For 1 im , let vi = a1ia2i ani . To challenge lin- ear independence of those m vectors, we consider the vector equation: c1v1 ++ cmvm = 0:
c1a11a21 an1 +++0c2a12a22 an2 cma1ma2m anm = 0 c1a11 + c2a12 ++cma1m c1an1 + c2an2 ++cmanm = 0 0 Equating coefficients brings us to the following homogeneous sys- tem of n equations in m unknowns: c1 c2 cm a c + a c + + a c =0 a a a 11 1 12 2 1m m 11 12 1n H: a21c1 + a22c2 + + a2ncn =0 coefS a a a S: . 21 .22 .2n . . . . . . an1c1 + an2c2 + + anmcm =0 an1 an2 anm Applying the Linearly Independent Theorem of page 22, we conclude that the above homogeneous system of equations has only the trivial
solution c1 ==c2 =cm =0 if and only if rref coefS has m leading ones. EXAMPLE 3.9 Determine if: 13– 21 2132 1314 4727 is a linearly independent set of vectors in 4 . SOLUTION: 1214 1 00 1 3137 rref 0 1 0 1 –3122 001 1 1247 000 0 Since the above matrix does not have 4 leading ones, the given set of vectors is not linearly independent.
CHECK YOUR UNDERSTANDING 3.7
Use Theorem 3.3 to show that there cannot exist a set of five linearly Answer: See page B-9. independent vectors in 4 . 3.2 Linear Independence 89
THEOREM 3.4 Any set of vectors containing the zero vector 0 is linearly dependent.
PROOF: Let 0 v1v2 vm be a subset of a vector space V. Since: 10 ++++0v2 0v3 0vm = 0 0 0 v1v2 vm is not linearly independent. The above theorem tells us that 0 is a linearly dependent set in any vector space V. As for the rest:
THEOREM 3.5 A finite set of vectors, distinct from 0 , is linearly independent if and only if no vec- tor in the set can be expressed as a linear combination of the rest.
PROOF: To establish the fact that linear independence implies that no CONTRAPOSITIVE PROOF vector in the set can be expressed as a linear combination of the rest, Let P and Q be two proposi- we show that if some vector in the set can be expressed as a linear tions. You can prove that: combination of the rest, then the set is not linearly independent (see PQ margin). Here goes: by showing that: Assume that one of the vectors in v1v2 vn can be Not-Q Not-P expressed as a linear combination of the rest. Since one can (After all if Not-Q implies Not-P, always reorder the given vectors we may assume, without loss of then you certainly cannot have P without having Q: think about it) generality, that v1 is a linear combination of the rest: v1 = a2v2 +++a3v3 anvn Leading us to: 1v1 – a2v2 – a3v3 – –0anvn = Since the coefficient of v1 is not zero, we conclude that v1v2 vn is linearly dependent. To establish the converse we again turn to a contrapositive proof: If v v v is not linearly independent, then we can find sca- v1v2 vn NOT linearly 1 2 n independent lars c1c2 cn , not all zero, such that: c1v1 +++c2v2 cnvn = 0 Assuming, without loss of generality, that c1 0 we find that we can express v1 as a linear combination of the rest: c2 c3 cn some v can be expressed as a i v1 = – -----v2 – -----v3 – – -----vn linear combination of the rest c1 c1 c1 The next theorem tells us that whatever can be built from a collection of linearly independent vectors, can only be built in one way:
In the exercises you are THEOREM 3.6 Let v1v2 vn be a linearly independent set. invited to establish the If a v ++ a v = b v ++ b v , then: converse of this theo- 1 1 n n 1 1 n n rem. ai = bi , for 1 in . 90 Chapter 3 Bases and Dimension
PROOF: n n a1v1 +++a2v2 anvn = b1v1 +++b2v2 bnvn
a v = b v a v +++a v a v – b v +++b v b v = 0 i i i i 1 1 2 2 n n 1 1 2 2 n n i = 1 i = 1 a1 – b1 v1 +++a2 – b2 v2 an – bn vn = 0
by linear Ind.: a1 – b1 ==0 a2 – b2 = 0 an – bn 0
ai = bi a1 = b1 a2 = b2 an = bn
CHECK YOUR UNDERSTANDING 3.8
Show that the vectors 213 502 11311 are linearly dependent in 3 , and that 8412 is in the space spanned by those vectors. Express 8412 as a linear combination of those Answer: See page B-9. vectors in two distinct ways. A linearly independent set of vectors S in a vector space V may not be able to accommodate additional vectors without losing its indepen- dence. This is not the case if V SpanS :
THEOREM 3.7 If v1v2 vn is a linearly independent set EXPANSION of vectors, and if v Spanv v v , then THEOREM 1 2 n v1v2 vn v is also linearly independent.
PROOF: Let S = v1v2 vn be linearly independent, and let v SpanS . We show that v1v2 vn v is linearly indepen- dent by showing that no vector in v1v2 vn v can be expressed as a linear combination of the rest. To begin with, we observe that v cannot be expressed as a linear com- bination of the rest, as v SpanS . Suppose then that some other vector in v1v2 vn v , say for definiteness the vector v1 , can be expressed as a linear combination of the rest:
v1 = c2v2 ++++c3v3 cnvn cv
Since S = v1v2 vn is linearly independent, c 0 (why?). But then: 1 c c c v = ---v – ----2-v – ----3-v – – ----n-v c 1 c 2 c 3 c n again contradicting the given condition that v SpanS .
CHECK YOUR UNDERSTANDING 3.9
Find a linearly independent set of four vectors in P3 which includes Answer: See page B-9. the two vectors x3 + x and – 7 . 3.2 Linear Independence 91
EXERCISES
Exercises 1-6. Determine if the given set of vectors is a linearly independent set in 3 . 1.215 4110 2. 000 –64– 5
5 3. 215 4110 4110 – 4. 32 – --- –64– –5 120 2 5.134 25– 1 010 220 6. 101 201 102 123
Exercises 7-12. Determine if the given set of vectors is a linearly independent set in M22 .
12 21 12 11 21 11– 7. 8. 34 34 43 34 23 –65 –
12 21 12 21 12 21 12 12 9. 10. 34 34 43 25 34 34 43 25
12 21 12 12 11 11 22 01 –22 00 11. 12. 34 34 43 25 11 11 21 20 22 01
Exercises 13-17. Determine whether the given set of vectors is a linearly independent set in P3 .
13. x + 1 x2 + x + 1 x3
14. x + 1 x2 + 13 x – 517
15. 3x3 3x2 ++3x 33 x3 +++3x2 6x 63 x + 3
16. 2x3 3x2 ++3x 33 x3 +++3x2 6x 63 x + 3
17. 2x3 3x2 ++3x 33 x3 +++3x2 6x 63 x + 5
Exercises 18-27. Determine if the given set of vectors in the function space FX of Theorem 2.4, page 44, is linearly independent.
18.5 sinx F 19. sin2x cos2x F
20.5sin 2x cos2x F 21. sin2x cos2x cos2x F
22. x2 sinx F 23. sin2x cos2x 2 F 24.ex e–x F 25. ex e2x F
26. 1ex + e–x ex – e–x F 27.lnx lnx2 F+ , where+ denotes the set of positive numbers. 92 Chapter 3 Bases and Dimension
28. For what real numbers a is 11a 1a 1 a11 a linearly dependent set in 3 ?
29. For what real numbers a is 2xa++1 x + 2 a linearly dependent set in P1x ?
x 1 x2 1 x2 x 30. For what real numbers a is ax2 – --- – ---– ----- + ax – --- – ----- – --- + a a linearly dependent 2 2 2 2 2 2 set in P2x ? 31. Find a value of a for which cosxa+ sinx is a linearly dependent set in the function space F ? 32. Find a value of a for which sinxa+ sinxx cos is a linearly dependent set in the func- tion space F ? 33. Let v be any nonzero vector in a vector space V. Prove that v is a linearly independent set.
34. Prove that every nonempty subset of a linearly independent set v1v2 vn is again lin- early independent.
35. Prove that if v1v2 vn is a linearly dependent set in a vector space V, then so is the set v1v2 vn w1w2 wm for any set of vectors w1w2 wm in V. 36. Establish the converse of Theorem 3.6.
37. Let S = v1v2 vm be a set of vectors in a space V. Show that if there exists any vector v SpanS which can be uniquely expressed as a linear combination of the vectors in S then S is linearly independent. 38. Show that 10 01 is a linearly independent set in the vector space of Example 2.5, page 47.
39. Let S = u1u2 un and T = v1v2 vm be linearly independent sets of vectors in a vector space V with SpanS SpanT = 0 . Prove that ST is also a linearly independent set. 3.2 Linear Independence 93
PROVE OR GIVE A COUNTEREXAMPLE
40. If uv is a linearly dependent set, then u = rv for some scalar r.
41. If uvw is a linearly dependent set, then u = rv + sw for some scalars r and s.
42. If uvw is a linearly independent set of vectors in a vector space V, then uu+ vuvw++ is also linearly independent.
43. If uvw is a linearly independent set of vectors in a vector space V, then uvvwwu– – – is also linearly independent.
44. For any three nonzero distinct vectors uvw in a vector space V, uvvwwu– – – is linearly dependent.
45. If uvw is a linearly independent set of vectors in a vector space V, and if a then auvw is also linearly independent.
46. If uvw is a linearly independent set of vectors in a vector space V, and if a is any non- zero number, then auu+ av uv++aw is also linearly independent.
47. If S = u1u2 un and T = v1v2 vm are linearly independent sets of vectors in a vector space V, then ST is also a linearly independent set.
48. If uv and vw are linearly independent sets of vectors in a vector space V, then uvw is also a linearly independent set. 94 Chapter 3 Bases and Dimension
3
§3. BASES
So far we have considered sets of vectors S = v1v2 vn in a space V that satisfy one of two properties: (1) S spans V: Every vector in V can be built from those in S. (2) S is linearly independent: No vector in S can be built from the rest. In a way, (1) and (2) are tugging in numerically opposite directions: From the spanning point of view: The more vectors in S the better. From the linear independence point of view: The fewer vectors in S the better. Sometimes, the set of vectors S is not too big, nor too small — it’s just right:
DEFINITION 3.4 A set of vectors = v1v2 vn in a BASIS vector space V is said to be a basis for V if: (1) spans V and: (2) is linearly independent.
In the exercises you are asked to verify the fact that: 2 S2 ==e1 e2 10 01 is a basis for ,
S3 ==e1e2 e3 100 010 001 is a basis for 3 , and that, in general: S = e e e n 1 2 n , where each entry in the n-tuple ei is 0 with the exception of the ith entry which equals 1, is a n STANDARD BASES IN basis for n , called the standard basis of n .
EXAMPLE 3.10 Show that 130 204 012 is a basis for 3 .
SOLUTION: Appealing to Definition 3.4, we challenge the given set of vectors on two fronts: spanning, and linear independence. Spanning. For any given abc 3 , we are to determine if there exist scalars x, y, and z such that: x130 ++y204 z012 = abc 3.3 Bases 95
Expanding the left side, and equating coefficients, we come to the following 33 system of equations: x + 2y = a 1 2 0 1 00 coefS rref S: 3xz+ = b 3 0 1 0 1 0 4y + 2z = c 0 4 2 001 the three given vectors
If you take the time to Figure 3.1 solve the system directly, Applying the Spanning Theorem (page 18) we conclude that you will find that: 130 204 012 spans 3 . 2a + 2bc– x = ------8 Linear independence. A consequence of: y = ------6a – 2b + c 1 2 0 1 00 16 rref 3 0 1 0 1 0 z = –2------6a ++b 3c- 8 0 4 2 001
the three given vectors and the Linear Independence Theorem for n (page 88). Since 130 204 012 is a linearly independent set which spans 3 , it is a basis for 3 .
EXAMPLE 3.11 Determine if the following set of matrices con- stitute a basis for the vector space M22 : 21 11 –03 04 30 22 –56 – 15
SOLUTION: The problem boils down to a consideration of the coeffi- cient matrix of a system of equations. What system? Well, if you take the spanning approach, then you will be looking at the vector equation:
21 11 –03 04 ab x ++y z +w = (*) 30 22 –56 – 15 cd
to see if it can be solved for any given matrix ab . cd On the other hand, if you take the linear independent approach, then you will consider the vector equation:
21 11 –03 04 00 x ++y z +w = (**) 30 22 –56 – 15 00 to see if it has only the trivial solution. 96 Chapter 3 Bases and Dimension
In either case, by equating entries on both sides of the vector equa- tions, you arrive at systems of equations: Form (*): From (**): 2xy++–03z w = a 2xy++–03z w = 0 xy++0z +4w = b xy++0z +4w = 0
3x ++2y – 6z w = c 3x ++02y – 6z w = 0x ++2y –55z w = d 0x ++02y –55z w = Yes, the two systems differ to the right of the equal signs, but both share a common coefficient matrix, which “twice” reveals the fact
that the four given vectors do not constitute a basis for M22 . Once by the Spanning Theorem of page 18, and then again by the Linear Independence Theorem of page 22:
only 3 leading ones and 4 variables 21–0 3 1 00– 1 so the four vectors are not linearly independent 1104 rref 0 1 05 32–1 6 001 1 02–5 5 0000 contains a zero row so the four vectors do not span
CHECK YOUR UNDERSTANDING 3.10 Determine if the following set of matrices constitute a basis for the
matrix space M22 : 21 11 –03 04 Answer: It is a basis. 30 22 –55 – 15
In Example 3.10 we found that the vector space 3 has a basis con- sisting of three vectors, and every fiber in your body is probably sug- gesting that each and every basis for 3 must also consist of 3 vectors. Those fibers are correct. Indeed we will prove that if a vector space V has a basis consisting of n vectors, then every basis for V must again contain n vectors. Our proof will make use of the following fun- damental result:
In words: There cannot THEOREM 3.8 If w w w spans V, and if be more lineally indepen- 1 2 m SPAN dent vectors than the v1v2 vn is a linearly independent sub- number of vectors in any VERSES spanning set. INDEPENDENT set of V, then nm .
PROOF: Assume that nm . (We will show that this assumption contradicts the given con- dition that v1v2 vn is a linearly independent set). 3.3 Bases 97
We begin by expressing each vi as a linear combination of the vec- tors in the spanning set w1w2 wm :
v1 = c11w1 +++c12w2 c1nwm
v2 = c21w1 +++c22w2 c2nwm . . . . (*) ...... vn = cn1w1 +++cn2w2 cnmwm Now Consider the following homogeneous system of n linear equa- tions in m unknowns, with nm : c x +++c x c x = 0 Note that the coefficients of the 11 1 21 2 n1 m ith equation coincide with those c x +++c x c x = 0 12 1 22 2 n2 m (**) th of the i column of (*).
c1nx1 +++c2nx2 cnmxm = 0 The Fundamental Theorem of Homogeneous Systems of Equations (page 20), tells us that (**) has a nontrivial solution: r1r2 rn (not all of the ris are zero). We now show that r1v1 +++r2v2 rnvn equals 0, contradicting the assumption that v1v2 vn is a linearly independent set (for some ri 0 ): From (*)
r1v1 +++r2v2 rnvn = r1c11w1 +++c12w2 c1nwm + r c w +++c w c w . 2 21 1 22 2 2n m . . + rncn1w1 +++cn2w2 cmnwm
Regrouping: = c11r1 +++c21r2 cn1rn w1 + c r +++c r c r w . 12 1 22 2 n2 n 2 Since r r r is a . 1 2 n . solution of (**): + c1nr1 +++c2nr2 cmnrn wm c11r1 +++c21r2 cn1rn = 0 . see margin: = 0w1 + 0w2 ++ 0wn = 0 c1nr1 +++c2nr2 cmnrn = 0 We are now in a position to show that all bases of a vector space must contain the same number of elements:
THEOREM 3.9 If w1w2 wn and v1v2 vm are bases for a vector space V, then nm= . 98 Chapter 3 Bases and Dimension
PROOF: Since w1w2 wn is a basis, it spans V, and since v1v2 vm is a basis, it is a linearly independent set. Applying Theorem 3.8, we have nm .
One more time: Since v1v2 vm is a basis, it spans V, and since w1w2 wn is a basis, it is linearly independent. Applying The- orem 3.8, we also have mn . Since nm and mn , nm= .
DEFINITION 3.5 A vector space with basis v1v2 vn is DIMENSION said to be finite dimensional of dimension n. The symbol dimV will be used to denote the dimension of the vector space V.
The trivial vector space V = 0 of CYU 2.6, page 48, has no basis In the exercises you are (see Theorem 3.4, page 89). Nonetheless, it is said to be of dimension asked to show that the zero. We also point out that a vector space that is not finite dimensional polynomial space of Exer- is said to be infinite dimensional. cise 22, page 50, is an infinite dimensional space. CHECK YOUR UNDERSTANDING 3.11
Prove that S = v1v2 vn is a basis for a vector space V if and only if every vector in V can uniquely be expressed as a linear com- Answer: See page B-10. bination of the vectors in S. Let V be a space of dimension n. The following theorem says that any So, if the number of vectors spanning set of n vectors in V must also be linearly independent, and equals the dimension of the that any linearly independent set of n vectors must also span V: space, then to show that those vectors is a basis you THEOREM 3.10 Let V be a vector space of dimension n 0 , and do not have to establish both let S be a set of n vectors in V. Then: linear independence and spanning, for either implies S spans V if and only if S is linearly independent. the other. PROOF: We first show that the assumption that S = v1v2 vn spans V and is not linearly independent leads to a contradiction: If S is not linearly independent, then some vector in S is a linear combination of the remaining elements in S (Theorem 3.5, page 89). Assume, without loss of generality, that it is the vector vn : vn = a1v1 +++a2v2 an – 1vn – 1 Let vV . Since S spans V:
v = c1v1 + c2v2 ++ cnvn
= c1v1 +++c2v2 cn – 1vn – 1 +cna1v1 +++a2v2 an – 1vn – 1
= c1 + cna1 v1 +++c2 + cna2 v2 cn – 1 + cnan – 1 vn – 1 3.3 Bases 99
The above argument shows that the set v1v2 vn – 1 spans V — a contradiction, since a space of dimension n, which neces- sarily contains a basis of n elements, and therefore a linearly inde- pendent set of n elements, cannot contain a spanning set of n – 1 vectors (Theorem 3.8).
We now show that the assumption that v1v2 vn is a linearly independent set which does not span V will also lead to a contradiction:
Let vn + 1 Spanv1v2 vn . The Expansion Theorem of
page 90 tells us that v1v2 vn vn + 1 is still a linearly inde- pendent set. This leads to a contradiction since a space of dimen- sion n, which necessarily contains a spanning set of n elements, cannot contain a linearly independent set of n + 1 vectors (Theo- rem 3.8). The following result is essentially a restatement of Theorem 3.10. It underlines the fact that you can show that a set of n vectors in an n- dimensional vector space is a basis by EITHER showing that they span the space, OR by showing that it is a linearly independent set—you DON’T have to do both:
THEOREM 3.11 Let v1v2 vn be a set of n vectors in a vector space V of dimension n. The following are equivalent:
(i)v1v2 vn is a basis.
(ii)v1v2 vn spans V.
(iii)v1v2 vn is linearly independent. The cycle: (i) PROOF: We can easily show that i ii iii i : i ii : By Definition 3.4. (iii) (ii) ii iii : By Theorem 3.10. iii i : By Theorem 3.10 and Definition 3.4. insures that the validity of any of the three prop- CHECK YOUR UNDERSTANDING 3.12 ositions implies that of Prove that the vector space of Example 2.5, page 47: the other two. Vxy= xy , under the operations: xy + x y = xx+ – 1 yy++ 1 rxy = rx–1 r + ry+ r – 1 Answer: See page B-10. has dimension 2. 100 Chapter 3 Bases and Dimension
STRETCHING OR SHRINKING TO A BASIS A basis has to be both a spanning and a linearly independent set of vectors. Help is on the way for any set of vectors that falls short on either of those two counts: THEOREM 3.12 Let V be a nonzero space of dimension n. Expansion Theorem: (a) Any linearly independent set of vectors in V can be extended to a basis for V. Reduction Theorem: (b) Any spanning set of vectors in V can be reduced to a basis for V.
PROOF: (a) Let v1v2 vm be a linearly independent set of vec- Procedure: Keep adding tors in V. If it spans V, then it is already a basis for V and we are done. vectors, while maintaining linear independence, till If not, then take any vector vm + 1 Spanv1v2 vm and add it you end up with n linearly to the set: v1v2 vm vm + 1 . This brings us to a larger set of independent vectors. vectors which, by the Expansion Theorem of page 90, is again lin- early independent. Continue this process until the set contains n lin-
early independent vectors: v1v2 vm vn , and then apply Theorem 3.11 to conclude that it is a basis for V.
(b) Let v1v2 vm be a spanning set of vectors in V. Since V Procedure: Keep throwing contains a linearly independent set of n elements (any basis will do), vectors away, while main- taining spanning, till you mn (Theorem 3.8). If mn= then we are done (Theorem 3.11). If end up with n spanning vectors. mn , then the spanning set v1v2 vm cannot be linearly independent; for if it were, then it would be a basis, and all bases have
n elements. Find a vector in v1v2 vm that is a linear combina- tion of the rest and remove it from that set to arrive at a smaller span- ning set. Continue this “tossing out” process until you arrive at a spanning set consisting of n elements—a basis for V (Theorem 3.11).
EXAMPLE 3.12 (a) Expand the linearly independent set: L = 1234 –4101 3312 to a basis for 4 . (b) Reduce the spanning set: Sx= + 12 x2 – 32 x – 3 x2 + 4 x2 –3x – to a basis for P2 . SOLUTION: (a) We are given that L is linearly independent, and need to comple- ment it with an additional 4-tuple, while maintaining linear indepen- dence. From earlier discussions, you know that if you randomly pick a 4-tuple, then the probability that it will be in Span(L) is nil. Let’s go with 2 – 517 : = 1234 –4101 3312 2 – 517 3.3 Bases 101
We, of course, have to demonstrate that the “gods were not against us,” and do so via the Linearly Independence Theorem for n of page 88:
14–32 1 000 rref 0 1 00 213– 5 3011 001 0 4127 0001 The above shows that is a linearly independent set. Since it con- sists of 4 vectors, and since 4 is of dimension 4 [Exercise 1(c)], is a basis for 4 . (b) It is easy to see that 1xx2 is a basis for the vector space P2x , and that consequently P2 is of dimension 3. Since Sx= + 12 x2 – 32 x – 3 x2 + 4 x2 –3x – contains five vectors, we have to throw two of them away in such a manner so as to
end up with three vectors that still span P2 ; or, equivalently, with three linearly independent vectors. We leave it for you to verify that x + 12 x2 – 32 x – 3 is a linearly independent set. As such, it
must be a basis for the three dimensional space P2 .
CHECK YOUR UNDERSTANDING 3.13
21 22 (a) Expand the linearly independent set L = to a 12 11 basis for M22 (b) Reduce the set: S = 312 – –93– 6 12– 2 –54– 6 624 – Answer: See page B-11. to a basis for SpanS . Dose S span 3 ?
The next example reveals a systematic approach that can be used to reduce a set of vectors S in n to a basis for SpanS : EXAMPLE 3.13 Find a subset of: S = 12– 1302 540 66– 1 which is a basis for SpanS .
SOLUTION: To determine which of the four vectors can be discarded, we challenge their linear independence, and turn to the equation: a12– 1 +++b302 c540 d66– 1 = 000 (*) Equating coefficients leads us to the following homogeneous system of equations: 102 Chapter 3 Bases and Dimension
a b c d a b c d a +++3b 5c 6d = 0 1356 1 0 2 3 coefS rref S: 2a +++0b 4c 6d = 0 2046 0 1 1 1 –2a ++b 0cd– = 0 –2011 – 0 0 0 0
equivalent systems Note that these are the four given vectors, but written in column form a ++2c 3d = 0 bcd+0+ = Figure 3.2 Let’s agree to call a vectors in S = 12– 1302 540 66– 1 that occupies the same column-location in coef(S) as that of a lead- ing-one-column in the rref-matrix, a leading-one vector [12– 1 and 302 are leading-one vectors]. We now proceed to show that those leading-one vectors constitute a basis for SpanS . Figure 3.2 tells us that system S will be satisfied for any a, b, c, and a ++2c 3d = 0 d for which: (**) bcd+0+ = a +02 = Note that c and d are the free Setting c = 1 and d = 0 in (**) leads us to , with variables in rref[coef (s)] b +01 = solution a = –2 and b = –1 . Substituting these values in (*) we have: –212– 1 – 1302 ++1540 066– 1 = 000
or: 540 = 212– 1 + 1302
Conclusion: the non-leading-one vector (5, 4, 0) can be expressed as a linear combination of the two leading-one vectors. . Setting d = 1 and c = 0 in (**) we find that a = –3 and b = –1 , bringing us to: –1312– 1 –0302 ++540 166– 1 = 000
or: 66– 1 = 312– 1 + 1302 Conclusion: the non-leading-one vector (6, 6, -1) can be expressed as a linear combination of the two leading-one vectors. Since 540 and 66– 1 are linear combinations of 12– 1 302 : Span12– 1302 540 66– 1 = Span12– 1 302 3.3 Bases 103
Covering up the two non-leading-one columns in the development of Figure 3.2: a b c d a b c d a +++3b 5c 6d = 0 1356 1 0 2 3 coefS rref S: 2a +++0b 4c 6d = 0 2046 0 1 1 1 –2a ++b 0cd– = 0 –2011 – 0 0 0 0
a ++2c 3d = 0 bcd+0+ = we see that the only solution of: a12– 1 + b302 = 000 is the trivial solution, and that the vectors 12– 1 302 are therefore linearly independent.
SUMMARIZING To find a basis for the space spanned by the vectors v1 = 12– 1, v2 = 302 , v3 = 540 , and v4 = 66– 1, we constructed the 34 matrix 1356 A = 2046 with columns the given four-tuples. We –2011 – then showed that the leading-one vectors, namely v1 = 12– 1 and v2 = 302 , constituted a basis for the space spanned by the four given vectors.
We state, without proof, a generalization of the above observation:
THEOREM 3.13 n Let v1v2 vm be vectors in . If A is the nm matrix whose ith column is the n-tuple vi , then the set consisting of those vectors vi , where the ith column of rrefA contains a leading one, is a basis for the space spanned by v1v2 vm.
CHECK YOUR UNDERSTANDING 3.14
Use the above theorem to address the problem of CYU 3.13(b). Answer: See page B-11. 104 Chapter 3 Bases and Dimension
‘ EXERCISES
1. (a) Prove that S ==e e 10 01 is a basis for 2 . Express –3 --5- as a linear 2 1 2 2 combination of the vectors in S2 . 3 (b) Prove that S3 ==e1e2 e3 100 010 001 is a basis for . Express
320 as a linear combination of the vectors in S3 .
n (c) Prove that Sn = e1e2 en is a basis for . 2. (a) Prove that = 205 0110 120 is a basis for 3 , and express 70– 5 as a linear combinations of the vectors in . (b) Show that S = 205 0110 120 111 is not a basis for 3 , and find two different representations for the vector 70– 5 as a linear combination of the vec- tors in S.
2 2 2 3. (a) Prove that = 2x + 3x – x x – 5 is a basis for P2 , and express x + 3x – 1 as a lin- ear combinations of the vectors in . 2 (b) Show that S = 2x 3x 5 x – 4 is not a basis for P2 , and find two different representa- tions for the vector x2 + 3x – 1 as a linear combination of the vectors in S.
Exercises 4-7. Determine if the given set of vectors is a basis for 3 . 4.215 4110 123 5. 000 –12– 5 215 – –
6. 215 4110 4310 7. 2 e e 2 e2
8. (a) Prove that the matrix space M22 has dimension 4.
(b) Prove that the matrix space Mmn has dimension mn .
Exercises 9-12. Determine if the given set of vectors is a basis for M22 .
12 23 34 41 –21 12– 12 12 9. 10. 34 41 12 23 34 34 –43 34–
12 12 10 02 2 --1- 13– 11 --1- –1 21– 11. 12. 2 3 30 04 34 34 13 1 31– ------4 01 --- 0 2 6 3.3 Bases 105
Exercises 13-15. Determine if the given set of vectors is a basis for M32 .
01 10 03 30 00 10 11 10 01 00 00 11 13.3003 01 10 01 00 14. 1100 00 11 11 00 01 10 01 10 03 00 11 00 00 00 00 11
12 22– 32 –21 –21 – 15. 31– –12 – 12– 31 31 –32 – 34– –13 – –32 23–
3 2 16. (a) Prove that the polynomial space P3 = a3x +++a2x a1xa0 ai R 0 i 3 is of dimension 4. (b) Prove that the polynomial space: n n – 1 Pn = anx +++an – 1x + a1xa0 ai R 0 in is of dimension n + 1 .
Exercises 17-20. Determine if the given set of vectors is a basis for P3 .
17. x3 +21 x3 –23x2 + x + 14 x3 –9x2 ++x 7
18. x3 +21 x3 –23x2 + x + 14 x3 –9x2 ++x 8
19. 2x3 ++3x2 x – 1 –9x3 –2x2 ++x 2 x3 –2x2 ++x 23 –2x3 ++x2 –1x
20. 2x3 ++3x2 x – 1 – x3 –2x2 ++x 2 x3 –2x2 ++x 23 –2x3 ++x2 –1x
Exercises 21-24. Extend the given linearly independent set of vectors to a basis for 4 . 21.1341 –1201 1122 22. 121 – 21212 1234
23.213– 1 1302 24. 2211
Exercises 25-28. Extend the given linearly independent set of vectors to a basis for P4 .
25.x4 + 1x3 +32x2 x3 +9x – 5 26. 3x –32x2 –32 x3 –32 x4 – 2
27.x4 ++x3 x2x3 ++x2 x x2 ++x 1 28. 2x4 ++++x3 x2 x 1 x4 – 5 106 Chapter 3 Bases and Dimension
Exercises 29-30. Does the given set of vectors span M22 ? If so, reduce the set to obtain a basis
for M22 .
13 20 33 01 21 12 29. –21 11– 01 12 01 34
–01 50 10 10 00 00 01 30. 20 –010 01 02 00 01– 10
Exercises 31-32. Does the given set of vectors span P3 ? If so, reduce the set to obtain a basis
for P3 .
31. 2x 3x2 6x2 + 4xx3 +31 x2 + x x3 +++3x2 x 1
32. 2x3 +4x – 1 x3 –642x2 – x2 + 25 x3 + 6x2 – 2 x3 + x – 1 x3 – 2x2 –2x3 – x2
Exercises 33-37. Use Theorem 3.13 to find a subset of the given set of vectors S in n which is a basis for SpanS . If necessary expand that basis for SpanS to a basis for the corresponding Euclidean space.
33.S = 214 –132 51– 6 428
34. S = 113 –132 158 32– 1
35. S = 113 –132 158 32– 1
36. S = 2130 0442 –2312 1158 3288
37. S = 13132 24142 11010 11202 22111 12345
Exercises 38-39. Find a subset of the given set of vectors S which is a basis for SpanS . If nec- essary, expand that basis for SpanS to a basis for P4 .
38. S = 2x4 ++x3 3x +1 – x4 – x3 + x2 x3 –22x3 ++x2 3x –22x4 –2x3 + x2 – 1
39. S = 5 – x3 – xx 4 ++++x3 x2 x 12 x4 – 2x2 2x4 + 2x2
Exercises 40-41. Find a basis for the space spanned by the given set of vectors in the function space vectors F of Theorem 2.4, page 44.
40.5 sinxx cos sin2x cos2x sin5 + x 41. sinxx cos sin2x cos2x cos2x sin2x
42. Show that abc a ++2bc= 0 is a subspace of 3 , and then find a basis for that sub- space. 3.3 Bases 107
43. Show that abcd ab+ = cd– is a subspace of 4 , and then find a basis for that sub- space.
2 44. Show that ax ++bx c abc= – is a subspace of P2 , and then find a basis for that sub- space.
3 2 45. Show that ax +++bx cx d abc== abd++ is a subspace of P3 , and then find a basis for that subspace.
46. Find all values of c for which 110 c210 0c 1 is a basis for 3 .
c 1 12c 01 c 0 47. Find all values of c for which is a basis for M22 . 00 10 –0c 0 c
48. Find a basis for the vector space of Example 2.5, page 47.
49. Suppose v1v2 v3 is a basis for a vector space V. For what values of a and b is av1 bv2 ab– v3 a basis for V? 50. Let S is a subspace of V with dimS ==dimV n . Prove that SV= .
51. Suppose that v1v2 vs is a linearly independent set of vectors in a space V of dimen- sion n, and that w1w2 wt spans V. Prove that snt . 52. A set of vectors S in a finite dimensional vector space V is said to be a maximal linearly independent set if it is not a proper subset of any linearly independent set. Prove that a set of vectors is a basis for V if and only if it is a maximal linearly independent set. 53. A set of vectors S in a finite dimensional vector space V is said to be a minimal spanning set if no proper subset of S spans V. Prove that a set of vectors is a basis for V if and only if it is a minimal spanning set. 54. Let H and K be finite dimensional subspace of a vector space V with HK = 0 , and let HK+ = hk+ h H and k K . Prove that dimHK+ = dimH + dimK . (Note: you were asked to show that HK+ is a subspace of V in Exercise 42, page 67.) 55. Let H and K be finite dimensional subspace of a vector space V, and let HK+ = hk+ h H and k K . Prove that:
dimHK+ = dimH + dimK – dimHK Suggestion: Start with a basis for HK and extend it to bases for H and K.
56. Prove that the polynomial space P of Exercise 22, page 50, is not finite dimensional by show- ing that it does not have a finite base. 57. (Calculus dependent) Show that SpP= p0 = 0 is a subspace of the polynomial space P of Exercise 22, page 50. Find a basis for S. 58. Prove that a vector space V is infinite dimensional (not finite dimensional) if and only if for any positive integer n, there exists a set of n linearly independent vectors in V. 108 Chapter 3 Bases and Dimension
PROVE OR GIVE A COUNTEREXAMPLE
59. If v1v2 v3 is a basis for a vector space V, and if c1 , c2 , and c3 are nonzero scalars, then
c1v1c2v2 c3v3 is also a basis for V.
60. If v1v2 vn – 1 is a linearly independent set of vectors in a space V of dimension n, and
if vn v1v2 vn – 1 , then v1v2 vn – 1 vn is a basis for V.
61. If v1v2 vn – 1 is a linearly independent set of vectors in a space V of dimension n, and
if vn Spanv1v2 vn – 1 , then v1v2 vn – 1 vn is a basis for V.
62. If v1v2 vn vn + 1 is a spanning set of vectors in a space V of dimension n, then
v1v2 vn is a basis for V.
63. If v1v2 vn vn + 1 is a spanning set of vectors in a space V of dimension n, and if
vn + 1 Spanv1v2 vn , then v1v2 vn is a basis for V.
64. If v1v2 v3 is a basis for a vector space V, then v1v1 + v2 v1 ++v2 v3 is also a basis for V.
65. It is possible to have a basis for the polynomial space P2x which consists entirely of poly- nomials of degree 2.
66. Let v1v2 vn + 1 be a spanning set for a space V of dimension n satisfying the property
that v1 ++v2 +vn + 1 = 0 . If you delete any vector from the set v1v2 vn + 1 , then the resulting set of n vectors will be a basis for V. Note: In set notation, an element cannot be repeated. In particular, no two of the vs in
v1v2 vn + 1 are the same. 67. If V is a space of dimension n, then V contains a subspace of dimension m for every integer 0 mn. Chapter Summary 109
CHAPTER SUMMARY
LINEAR A vector v in a vector space V is said to be a linear combination COMBINATION of vectors v1v2 vn in V, if there exists scalars c1c2 cn such that:
v = c1v1 +++c2v2 cnvn
SPANNING The set of linear combinations of v1v2 vn is a subspace of V. It is said to be the subspace of V spanned by those vectors,
and is denoted by Spanv1v2 vn :
Spanv1v2 vn = c1v1 +++c2v2 cnvn
If V = Spanv1v2 vn , then v1v2 vn is said to span the vector space V. If every vector in a set S is 1 Let the set of vectors v v v and w w w be contained in the space 1 2 n 1 2 m such that wi Spanv1v2 vn for 1 im . Then: spanned by another set S2 , then SpanS1 is a subset Spanw1w2 wm Spanv1v2 vn of SpanS2 .
If v1v2 vn Spanw1w2 wm and w1w2 wm Spanv1v2 vn
then: Spanv1v2 vn = Spanw1w2 wm
LINEARLY INDEPENDENT The vectors v1v2 vn are linearly independent if: SET a1v1 ++anvn ==0 ai 01 in
If the vectors v1v2 vn are not linearly independent then they are said to be linearly dependent. Unique representation. The vectors v1v2 vn are linearly independent if and only if
a1v1 ++ anvn = b1v1 ++ bnvn
implies that ai = bi , for 1 in . No vector can be built from A collection of two or more vectors is linearly independent if the rest. and only if none of those vectors can be expressed as a linear combination of the rest. Expanding a linearly inde- Let S = v1v2 vn be a linearly independent set. If pendent set. v SpanS , then v1v2 vn v is again a linearly indepen- dent set. 110 Chapter 3 Bases and Dimension
Linear Independence Theo- A homogeneous system of m linear equations in n unknowns S has rem. only the trivial solution if and only if rref coefS has n leading ones. n n Linear independence in R . A set of vectors v1v2 vm in is linearly independent if and only if the row-reduced-echelon form of the nm matrix th whose i columns is the (vertical) n-tuple vi has m leading ones.
BASIS A set of vectors = v1v2 vn in a vector space V is said to be a basis for V if: (1) spans V and: (2) is linearly independent.
A spanning set can’t have If w1w2 wn spans a space V and if v1v2 vm is a fewer elements than the linearly independent subset of V, then nm . number of elements in any linearly independent set.
All bases for a vector space If w1w2 wn and v1v2 vm are bases for a vector contain the same number of space V, then nm= . vectors.
You can show that a set of Let v1v2 vn be a set of n vectors in a vector space V of n vectors in an n-dimen- dimension n. The following are equivalent: sional vector space is a basis by EITHER showing (i)v1v2 vn is a basis. that they span the space, OR by showing that it is a lin- (ii)v1v2 vn spans V. early independent set—you (iii)v v v is linearly independent. DON’T have to do both: 1 2 n
Expansion Theorem Any linearly independent set of vectors in V can be extended to a basis for V. Reduction Theorem Any spanning set of vectors in V can be reduced to a basis for V.
Reducing a set of vectors S n Let v1v2 vm be vectors in . If A is the nm matrix n in to a basis for th Span(S) whose i column is the n-tuple vi , then the set consisting of th those vectors vi , where the i column of rrefA contains a leading one, is a basis for the space spanned by v1v2 vm . 4.1 Linear Transformations 111
4115 CHAPTER 4 LINEARITY In this chapter we turn our attention to functions from one vector space to another which, in a sense, preserve the algebraic structure of those spaces. Such functions, called linear transformations, are introduced in Section 1. As you will see, each linear transformation T: VW gives rise to two important subspace; one, called the kernel of T, resides in the vector space V, while the other, called the image of T, lives in W. Those two subspaces are investigated in Section 2. Linear transformations which are also one-to-one and onto are called isomorphisms, and they are considered in Section 3.
§1. LINEAR TRANSFORMATIONS Up to now we have focused our attention exclusively on the internal nature of a given vector space. The time has come to consider functions from one vector space and another: A linear transformation is also called a linear func- DEFINITION 4.1 A function T: VW from a vector space V to tion, or a linear map. A LINEAR a vector space W is said to be a linear trans- linear map T: VV from a formation if for every vv V and r : vector space to itself is said TRANSFORMATION to be a linear operator. (1): Tvv+ = Tv + Tv and (2): Trv = rTv Let’s focus a bit on the statement: Tvv+ = Tv + Tv for T: VW . There are two plus signs in the equation, but the one on the left is happening in the space V, while the one on the right takes place in the space W. What the statement is saying is that you can per- form the sum in the vector space V and then carry the result over to the vector space W via the transformation T, or you can first carry those two vectors v and v over to W (via T) and then perform their sum in the space W. In the same fashion, Trv = rTv is saying that you can perform scalar multiplication in V and then carry the result over to W, or you can first carry the vector v to W and then perform the scalar multiplication in that vector space.
EXAMPLE 4.1 Show that the function T: 3 2 given by Tabc = 2ab+ –c is linear. 112 Chapter 4 Linearity
SOLUTION: Let abc a b c 3 and r . T preserves sums: A smoother approach: T abc + a b c Tabc + a b c = Taa+ bb+ cc+ = Taa+ bb+ cc+ = 2aa+ + bb+ –cc+ Tabc = 2ab+ –c : = 2aa+ + bb+ –cc+ = 2a +++2a bb – c – c = 2a +++2a bb – c – c
= 2ab+ –c + 2a + b –c same = Tabc + Ta b c and: Tabc +2Ta b c = ab+ –c + 2a + b –c = 2ab++2a +b – c – c T preserves scalar products: Trabc ==Tra rb rc 2ra+ rb –rc ==r2ab+ –c rTabc
EXAMPLE 4.2 Determine if the function f: 3 3 given by fabc = 2cb2 –a is linear.
SOLUTION: If you are undecided on whether or not f is linear, you may want to challenge its linearity with a couple of specific voters ? f 213 + 401 = f213 + f401 ? f614 = f213 + f401 ? fabc = 2cb2 –a : 81– 6 = 61– 2 + 20– 4 81– 6 = 81– 6 Yes! The above establishes nothing. It certainly does not show that the function f is not linear, nor does it establish its linearity since we have but demonstrated that it “works” for the two chosen vectors 213 and 401 . Let’s try another pair: ? f 32– 5 + 49– 3 = f32– 5 + f49– 3 ? f772 = f32– 5 + f49– 3 ? 4497– = 10 4– 3 + –681– 4 ? 4497– = 4857– No! Since f 32– 5 + 49– 3 f32– 5 + f49– 3, the func- tion is not linear. You can also show that the above function is not linear by dem- onstrating, for example, that f2123 2f123 . To show that a function is not linear you need only come up with a specific counterexample which “shoots down” EITHER (1) or (2) of Definition 4.1. 4.1 Linear Transformations 113
CHECK YOUR UNDERSTANDING 4.1
Is the function f: 2 3 given by fab = ab+ 2ba – b Answer: Ye. linear? Justify your answer.
Given that a linear transformation preserves vector space structures, it may come as no surprise to find that it maps zeros to zeros, and inverses to inverses: THEOREM 4.1 If T: VW is linear, then: In order to distinguish where the different zero in the space W zeros preside, we are (a) T0V = 0W using 0V and 0W to indicate the zero is in zero in the space V the vector space V and W, respectively. inverse of the vector Tv in the space W (b) T–v = –Tv inverse of v in the space V
PROOF: (a) Note that three different zeros are featured below: by linearity
T0V ===T0 0V 0T0V 0W Theorem 2.7, page 53
by linearity
(b) T–v ===T–1 v –1 Tv –Tv Theorem 2.8, page 54
CHECK YOUR UNDERSTANDING 4.2
2 Use Theorem 4.1(a) to show that the function f: P2x given Answer: See page B-12. by fab = ax2 ++bx 1 is not linear. The following Theorem compresses the two conditions of Definition 4.1 into one: You can perform the vector operations in V THEOREM 4.2 T: VW is linear if and only if: and then apply T to that Trvv+ = rTv + Tv result: Trvw+ , or you can first apply T for all vv V and r and then perform the vector operations in W: PROOF: If T is linear, then for all vv V and r : rTv + Tw . Either way, you will end up at Trvv+ ==Trv + Tv rTv + Tv the same vector in W. (1) of Definition 4.1 (1) of Definition 4.1 114 Chapter 4 Linearity
Conversely, if Trvv+ = rTv + Tv for all vv V and r , then: (1): Tvv+ ==T1vv+ 1Tv + Tv =Tv + Tv Axiom (viii), page 40
Theorem 4.1(a) (2): Trv ==Trv + 0 rTv + T0 ==rTv + 0 rTv Axiom (iii), page 40
You are invited to establish the following generalization of the above result in the exercises:
THEOREM 4.3 Let T: VW be linear. For any vectors
v1v2 vn in V, and any scalars
a1a2 an :
Ta1v1 ++ anvn = a1Tv1 ++anTvn
CHECK YOUR UNDERSTANDING 4.3 Answer: See page B-12. Use Theorem 4.2 to establish the linearity of the function of CYU 4.1. The following theorem tells us that linear transformations map sub- spaces to subspaces: THEOREM 4.4 Let T: VW be linear. If S is a subspace of V, then TS= TssS is a subspace of W.
See Theorem 2.13, page 61 PROOF: For Ts1 Ts2 TS , and r :
rT s1 + Ts2 = Trs1 + s2 TS since S is a subspace, since T is linear rs1 + s2 S
CHECK YOUR UNDERSTANDING 4.4
PROVE OR GIVE A COUNTEREXAMPLE: Let T: VW be linear, and SV . If TS is a subspace of W, then S must be a subspace of V.
A LINEAR MAP IS COMPLETELY DETERMINED BY ITS ACTION ON A BASIS
Suppose you have a run-of-the-mill function f: 2 3 , and you know that f10 = 123 and that f01 = 23– 5 . What can you say about f35 ? Nothing. But if T: 2 3 is linear, with T10 = 123 and T01 = 23– 5 , then: 4.1 Linear Transformations 115
T35 = T 310 + 501 by linearity: = 3T10 + 5T01 ==3123 + 523– 5 13 21– 16
Yes: A linear transformation T: VW is COMPLETELY DETERMINED by its action on a basis of V In particular, if you have two linear transformations that act identi- cally on a basis, then those two transformations must, in fact, be one and the same: THEOREM 4.5 Let V be a finite dimensional space with basis v1v2 vn . If T: VW and L: VW
are linear maps such that Tvi = Lvi for 1 in, then Tv = Lv for every v V .
PROOF: Let v V . Since v1v2 vn is a basis for V, there exist
scalars a1a2 an such that:
v = a1v1 + a2v2 ++ avn We then have: linearity of T
Tv ==Ta1v1 + a2v2 ++ anvn a1Tv1 +++a2Tv2 anTvn Since Tv= Lv: i i = a1Lv1 +++a2Lv2 anLvn linearity of L: ==La1v1 + a2v2 ++ anvn Lv
CHECK YOUR UNDERSTANDING 4.5
PROVE OR GIVE A COUNTEREXAMPLE: Theorem 4.5 still holds if
Answer: See page B-12. v1v2 vn is a spanning set for V (not necessarily a basis).
The next theorem gives a method for constructing all linear maps from one vector space to another:
THEOREM 4.6 Let v1v2 vn be a basis for a vector LINEAR space V, and let w1w2 wn be n arbitrary CONSTRUCTION vectors (not necessarily distinct) in a vector space W. There is a unique linear transforma-
tion T: VW which maps vi to wi for 1 in; and it is given by:
Ta1v1 ++ anvn = a1w1 ++ anwn 116 Chapter 4 Linearity
PROOF: From Theorem 4.4, we know that there can be at most one
linear transformation T: VW such that Tvi = wi for 1 in . We complete the proof by establishing the fact that the above func- tion T is indeed linear:
For v = a1v1 ++ anvn and v = b1v1 ++ bnvn in V, and r :, n n Trv+ v = Trav + b v Trv+ v i i i i i = 1 i = 1 n n definition of ==Tra + b v ra + b w i i i i i i i = 1 i = 1 n n
==raiwi + biwi rT v + Tv T rT v + Tv i = 1 i = 1
EXAMPLE 4.3 Let T: 2 3 be the linear transformation which maps 11 to 324 and 41 to 10– 5. Determine: (a) T96 (b) Tab
SOLUTION: (a) It is easy to see that 11 41 is a basis for 2 . We express 96 as a linear combination of the vectors in that basis: 9 = r + 4s 96 = r11 + s41 6 = rs+ _ : 3== 3ss 1 6 ==r +51 r Applying the linear transformation to 96 = 511 + 141 we have: T96 ==T 511 + 141 5T11 + 1T41 = 5324 + 10– 5 = 15 10 20 + 10– 5 = 16 10 15 (b) Repeating the above process for ab , we have: ab = r11 + s41 ar= + 4s brs= + ab– _: ab–== 3ss ------3 ab– 4ba– br==+ ------ r ------3 3 4.1 Linear Transformations 117
Consequently: 4ba– ab– Tab ==rT11 + sT41 ------324 + ------10– 5 3 3 – 2a + 11b – 2a + 8b – 9a + 21b details omitted: = ------------3 3 3
CHECK YOUR UNDERSTANDING 4.6
3 Answer: Let T: R P2x be the linear transformation which maps (a) 4x2 + 4x – 10 100 and 020 to 2x2 + x , and maps 111 to x – 5 . 2ab+ – 3c x2 (b) Determine: c + a + --b- – --- x – 5c 2 2 (a) T342 (b) Tabc .
COMPOSITION OF LINEAR FUNCTIONS
As you may recall from earlier courses, for given functions f: XY and g: YZ , the composition of f followed by g is that function, denoted by gf , that is given by gf x = gfx (first apply f, and then apply g): f g X Y Z x z . .y .
gf The following theorem tells us that the composition of linear functions is again a linear function:
THEOREM 4.7 If T: VW and L: WZ are linear, then the composition LT: VZ is also linear. PROOF: For vv V and r we show that the function LT sat- isfies the condition of Theorem 4.1: linearity of T L T rv+ v LT rv+ v ==LTrvv+ LrTv + Tv linearity of L: = rL Tv + LTv
rLT v + LT v Definition of composition: = rLT v + LT v 118 Chapter 4 Linearity
EXAMPLE 4.4 3 Let T: M22 and L: M22 P2 be given by:
Tabc = ab+0 cb– and: ab L= bx2 ++ac– xc cd (a) Show that T and L are linear. (b) Show directly that the composite function 3 LT: P2 is again linear.
SOLUTION: Linearity of T: Trabc + a'b' c' = Tra+ a' rb+ b' rc+ c'
= ra+ a' +0rb+ b' rc+ c' –rb+ b'
= r ab+0+ a' +0b' cb– c' –b' = rTabc + Ta'b' c' Linearity of L:
ab a' b' ra+ a' rb+ b' Lr+ = T cd c' d' rc+ c' rd+ d' = rb+ b' x2 ++ra+ a' – rc+ c xrcc+ ' = rbx2 ++ac– xc+ b'x2 ++a' – c' xc'
ab a' b' = rT+ T cd c' d'
3 The composite function LT: P2 : ab+0 LT abc ==LTabc L cb– = 0x2 ++ab+ – c xc = abc+ – xc+ 4.1 Linear Transformations 119
3 Linearity of LT: P2 :
LT rabc + a'b' c' = LT ra+ a' rb+ b' rc+ c' = r aa++' r bb +' – rc+ c' xrcc+ + ' = ra+ b– cxc+ + a' + b' – c' xc+ ' = rLT abc + LT a'b' c'
CHECK YOUR UNDERSTANDING 4.7
(a) Show that 10 11 and 010 110 011 are bases for 2 and 3 , respectively. (b) Let T: 2 3 be the linear transformation which maps 10 to 020 and 11 to 101 . Let L: 3 2 be the linear transformation which maps 010 to 01 and both 110 and Answer: (a) See page B-13. 011 to 10 . Determine the map L T: 2 2 . (b) 2b 2a – 4b 120 Chapter 4 Linearity
EXERCISES
Exercises 1-18. Determine whether or not the given function f is a linear transformation.
1.f: , where fx = –5x . 2.f: , where fx = 0 . 3.f: , where fx = 1 .
4.f: 2 , where fxy = 2x – 3y .
5.f: 2 , where fxy = xy .
6.f: 2 2 , where fxy = xyxy+ – .
7.f: 2 3 , where fxy = xxyy .
x y 8.f: 2 3 , where fxy = ---y --- . 2 2
3 2 9.f: P2x , where fabc = ax ++bx c .
2 10.f: P2 , where fax++bx c = abc++ .
2 11.f: P2 , where fax++bx c = abc .
2 2 12.f: P2 P2 , where fax++bx c = ax+ 1 + bx+ 1 + c .
2 3 2 13.f: P2 P3 , where fax++bx c = ax ++bx cx .
3 0 ab 14.f: M23 , where fabc = . cba
ab 15.f: M22 , where f = ad– bc . cd
ab ab–0 16.f: M22 M22 , where f = . cd 0 cd
17.f: , where fx = sinx .
18.f: V , where fx = 2x , and V is the vector space of Example 2.5, page 47. 4.1 Linear Transformations 121
19. Let the linear map T: 2 3 be such that: T11 ==120 T02 102 Find: (a) T53 (b) Tab
20. Let the linear map T: 3 2 be such that: T100 ==32 T010 23 and T001 =11 Find: (a) T53– 2 (b) Tabc
2 21. Let the linear map T: M22 be such that:
T10 ==10 T11 21 21 32 Find: (a) T53 (b) Tab
2 22. Let the linear map T: M22 be such that:
T11 ==10 T31 21 21 32 Find: (a) T53 (b) Tab
23. Show that there cannot exist a linear transformation T: R2 R2 such that: T12 ==53 T53 12 and T11 =22
2 24. Show that there cannot exist a linear transformation T: R P2 such that: T12 ==x2 + 2 T53 5x +3 and T11 =x2 ++x 1
25. Show that the identity function I: V V , given by Iv = v for every v in V, is linear.
26. Show that the zero function Z: VW , given by Zv = 0 for every v in V, is linear. (Refer- ring to the equation Zv = 0 , where does 0 live?)
27. In precalculus and calculus, functions of the form yaxb= + are typically called linear functions. Give necessary and sufficient conditions for a function of that form to be a linear operator on the vector space .
28. Show that for any a the function Ta: F , where Taf = fa is linear. (See Theorem 2.4, page 44.)
29. (Calculus Dependent) Let D be the subspace of the function space F consisting of all differentiable functions. Let T: D F be given by Tf= f , where f denotes the derivative of f. Show that T is linear. 122 Chapter 4 Linearity
30. (Calculus Dependent) Show that the function T: P2 P4 , given by d Tpx = 5x3 px + 5 is linear. dx
31. (Calculus Dependent) Show that if the linear function T: P2 P1 is such that Tx2 ==2xTx 1 , and T1 = 0 , then T is the derivative function.
32. (Calculus Dependent) Let Iab denote the vector space of all real-valued functions that are integrable over the interval ab . Let T: Iab be given by Tf= bfxdx . a Show that T is linear.
33. Let T: VW be linear and let S be a subspace of V. Show that Tv v S is a subspace of W. 34. (PMI) Use the Principle of Mathematical Induction to prove Theorem 4.3. 35. Let LVW = T: VW T is linear , with addition and scalar multiplication given by:
T1 + T2 v ==T1v + T2v and rT v rTv Show that LVW is a vector space.
36. (a) Show that if a function f: satisfies the property that frx = rfx for every r and x , then f is a linear function: which is ti say, that it must also satisfy the
property that fx1 + x2 = fx1 + fx2 for every x1x2 . (b) Give an example of a function f: 2 2 satisfying the property that frab = rfab for every r and every ab 2 but which does not satisfy the property that f: ab + cd = fab + fcd . (Note: It is by no means a trivial task to establish the existence of a non-linear function f: VW which
satisfies the property that fv1 + v2 = fv1 + fv2 for every v1 v2 V .)
37. Let f: VW satisfy the condition that fv1 + v2 = fv1 + fv2 for every v1 v2 V . Show that: (a)f3v = 3fv for every v V . (b) f --2-v = --2- fv for every v V . 3 3
38. Let f: VW satisfy the condition that fv1 + v2 = fv1 + fv2 for every v1 v2 V . Show that: (a) (PMI) fnv = nfv for every v V , and for every integer n. a a (b)f---v = --- fv for every v V , and for every rational number --a- . b b b 4.1 Linear Transformations 123
Exercises 39-44. Determine the indicated composition of the given linear functions, and then directly verify its linearity. 39.LT: 2 2 , where T: 2 is given by Tab = ab+ and L: 2 by La = 2aa – .
40.LT: 2 , where T: 2 2 is given by Tab = –a ab+ and L: 2 by Lab = a + 2b
2 2 41.LT: P2 , where T: is given by Tab = 3ab+ and L: P2 by La = ax2 + 2a .
3 2 42.LT: P2 M22 , where T: P2 is given by Tax++bx c = abc and
3 2ab L: M22 by Labc = . 0 –c
43.KLT: 3 , where T: 2 is given by Ta = aa – , L: 2 2 by Lab = 2aa + b, and K: 2 3 by Kab = –a 2ba + b .
3 3 2 44.KLT: P2 , where T: is given by Tabc = abac+ – , 2 2 L: by Lab = ab– , and K: P2 by Ka = ax ++2ax 3a . 45. (PMI) Let Li: Vi Vi + 1 be linear, for 1 in . Show that Ln L2L1: V1 Vn + 1 is linear.
PROVE OR GIVE A COUNTEREXAMPLE
46. For any a the function Ta: VV given by Tav = av is linear.
47. For any v V the function T : VV given by T v = vv+ is linear. 0 v0 v0 0
48. Let T: VW be linear. If Tv1 Tv2 Tvn is a linearly independent subset of W
then v1v2 vn is a linearly independent subset of V.
49. Let T: VW be linear. If v1v2 vn is a linearly independent subset of V then
Tv1 Tv2 Tvn is a linearly independent subset of W.
50. If for given functions f: VW and g: WZ the composite function gf: VZ is lin- ear, then both f and g must be linear.
51. If for given functions f: VW and g: WZ the composite function gf: VZ is lin- ear, then f must be linear.
52. If, for given functions f: VW and g: WZ , the composite function gf: VZ is lin- ear, then g must be linear. 124 Chapter 4 Linearity
4
§2. KERNEL AND IMAGE For any given transformation T: VW , we define the kernel of T to be the set of vectors in V which map to the zero vector in W [see Figure 4.1(a)], and we define the set of all vectors in W which are “hit” by some Tv to be the image of T [see Figure 4.1(b)].
V W V W T . T 0
Kernel of T Image of T (a) (b) Figure 4.1 More formally:
DEFINITION 4.2 Let T: VW be linear. The kernel (or null KERNEL space) of T is denoted by KerT and is given by: KerT = v VTv = 0 The image of T is denoted by ImT and is IMAGE given by: ImT = w WTv = w for some v V
Both the kernel and image of a linear transformation turn out to be subspaces of their respective vector space:
THEOREM 4.8 Let T: VW be linear. Then: (a)KerT is a subspace of V. (b)ImT is a subspace of W.
PROOF: We employ Theorem 2.13, page 61 to establish both parts of the theorem. v1 v2 KerT and rR (a) Since T0 = 0 , KerT is nonempty. Let v1 v2 KerT and r . Then: linearity
Trv1 + v2 ===rTv1 + Tv2 r 0 + 0 0 since v v kerT rv1 +Kerv2 T 1 2
Since T maps rv1 + v2 to 0, rv1 +Kerv2 T . 4.2 Kernel and Image 125
(b) Since T0 = 0 , ImT is nonempty. w w ImT and rR 1 2 Let w1 w2 ImT and rR . Choose vectors v1 v2 such that Tv1 = w1 and Tv2 = w2 (how do we know that such vec- tors exist?). Then:
Trv1 + v2 ==rTv1 + Tv2 rw1 + w2 Since we found a vector in V which maps to rw1 + w2 , rw +Imw T 1 2 rw1 +Imw2 T .
DEFINITION 4.3 Let T: VW be linear. NULLITY The dimension of KerT is called the nul- lity of T, and is denoted by nullityT . The dimension of ImT is called the rank RANK of T, and is denoted by rankT . The following theorem will be useful in determining the rank of a lin- ear transformation.
In particular, if THEOREM 4.9 Let T: VW be linear. If = v1v2 vn is a basis for V, then Spanv1v2 vn = V Then: Tv1 Tv2 Tvn will span ImT . SpanTv1 Tv2 Tvn = ImT
PROOF: We show that any w ImT can be written as a linear com-
binations of the vectors Tv1 Tv2 Tvn : Let w ImT , and let v V be such that Tv = w . Since w ImT S = v1v2 vn spans V, we can express v as a linear combination of the vectors in S:.
v = a1v1 +++a2v2 anvn n By linearity: w = a Tv Tv ==w a Tv +++a Tv a Tv i i 1 1 2 2 n n i = 1
EXAMPLE 4.5 3 (a) Show that the functionT: P2 given by: Tabc = ab+ x2 ++cx c is linear. (b) Determine ImT and rankT . (c) Determine KerT and nullityT . 126 Chapter 4 Linearity
SOLUTION: (a) For abc a b c 3 and r : Trabc + a b c = Tra+ arb+ b rc+ c = ra+ a + rb+ b x2 ++rc+ c xrcc+ regrouping: = ra+ bx2 ++cx c + a + b x2 ++cxc = rTabc + Ta b c (b) By Theorem 4.9, we know that the vectors: T100 = x2 T010 ==x2 T001 x + 1 span the image of T. Since x2 and x + 1 are linearly independent, x2 x + 1 is a basis for ImT . Consequently, rankT = 2 .
(c) By definition: KerT = abc Tabc = 0 = abc ab+ x2 ++cx c = 0x2 ++0x 0 This leads us to the following system of equations: ab+0= c ==0 and ab– c = 0 Thus: KerT = aa – 0 aR . Since 110 – is a basis for KerT , nullityT = 1 .
CHECK YOUR UNDERSTANDING 4.8
(a) Show that the function T: P2 M22 given by: Tax2 ++bx c = ab is linear. Answer: (a) See page B-13. ca (b) rankT = 3 , nullityT = 0 (b) Determine rankT and nullityT .
THEOREM 4.10 Let V be a vector space of dimension n, and DIMENSION let T: VW be linear. Then: THEOREM rankT + nullityT = n
PROOF: Start with a basis v1v2 vk for KerT and extend it (if
necessary) to a basis v1v2 vk vk + 1 vkt+ for V [see Theo- rem 3.12(a), page 100], where: kt+ = n —the dimension of V.
We will show that the t vectors Tvk + 1 Tvkt+ constitute a basis for ImT . This will complete the proof, for we will then have:
rankT + nullityT ==k + t n 4.2 Kernel and Image 127
As you know, to establish the fact that Tvk + 1 Tvkt+ is a basis for ImT we have to verify that:
(1) The vectors Tvk + 1 Tvkt+ span the space ImT .
Why can’t you simply And that: show just one of the two? (2) The vectors Tvk + 1 Tvkt+ are linearly independent. Let’s do it: (1) Let w ImT . Choose v V such that Tv = w .
Since v1v2 vk vk + 1 vkt+ is a basis for V, we can express v as a linear combination of those kt+ vectors: v = a1v1 ++++a2v2 akvk ak + 1vk + 1 ++akt+ vkt+ By linearity: w ==Tv a1Tv1 ++++a2Tv2 akTvk ak + 1Tvk + 1 ++akt+ Tvkt+
Since v1v2 vk KerT
==0 +++ak + 1Tvk + 1 akt+ Tvkt+ ak + 1Tvk + 1 ++akt+ Tvkt+
a linear combination of the vectors Tvk + 1 Tvkt+
(2): Assume that: b Tv ++b Tv = 0 1 k + 1 t kt+ b1Tvk + 1 + b2Tvk + 2 ++btTvkt+ = 0 By linearity:
Tb1vk + 1 + b2vk + 2 ++btvkt+ = 0 And therefore:
b1vk + 1 +Kerb2vk + 2 ++btvkt+ T
Since v1v2 vk is a basis for KerT , we can then find scalars
c1 ck such that:
b1vk + 1 + + btvkt+ = c1v1 ++ckvk Or:
–c1 v1 ++–ck vk +b1vk + 1 + + btvkt+ = 0
Since v1v2 vk vk + 1 vkt+ is a basis, it is a linearly inde- pendent set of vectors. Consequently those c’s and b’s must all be b ==0 b 0 b =0 1 2 t zero; in particular: b1 ==0 b2 0 bt =0 . 128 Chapter 4 Linearity
EXAMPLE 4.6 Determine bases for the kernel and image of the linear function T: P3 M22 given by:
Tax3 +++bx2 cx d = ab– c 2ca+ d
SOLUTION: We go for the kernel, as it is generally easier to find than the image apace: To say that: Tax3 +++bx2 cx d = 0
Is to say that: ab– c = 00 2ca+ d 00 Equating coefficients leads us to a homogeneous system of equations: System S is certainly easy each of these four enough to solve directly. equations is equal to 0. Still: a b cd a b cd ab–0= ad= – 11–00 1 001 c = 0 coef[S] 0010 0 1 01 bd= – S: rref 2c = 0 0020 001 0 c = 0 ad+0= 1001 0000 d is free From the above, we see that: KerT ==– dx3 – dx2 + dd R dx– 3 –1x2 + dR Conclusion: – x3 –1x2 + is a basis for KerT . Knowing that nullityT = 1 , we turn to Theorem 4.10 and con- clude that
rankT ===dimP3 –41nullityT –3 See Exercise 16, page 105 At this point, the easiest way to find a basis for ImT is to apply T to
3 vectors in P3 making sure that you end up with 3 linearly indepen- dent vectors in ImT —a basis for ImT . Which 3 vectors should we start with? Basically, you can take any 3 randomly chosen vec- tors, and the chances are that they will do fine (think about it); we will go with the 3 vectors x3 , x2 , and x + 1 : Recall that: 1x3 +++0x2 0x 0 0x3 +++1x2 0x 0 3 2 Tax3 +++bx2 cx d = 0x +++0x 1x 1
ab– c 3 10 2 –01 11 2ca+ d Tx == Tx Tx + 1 = 01 00 21 4.2 Kernel and Image 129
We leave it to you to verify that above three vectors, which are cer- tainly in Im(T) (why?), are linearly independent, and therefore consti- tute a basis for the 3-dimensional space Im(T).
CHECK YOUR UNDERSTANDING 4.9
Determine bases for the kernel and image of the linear function Answer: See page B-14. T: 3 4 given by Tabc = 2ab + ccb .
ONE-TO-ONE AND ONTO FUNCTIONS As you may recall: ...... DEFINITION 4.4 A function f from a set A to a set B is said . . . . ONE-TO-ONE to be one-to-one if: A B A B fa==fa a a one-to-one not one-to-one A function f from a set A to a set B is said to ...... ONTO ...... be onto if for every bB there exist aA such that fa= b . A B A B onto not onto When dealing with a linear transformation, we have:
The first part of this theorem THEOREM 4.11 (a) A linear transformation T: VW is one- is telling is that if a linear map is “one-to-one at 0,” to-one if and only if KerT = 0 . then it is one-to-one every- where. Certainly not true for (b) A linear transformation T: VW is onto other functions: if and only if ImT = W .
PROOF: (a) Let T be one-to-one, and suppose that Tv = 0 . Since T0 = 0 [Theorem 4.1(a), page 113], and since there can be but one v V that is mapped to 0 W (T is one-to-one), v = 0 . Conversely, assume that Tv = 0 v = 0 . Then:
Tv1 = Tv2
Tv1 – Tv2 = 0
T is linear: Tv1 – v2 = 0
Tv = 0 v = 0: v1 – v2 = 0
v1 = v2 (b) Follows directly from the definition of onto (Definition 4.4) and the definition of ImT (Definition 4.2). 130 Chapter 4 Linearity
EXAMPLE 4.7 Show that the linear function T: 3 3 given by Tabc = ac+ 3bc – b is one-to-one.
SOLUTION: We show that Tv ==0 v 0 : Tabc = 0 Tabc ==ac+ 3bc – b 000 equating coefficients: 3b ==0 b 0 cb–0=== c b c 0 abc = 0 ac+0== a –c a =0 Applying Theorem 4.11, we conclude that T is one-to-one.
Consider the adjacent graph of the func- tion fx= x3 – x . As you can see, there 2. are some y’s which are only “hit by one x” 1 (like y = 2 ), and there are some y’s . . . which are “hit by more than one x” (like -1 1 y = 0 ) — the function is kind of one-to- -1 one in some places, and not one-to-one in -2 other places. Linear transformations are not so fickle; if a linear transformation is “one-to-one anywhere” (not just at 0) then it is one-to-one everywhere:
CHECK YOUR UNDERSTANDING 4.10
Let T: VW be linear. Show that if there exists v V such that Answer: See page B-14. Tv ==Tv v v then T is one-to-one. 4.2 Kernel and Image 131
EXERCISES
Exercises 1-18. Determine if the given function is linear. If it is, find a basis for its kernel and its image space.
1.f: , where fx = –5x 2.f: 2 , where fx = x 2x
3.f: 2 , where fx = xx – 4.f: 2 , where fa = aa 2
5.f: 2 2 , where fab = ab a 6.f: 2 2 , where fab = –2ba
2 3 3 7.f: P1 , where fab = ax+ b 8.f: , where f abc = 0bc
ab 2 9.f: M22 P2 , where f = ab– ++bc– xca– x cd
2 ab 10.f: M22 , where fab = ab+ ab–
4 a 2b 11.f: M22 , where fabcd = cd c+ d
3 12.f: P2 , where fpx = p1 p2 p3
2 13.f: P2 , where fpx = p0 p1
14.f: P3 P4 , where fpx = xp x
15.f: P3 P3 , where fpx = px+ p1
p1 p2 16.f: P3 M22 , where fpx = p3 p4
3 ab– bc– 17.f: M22 , where fabc = ab+ bc+
4 2ac+ d 18.f: M22 , where fabcd = cd– b 132 Chapter 4 Linearity
Exercises 19-27. (Calculus Dependent) Show that the given function is linear. Find a basis for its kernel and its image space.
19.f: P3 P3 , where fpx = px
20. f: P3 P3 , where fpx = px
21. f: P3 P3 , where fpx = px + px
22. f: P2 P , where fpx = px
23.f: P2 P3 , where fpx = 2px– 3px
1 24.f: P1 where fpx = pxdx 0
1 25.f: P2 where fpx = pxdx 0
x 26.f: P1 P2 where fpx = ptdt 0
x 27.f: P2 P3 where fpx= ptdt 0
28. Let T: 3 3 be given by Tabc = ab– ac+ c . (a) Which, if any of the following 3-tuples are in the kernel of T? (i) –330 (ii) 033 – – (iii) 300 (b) Which, if any of the following 3-tuples are in the image T? (i) –330 (ii) 033 – – (iii) 300
3 3 29. Let T: P3 be given by Tabc = ax ++ax b+ c . (a) Which, if any of the following 3-tuples are in the kernel of T? (i) 100 (ii) 033 – (iii) 001 (b) Which, if any of the following 3-tuples are in the image of T? (i) x3 + x (ii) 5 (iii) x3 + 5 30. Determine a basis for the kernel and image of the linear transformation T: 3 3 which maps 100 to 111 , 010 to 325 – , and 001 to –23– 4.
3 31. Determine a basis for the kernel and image of the linear transformation T: M22
which maps 100 to 21 , 010 to 22 , and 001 to 01 . 01 13 10 4.2 Kernel and Image 133
32. Determine a basis for kernel and image of the linear transformation T: M22 P4 which
maps 10 to 2x4 + 1 , 11 to x4 – x , 11 to 5, and 11 to x2 . 00 00 10 11
33. Find, if one exists, a linear transformation T: 2 3 such that: (a) 913 is a basis for ImT . (b)900 010 is a basis for ImT . (c)913 319 931 is a basis for ImT . 34. Find, if one exists, a linear transformation T: 3 3 such that: (a) 913 is a basis for KerT . (b)900 010 is a basis for KerT . (c)913 319 931 is a basis for KerT . 35. Let dimV = n , and let T: VV be the linear operator Tv = rv , for r . Enumer- ate the possible values of nullityT and rankT . 36. Let T: VW be linear with dimV = 2 and dimW = 3 . Enumerate the possible val- ues of nullityT and rankT . 37. Let T: VW be linear with dimV = 3 and dimW = 2 . Enumerate the possible val- ues of nullityT and rankT . 5 38. Let T: P3 be a one-to-one linear map. Determine the rank and nullity of T. 5 39. Let T: P3 be an onto linear map. Determine the rank and nullity of T. 40. Let T: VV be a linear operator, with dimV = n . Prove that ImT = V if and only if T is one-to-one.
41. Let T: VW be linear, with dimV = dimW . Prove that v1v1 vn is a basis for
V if and only if Tv1 Tv2 Tvn is a basis for W. 42. Give an example of a linear transformation T: VW such that ImT = KerT . 43. Let T: VW be a linear transformation, with dimV = n . Prove that T is one-to-one if and only if rankT = n . 44. Let T: VW be linear, with dimV = dimW . Prove that T is one-to-one if and only if T is onto. 45. Let T: VW and L: WZ be linear. (a) Prove that KerT KerLT . (b) Give an example for which KerT = KerLT . (c) Give an example for which KerT KerLT . 134 Chapter 4 Linearity
PROVE OR GIVE A COUNTEREXAMPLE
46. If spanTv1 Tv2 Tvn = ImT then spanv1v2 vn = V .
47. There exists a one-to-one linear map T: M23 P4 .
48. There exists a one-to-one linear map T: P4 M23 .
49. There exists an onto linear map T: M23 P4 .
50. There exists an onto linear map T: P4 M23 . 51. If T: VW is linear and dimV dimW , then T cannot be onto. 52. If T: VW is linear and dimV dimW , then T cannot be one-to-one. 53. If T: VW is linear and dimV dimW , then T cannot be onto. 54. If T: VW is linear and dimV dimW , then T cannot be one-to-one. 55. There exists a linear transformation T: 2 4 such that rankT = nullityT . 56. There exists a linear transformation T: 3 4 such that rankT = nullityT . 57. There exists a linear transformation T: 2 4 such that rankT nullityT . 58. There exists a linear transformation T: 2 4 such that rankT nullityT . 59. If T: VW is linear and if W is finite dimensional, then V is finite dimensional. 60. If T: VW is linear and if V is finite dimensional, then W is finite dimensional. 61. If T: VW is linear and if V is finite dimensional, then ImT is finite dimensional. 62. If T: VW is linear and if ImT is finite dimensional, then V is finite dimensional.If T: VW is linear and if KerT is finite dimensional, then either V or W is finite dimen- sional. 63. Let T: VW and L: WZ be linear. If dimV = 3dim W = 2 , and nullityT = 1 , then rankLT 1 . 64. Let T: VW and L: WZ be linear. If dimV = 3dim W = 3 , rankT = 1 , and nullityL = 2 , then rankLT = 1 . 65. Let T: VW and L: WZ be linear. If dimV = 3dim W = 3 , rankT = 1 , and nullityL = 2 , then rankLT 1 . 66. Let T: VW and L: WZ be linear, with dimW = n and dimZ = m . If T is one- to-one and L is onto, then dimV = nm– . 4.3 Isomorphisms 135
4
§3. ISOMORPHISMS
We can all agree that there is little difference between the vector n space with its horizontal n-tuples, and the space Mn 1 of “vertical n-tuples”. In this section, we show that there is, in fact, little difference between the vector space n and any n dimensional vector space what- soever.
BIJECTIONS AND INVERSE FUNCTIONS One-to-one and onto functions were previously defined on page 129. Of particular interest are functions that satisfy both properties:
DEFINITION 4.5 A function f: A B that is both one-to-one BIJECTION and onto is said to be a bijection. . . Roughly speaking: ...... A bijection f: A B serves to pair of each A B elements of A with those of B (see margin). a bijection THEOREM 4.12 If f: A B and g: BC are bijections, then the composite function gf: AC is also a bijection. PROOF: gf is one-to-one: gf a = gf b gf a = gf b Definition of composition: gfa = gfb Since g is one-to-one: fa= fb ab= Since f is one-to-one: ab=
f g gf is onto: Let cC be given (see margin). Since g is onto there . c exists bB such that gb= c . Since f is onto, there a. b . exists aA such that fa= b . We then have: A B C g f gf a ===gfa gb c this a “does the trick”
Figuratively speaking, if you reverse the arrows of the bijection f: 1234 6510 of Figure 4.2(a), you end up with the bijection f –1: 6510 1234 of Figure 4.2(b), which is called the inverse of f. 136 Chapter 4 Linearity
–1 Only bijections f f 1. .6 1 6 f: X Y . . 2 2 . 5 . 5 have inverses 3. . 1 3 . 1 –1 . . . f : Y X 0 4. . 4. .0 (a) (b) Figure 4.2 The relationship between the functions f and f –1 depicted in the mar- gin reveals the fact that each function “undoes” the work of the other. f For example: 1. 6 . –1 –1 –1 2. f f 2 ===f f2 f 5 2 .5 3 1 and .4 0. . f –1 . f f –1 5 ===ff–15 f2 5 In general: DEFINITION 4.6 The inverse of a bijection f: XY is the –1 INVERSE function f : YX such that: –1 FUNCTIONS f f x = x for every x in X and –1 ff y = y for every y in Y.
CHECK YOUR UNDERSTANDING 4.11
–1 Answer: See page B-14. Prove that if f: XY is a bijection, then so is f : YX .
BACK TO LINEAR ALGEBRA
THEOREM 4.13 If the bijection T: VW is linear, then its inverse, T –1: WV is also linear. PROOF: Let ww W and r be given. Let vv V be such that Tv = w and Tv = w . Then: linearity of T T –1rww+ ==T –1rTv + Tv T –1Trvv+ ===T–1T rvv+ rvv+ rT–1w + T–1w Definition of composition Definition 4.6 since Tv ==w and Tv w 4.3 Isomorphisms 137
EXAMPLE 4.8 Show that the linear map T: 3 3 given by: Tabc = ac+ 3bc – b is a bijection. Find its inverse and show, directly, that T –1 is linear.
SOLUTION: T IS ONE-TO-ONE. We start with Tabc = Tab c and go on to show that abc = ab c (see Definition 4.4, page 129): Tabc = Tab c ac+ 3bc – b = a + c 3b c – b
ac+ = a + c aa= 3b = 3b bb= then then cb– = c – b cc= T IS ONTO. We start with ab c and find an abc such that T Tabc = ab c (see Definition 4.4, page 129):
abc Tabc ==ac+ 3bc – b ab c . .abc Equating coefficients brings us to the system of equations: b a = a – c – ---- ac+ = a 3 b 3bb= b = ---- then 3 then cb– = c b c = c + ---- 3 b b b Let’s make sure that abc = a – c – ---- ---- c + ---- works: 3 3 3 b b b Tabc = Ta – c – ---- ---- c + ---- 3 3 3 b b b b b ==a – c – ---- +3c + ---- ---- c + ---- – ---- ab c 3 3 3 3 3
FINDING THE INVERSE OF T. Since b b b Ta – c – -------- c + ---- = ab c : 3 3 3
–1 b b b T ab c = a – c – ---- ---- c + ---- 3 3 3 Dropping the primes, we show, directly, that T–1: 3 3 given by b b b T–1abc = ac– – ------ c + --- IS LINEAR: 3 3 3 138 Chapter 4 Linearity
–1 –1 T –1rabc + ab c T rabc + ab c = T ra+ a rb+ b rc+ c rb b rb b rb b = ra+ a – rc – c – ----- – ---- ----- + ---- rc++ c ----- +---- 3 3 3 3 3 3 b b b b b b = = rac– – ------ c + --- + a – c – ---- ---- c + ---- 3 3 3 3 3 3
–1 –1 –1 rT abc + T –1a b c = rT abc + T a b c
CHECK YOUR UNDERSTANDING 4.12
2 Show that the linear map T: P1 given by: Tab = ab+ xa– Answer: See page B-14. is a bijection. Find its inverse, and show directly that it is linear. Roughly speaking, two vector spaces will be considered to be the “same” if the vectors of one space can be pared off with those of the other, while preserving the vector space structures of those spaces. More formally:
DEFINITION 4.7 A linear map T: VW which is one-to-one ISOMORPHISM and onto is said to be an isomorphism from the vector space V to the vector space W.
3 EXAMPLE 4.9 Show that the function T: P2 given by: Tabc = – ax2 ++bc+ x 3c is an isomorphism.
SOLUTION: We start off by establishing linearity, for we can then take advantage of previously established theory to show that the function is one-to-one and onto: Linearity: Trabc + ab c = Tra + a rb + b rc + c' = – ra + a x2 ++rb + b + rc + c x 3rc + c = ra– x2 ++bc+ x 3c + –ax2 ++b + c x 3c = rTabc + Tab c One-to-one: We show that KerT = 0 [see Theorem 4.11(a), page 129]: Tabc ==0 – ax2 ++bc+ x 3c 0x2 ++0x 0 Equation coefficients: –0a === bc+ 03 c 0 or: abc===0 4.3 Isomorphisms 139
Onto: We show that ImT = P2 [see Theorem 4.11(b)]: The Dimension Theorem of page 126, tells us that: rankT +3nullityT = Knowing that the nullity is 0, we conclude that rankT = 3 . Since P2 is of dimension 3: ImT = P2 (Exercise 50, page 107).
THEOREM 4.14 (a) Every vector space is isomorphic to itself. This theorem asserts that “isomorphic” is an equiv- (b) If V is isomorphic to W, then W is isomor- alence relation on any set phic to V. of vector spaces. See Exer- cises 37-39. (c) If V is isomorphic to W, and W is isomor- phic to Z, then V is isomorphic to Z.
PROOF: (a) The identity map IV : VV is easily seen to be an iso- morphism. (b) To say that V is isomorphic to W is to say that there exists an iso- morphism T: VW . Since T –1:WV is also a linear bijection (CYU 4.12 and Theorems 4.13), W is isomorphic to V. (c) Let T: VW and L: WZ be isomorphisms. Since LT: VZ is both a bijection (Theorem 4.12) and linear (Theorem 4.7, page 117), V is isomorphic to Z. Theorem 4.14(b) enables us to formulate the following definition: DEFINITION 4.8 Two vector spaces V and W are isomorphic, ISOMORPHIC written VW , if there exists an isomor- SPACES phism from one of the vector spaces to the other.
CHECK YOUR UNDERSTANDING 4.13
4 Prove that M22 . (You have to exhibit an isomorphism from Answer: See page B-15. one of the spaces to the other, whichever you prefer). The following lovely result says that all n dimensional vector spaces are isomorphic to the Euclidean n-space. THEOREM 4.15 If V is a vector space of dimension n, then: V n
PROOF: Let v1v2 vn be a basis for V, and let e1e2 en be the standard basis for n (see page 94). Consider the function n T: V , given by: Ta1v1 ++ anvn = a1e1 ++ anen . Theorem 4.6, page 115 assures us that T is linear. We complete the proof by showing that T is both one-to-one and onto (and therefore an isomorphism). 140 Chapter 4 Linearity
Tv = Tv T is one-to-one: Let Tv = Tv , for: v ==a1v1 ++ anvn and v b1v1 ++ bnvn Then: Ta1v1 ++ anvn = Tb1v1 ++ bnvn
a1Tv1 ++anTvn = b1Tv1 ++bnTvn
a1e1 ++ anen = b1e1 ++ bne1n Since e1e2 en is a linearly independent set of vectors we have, by Theorem 3.6, page 89, that ai = bi , for vv= 1 in; in other words: vv= . n T is onto: For X = a1e1 ++ anen : For X Rn Ta1v1 ++ anvn = a1Tv1 ++anTvn
==a1e1 ++ anen X Tv = X CHECK YOUR UNDERSTANDING 4.14 Let V and W be finite-dimensional vector spaces. Show that VW if Answer: See page B-15. and only if dimV = dimW .
A ROSE BY ANY OTHER NAME
Let T: VW be an isomorphism. Being a bijection it links every element in V with a unique element in W (every element in V has its own W-counterpart, and vice versa). Moreover, if you know how to T–1 v1 w1 function algebraically in V, then you can also figure out how to function T–1 v w algebraically in W. Suppose, for example, that you forgot how to add or 2 2 scalar multiply in the space W, but remember how to add and scalar rw1 + w2 rv + v –1 1 2 multiply in V. To figure out rw1 + w2 in W you can take the “T bridge” back to V and find the vectors v1 and v2 such that Tv1 = w1 T and Tv2 = w2 . Do the calculations rv1 + v2 in V and then take the “T bridge” back to W to find the value of the vector rw1 + w2 , for it coincides with the vector Trv1 + v2 :
Trv1 + v2 ==rTv1 + Tv2 rw1 + w2 Indeed, the intimacy between isomorphic vector spaces is so great that isomorphic spaces are said to be equal up to an isomorphism. Basi- cally, if a vector space W is isomorphic to V, then the two spaces can only differ from each other in “appearance.” For example, M22 looks different than 4 , but you can easily link its elements with those of V:
ab link abcd cd And that linkage preserving the algebraic structure:
r ab ++a b link rabcd a b c d cd c d 4.3 Isomorphisms 141
We note that all vector space properties are preserved under an iso- morphisms. In particular, as you are asked to verify in the exercises, a linear map T: VW maps: Linearly independent sets in V to linearly independent sets in W. Spanning sets in V to spanning sets in W. CHECK YOUR UNDERSTANDING 4.15
Let L: VW be an isomorphism. Prove that if v1v2 vn is a Answer: See page B-16. basis for V, then Lv1 Lv2 Lvn is a basis for W. At times, one can take advantage of established properties of Euclid- ean spaces to address issues in other vector spaces: EXAMPLE 4.10 Find a subset of the set: 12 –23 11– 2 36 –146 S = 32– 13 313– 96– –46 which is a basis for SpanS .
SOLUTION: Lets move the elements of S over to 4 via the isomor- ab phism T= abcd : cd
For f: XY , and SX : 123– 2–3 213 11–313 2 – TS= fS= fssS 369– 6 –1466–4 Applying Theorem 3.13, page 103, to the vectors in TS we see that the first, second, and fifth vector in TS constitute a basis for SpanTS:
13–1136– 1 0230 22–614 2 0 1 –003 rref 31 3 9– 6 00 0 01 –3132 –6–4 00 0 00
Utilizing the result of the CYU 4.16, we conclude that the corre- sponding first, second, and fifth matrix in S constitute a basis for SpanS .
CHECK YOUR UNDERSTANDING 4.16 Proceed as in Example 4.10 to find a subset of the set
S = 2x3 –53x2 + x – 1x3 –8x2 + x – 3 x2 + 11x – 5 –2x3 ++x2 3x – 2
Answer: See page B-16. in P3 which is a basis for Span (S). 142 Chapter 4 Linearity
Let f be a bijection from a vector space V (with addition denoted by v1 + v2 and scalar multiplication by rv ) to a set X. Just as the bijection f can be thought of as simply “renaming” each v V with its counter- part in X, so then can f be used to “carry” the vector space structure of V onto the set X; specifically: THEOREM 4.16 Let f be a bijection from a vector space V to a set X. With addition and scalar multiplication on X defined as follows: (*) (**) –1 –1 –1 x1 x2 = ffx1 + f x2 and rx = frfx Go back to V and and sum in V. Similarly Then carry the sum back to X. the set X evolves into a vector space. Moreover f itself turns out to be an isomorphism from the vector space V to the vector space X. PROOF: The set X is clearly closed with respect to both (*) and (**) operations. We will content ourselves by verifying the zero and inverse axioms of Definition 2.6, page 40: Zero Axiom. Let 0 be the zero vector in V. We show that f0 is the f0 turns out to zero vector in X. be the zero in X. For xX , let v be the vector in V such that fv = x . Then: f0 x ==f0 fv ff–1f0 + f –1fv ===f0 + v fv x
fv– turns out to be Inverse Axiom: For xX , let v be the vector in V such that the inverse of fv . fv = x . We show that f–v is the inverse of xf= v : . xf –v ==fv f–v ff–1fv+ f –1fv–
==fv+ – v f0 the zero in X
In the exercises you are invited to verify that the remaining axioms of Definition 2.6 are also satisfied, thereby establishing the fact that X with the above specified addition and scalar multiplication is indeed a vector space. We now show that the given bijection f from the vector space V to the above vector space X is an isomorphism. Actually, since f was given to be a bijection, we need only establish the linearity of f. Let’s do it:
For v1 v2 V , let x1 ==fv1 and x2 fv2 . Then, for any r : frv1 + v2 ==rx 1 x2 rfv1 + fv2 by (*) and (**) 4.3 Isomorphisms 143
EXAMPLE 4.11 Let 2 denote the Euclidean two-space, and let X be the set Xxy= xy R . (a) Show that the function f: 2 X given by fxy = x – 1 xy+ is a bijection, and find its inverse f –1: X 2 . (b) Determine the vector space structure on X induced by the function f, as is described in Theorem 4.16. (c) Identify the zero in the above vector space X, and the inverse of the vector xy X .
SOLUTION: (a) f is one-to-one:
fx1 y1 = fx2 y2
x1 – 1 = x2 – 1 x1 = x2 x1 – 1 x1 + y1 = x2 – 1 x2 + y2 x1 + y1 = x2 + y2 y1 = y2
f is onto: Let xy X . Then: fx+ 1 yx–1– ==x + 11– x ++1 yx–1– xy From the above we can easily see that: f –1xy = x + 1 yx–1– (b) Theorem 4.16 assures us that the set X achieves “vector space- hood” when it is augmented with the following operations:
x1 y1 x2 y2 = fx1 + 1 y1 –1x1 – + x2 + 1 y2 –1x2 –
= fx1 ++x2 2 y1 + y2 –2x1 – x2 – fxy = x – 1 xy+ : = x1 ++x2 1 y1 + y2 rxy = frx+ 1 – x + y – 1 ==frx+ r – rx + ry– r rx+ r – 1 ry
(c) Theorem 4.16 assures us that fxy = x – 1 xy+ is a linear map (in fact an isomorphism). That being the case, f00 = –10 must be the zero vector in X. Less there be any doubt: –10 xy = ff–1–10 + f –1xy –1 f xy = x + 1 yx–1– : = f 00 + x + 1 – x + y – 1 ==fx+ 1 – x + y – 1 xy fxy = x – 1 xy+ 144 Chapter 4 Linearity
As for the inverse of xy X : –xy ==ff– –1xy fx–+ 1 yx–1– –1 f xy = x + 1 yx–1– : ==fx–1– – y ++x 1 –2x – –y
fxy = x – 1 xy+
Let’s challenge the above formula with the vector 32 X . The formulas tells us that –332 ==–2–2 – –25 – . If that is correct, then 32 –25 – has to be the zero vector –01 , and it is: 32 –25 – = ff–132 + f –1–25 –
===f42 – + –24 f00 –01
CHECK YOUR UNDERSTANDING 4.17
Let 3 denote the Euclidean three-space, and let X be the set X= xyz xy R . (a) Show that the function f: 3 X given by fxyz = 2xx + z – z is a bijection, and find its inverse. (b) Determine the vector space structure on X induced by the func- tion f, as is described in Theorem 4.16. (c) Identify the zero in the above vector space X, and the inverse of Answer: See page B-17. the vector xyz X .
It can be shown that there exists a bijection from n to R for any positive integer n. Consequently:
THEOREM 4.17 Every Euclidean vector space n (and there- fore every finite dimensional vector space) sits (isomorphically) in the set R of real numbers. PROOF: A direct consequence of Theorem 4.16. 4.3 Isomorphisms 145
EXERCISES
Exercises 1-6. Determine if the given linear function f is a bijection. If so, find its inverse f –1 and show directly that it is also linear. 1.f: , where fx = –5x .
2.f: 2 , where fx = xx – .
3.f: 2 2 , where fab = –2ba .
2 4.f: P1 , where fab = ax+ b .
3 2 5.f: P2 , where fabc = bx + cx– a .
2 ab 6.f: M22 , where fab = . ab+ ab–
Exercises 7-17. Determine if the given function is an isomorphism. 7.f: , where fx = –5x . 8.f: , where fx = x + 1 .
2 9.f: P1 , where fab = ab+ x .
2 10.f: P1 , where fab = ax+ b .
3 2 11.f: P2 ,where fabc = cx + bx– a.
12.f: P2 P2 , where fpx = px+ 1 .
13.f: P2 P3 , where fpx = xp x .
3 14.f: P2 , where fpx = p1 p2 p3 .
15.f: P3 P3 , where fpx = px+ p1 .
p1 p2 16.f: P3 M22 , where fpx = . p3 p4
4 2ac 17.f: M22 , where fabcd = . db 146 Chapter 4 Linearity
Exercises 18-22. (Calculus dependent) Determine if the given function is an isomorphism.
18. f: P2 P2 , where fpx = px .
19. f: P2 P2 , where fpx = px + px .
20. f: P2 P2 , where fpx = px .
21.f: P2 P2 , where fpx = 2px– 3 .
1 22.f: P2 P2 , where fpx = px+ pxdx . 0
Exercises 23-24. As is done in Example 4.11, show that the given function f is a bijection and find its inverse. Determine the vector space structure on the given set X induced by f and identify the zero and the inverse of the vector xy X in the resulting space X.
23. f: 2 X = xy xy R given by fxy = xy++3 x – 4 .
24. f: 3 X = xyz xyz R given by fxyz = x +21y –32 z + 4 .
25. Show that if the functions f: XY and g: YZ have inverses, then the function –1 –1 –1 gf: XZ also has an inverse and that gf = f g .
26. For r , let fr: VV be given by frv = rv . For what values of r is fr an isomor- phism? 27. For v a vector in the space V let f : VV be given by f v = vv+ . 0 v0 v0 0 (a) Show that f is a bijection. v0 (b) Give necessary and sufficient conditions for f to be an isomorphism. v0 28. Find a specific isomorphism from Vax= 3 ++ab+ x cabc to 3 .
29. Show that the vector space + of Example 2.4, page 46, is isomorphic to the vector space of real numbers, .
30. Find an isomorphism between the vector space of Example 2.5, page 47 and 2 .
31. Suppose that a linear transformation T: VW is one-to-one, and that v1v2 vn is a linearly independent subset of V. Show that Tv1 Tv2 Tvn is a linearly indepen- dent subset of W. (In particular, the above holds if T is an isomorphism.)
32. Suppose that a linear transformation T: VW is onto, and that v1v2 vn is a span- ning set for V. Show that Tv1 Tv2 Tvn is a spanning set for W. (In particular, the above holds if T is an isomorphism.) 4.3 Isomorphisms 147
33. Prove that a linear transformation T: VW is an isomorphism if and only if for any given basis v1v2 vn for V, Tv1 Tv2 Tvn is a basis for W. 34. Let V be a vector space of dimension n, and let LV be the vector space of linear trans- formations from V to (see Exercise 35, page 122). Prove that LV is also of dimen- sion n and is therefore isomorphic to V. (The space LV is called the dual space of V.)
Suggestion: For v1v2 vn a basis for V, show that T1T2 Tn is a basis for
1ifij= LV , where Ti is the linear transformation given by: Tivj = . 0ifij 35. Let V be a vector space of dimension n, and let W be a vector space of dimension m. Let LVW be the vector space of linear transformations from V to W (see Exercise 35, page 122). Prove
that LV Mmn .
Suggestion: For v1v2 vn a basis for V, and w1w2 wn a basis for W, show that Tij is a basis for LVW , where Tij is the linear transformation given by:
wj if ij= Tijvj = . 0ifij Exercises 37-39. (Equivalence Relation) A relation on a set X is a set Sab= ab X . If ab S , then we will say that a is related to b, and write ab . An equivalence relation on a set X is a relation which satisfies the following three properties: (i) Reflexive property: aa . (ii) Symmetric property: If ab , then ba . (iii) Transitive property: If ab and bc , then ac . 36. A partition of a set X is a collection of mutually disjoint (nonempty) subsets of X whose union equals X. (In words: a partition breaks the set X into disjoint pieces.) (a) Let be an equivalence relation of a set X, and for any given xX define the equivalence class of x to be the set of all elements of X that are related to x: x = x Xx x . Prove that x xX is a partition of X. (b) Let S be a partition of a set X. Prove that there exists an equivalence relation on X such that the Sx= xX , where x = x Xx x . a c 37. Show that the relation defined by --- --- if and only if ad= bc is an equivalence relation b d on the set Q of rational numbers (“fractions”). 38. Show that the relation VW if the vector space V is isomorphic to the vector space W is an equivalence relation on any set of vector spaces. 148 Chapter 4 Linearity
PROVE OR GIVE A COUNTEREXAMPLE
39. (a) If f: XY is an onto function, then so is the function gf: XZ onto for any function g: YZ .
(b) If g: YZ is an onto function, then so is the function gf: XZ onto for any func- tion f: XY .
40. (a) Let f: XY and g: YZ . If gf: XZ is onto, then f must also be onto. (b) Let f: XY and g: YZ . If gf: XZ is onto then g must also be onto.
41. (a) If f: XY is a one-to-one function, then so is the function gf: XZ one-to-one for any function g: YZ .
(b) If g: YZ is a one-to-one function, then so is the function gf: XZ one-to-one for any function f: XY . 42. If T: VW and L: VW are isomorphisms, then TL= .
43. If T: VW is an isomorphism, and if a 0 , then Ta: VW given by Tav = aTv is also an isomorphism.
44. Let T: VW and L: WZ be linear. If gf: VZ is an isomorphism, then T and L must both be isomorphisms. 45. If T: VW and L: VW are isomorphisms, then so is the function TL+ : V W given by TL+ v = Tv + Lv an isomorphism. 149 Chapter 4 Linearity
CHAPTER SUMMARY
LINEAR A function T: VW from a vector space V to a vector space W is TRANSFORMATION said to be a linear transformation if for all vw V and r : Tvw+ ==Tv + Tw and Trv rTv
The two conditions for T: VW is linear if and only if: linearity can be incor- Trvw+ = rTv + Tw porated into one state- ment. for all vw V and r . The above result can be Let T: VW be linear. For any vectors v1v2 vn in V, and any extended to encompass scalars a1a2 an : n-vectors and scalars. Ta1v1 ++ anvn = a1Tv1 ++anTvn Linear transformations If T: VW is linear, then: map zeros to zeros and T0 = 0 and T–v = –Tv inverses to inverses. A linear transformation Let V be a finite dimensional space with basis v1v2 vn . If is completely deter- T: VW and L: VW are linear maps such that Tv = Lv mined by its action on a i i basis. for 1 in , then Tv = Lv for every v V . A method for construct- Let v1v2 vn be a basis for a vector space V, and let ing all linear transfor- w w w be n arbitrary vectors (not necessarily distinct) in a mations from a finite 1 2 n dimensional vector vector space W. There is then a unique linear transformation space to any other vec- L: VW which maps vi to wi for 1 in ; and it is given by: tor space. La1v1 ++ anvn = a1w1 ++ anwn The composition of lin- If T: VW and L: WZ are linear, then the composition ear maps is linear. LT: VZ is also linear. KERNEL Let T: VW be linear. The kernel (or null space) of T is denoted by KerT and is defined by: KerT = v VTv = 0 IMAGE The image of T is denoted by ImT and is defined by: ImT = w WTv = w for some v V
Both the kernel and Let T: VW be linear. Then: image of a linear transformation are sub- KerT is a subspace of V spaces. ImT is a subspace of W Chapter Summary 150
NULLITY Let T: VW be linear. The dimension of KerT is called the nul- lity of T, and is denoted by nullityT .
RANK The dimension of ImT is called the rank of T, and is denoted by rankT .
The Dimension Let V be a vector space of dimension n, and let T: VW be linear. Theorem. Then: rankT + nullityT = n
ONE-TO-ONE A function f from a set A to a set B is said to be one-to-one if: fa==fa a a
ONTO A function f from a set A to a set B is said to be onto if for every bB there exist aA such that fa= b .
A linear transformation T: VW is one-to-one if and only if KerT = 0 . A linear transformation T: VW is onto if and only if ImT = W .
BIJECTION A function f: A B that is both one-to-one and onto is said to be a bijection.
The composite of bijec- If f: A B and g: BC are bijections, then the composite function tions is again a bijec- gf: AC is also a bijection. tion.
INVERSE FUNCTION The inverse of a bijection f: XY is the function f –1: YX such that f –1 f x = x for every x in X, and ff –1 y = y for every y in Y.
The inverse of a linear If the bijection T: VW is linear, then its inverse, T –1: WV is bijection is again lin- also linear. ear.
ISOMORPHISM A bijection T: VW that is also a linear transformation is said to be an isomorphism from the vector space V to the vector space W. If there exists an isomorphism from V to W we then say that V and W are isomorphic, and write: VW .
Every vector space is isomorphic to itself. If V is VV isomorphic to W, then W is isomorphic to V. If V is VW WV isomorphic to W, and W is isomorphic to Z, then V is isomorphic to Z. VW and WZ VZ All n-dimensional vector spaces are isomorphic to If dimV = n , then V Rn . Euclidean n-space. 5.1 Matrix Multiplication 151
5 CHAPTER 5 MATRICES AND LINEAR MAPS It turns out that linear transformations can, in a sense, be represented by matrices. Such representations are developed and scrutinized in Sec- tions three and four — a development that rests on the matrix theory presented in the first two sections of the chapter.
§1. MATRIX MULTIPLICATION
The matrix space Mmn of Theorem 2.2, page 42, comes equipped with a scalar multiplication. We now turn our attention to another form of multiplication, one that involves a pair of matrices (instead of a sca- lar and a matrix). Left to one’s own devices, one would probably define the product of matrices in the following fashion: Take two matrices of equal dimension, and simply mul- As with: 23 53 = 10 9 tiply corresponding entries 54 06 024 to obtain their product.
The above might be “natural,” but as it turns out, not very useful. Here, as you will see, is a most useful definition: In general, we will use: A = a a DEFINITION 5.1 If Amr = aij and Brn = bij , then: mn ij or ij mn MATRIX to denote an m by n matrix Amr Brn ==Cmn cij with entries a . ij MULTIPLICATION where: cij = ai1b1j ++++ai2b2j ai3b3j airbrj Using Sigma notation, we have: r
cij = aibj
b 1 j ai1air = 1 b rj th IN WORDS: To get cij of CAB= , run across the i row of A and r down the j th column of B, multiplying and adding along the way (see c = a b ij i j margin). = 1 Note: The above is meaningful only if the number of columns of the matrix on the left equals the number of rows of the matrix on the right. 152 Chapter 5 Matrices and Linear Maps
EXAMPLE 5.1 Find the product CAB= , if: 1235 201 A == and B 2043 524 1532
SOLUTION: Since the number of columns of A equals the number of rows of B, the product AB is defined:
A2 3 B3 4 = C24
same
Below, we illustrate the process leading to the value of c11 (run
across the first row of A and down the first column of B), and for c23 (run across the second row of A and down the third column of B):
1 235 123 5 201 3 201 3 2 043 == 204 3 524 524 35 1 532 153 2 21 ++02 11 53 ++24 43 At this point, we are confident that you can find the remaining entries:
1235 201 3 9912 2043 = 524 13 30 35 39 1532 In any event: GRAPHING CALCULATOR GLIMPSE 5.1
CHECK YOUR UNDERSTANDING 5.1
(a) Perform the product AB for: 33 37 35 Answer: (a) 30 26 64 54 36 A ==42 and B 35 (b) Number of columns 90 in A does not equal the number of rows in B. (b) Explain why the product BA is not defined. 5.1 Matrix Multiplication 153
In some ways, matrices behave differently than numbers. For one thing, even when both products AB and BA are defined, as is the case when the matrices A and B are square matrices of the same dimension, those products need not be equal: 12 56 19 22 56 12 23 34 Matrix multiplication is == while not commutative. 34 78 43 50 78 34 31 46
Some familiar multiplication properties do however carry over to matrices: THEOREM 5.1 Assuming that the matrix dimensions are such that the given operations are defined, we have: (i) AB+ C = AB+ AC Distributive Properties { (ii) AB+ CACBC= + (iii) Associative ABC= AB C Properties { (iv) rAB==rA BArB
PROOF: We establish (i): AB+ C = AB+ AC, and invite you to verify the rest in the exercises. Let: The properties of this A = a B = b C =c theorem are not particu- mr ij rn ij rn ij larly difficult to estab- lish. The trick is to Dmn ==AB+ C dij Emn ==AB+ AC eij carefully keep track of the entries of the matri- We need to show that d = e . Let’s do it: ces along the way, ij ij r distribute r r
dij == aitbtj + ctj aitbtj + aitctj =eij t = 1 t = 1 t = 1 down the down the across the th the across run across the i row of A th and down the j column of BC+ j j th i th i th th column of column of colum of colum of A A C B
POWERS OF SQUARE MATRICES Why we are restricting this 2 discussion to square matrices? If A is a square matrix, what should A represent? That’s right: Because: A2 = AA. In a more general setting, we have: 2 21 35 = ? DEFINITION 5.2 For A a square matrix: 42 POWERS A2 = AA A3 ==AAA and An AAn – 1 154 Chapter 5 Matrices and Linear Maps
While the definition of An mimics that of an for aR , not all of the familiar properties of exponents hold for matrices, even when the matrix expressions are well-defined. The property ab n = anbn , for example, does not carry over to matrices: 2 2 12 22 42 42 42 20 8 == = 02 10 20 20 20 84
2 2 12 22 12 12 22 22 while: = 02 10 02 02 10 10
==16 64 18 16 04 22 88
In = M , ab= ba . 2 2 2 11 12 22 12 22 Why not for M22 ? And we see that: 02 10 02 10
The following familiar properties are, however, a directly consequence of Definition 5.2: THEOREM 5.2 For A a square matrix, and positive integers n and m: AnAm ==Anm+ and An m Anm
CHECK YOUR UNDERSTANDING 5.2
PROVE OR GIVE A COUNTEREXAMPLE:
For any two matrices AB M22 : Answer: See page B-18. AB+ 2 =2A2 ++AB B2
COLUMN AND ROW SPACES
DEFINITION 5.3 The columns space of a matrix AM mn , m COLUMN AND is the subspace of spanned by the col- umns of A. ROW SPACE The row space of a matrix AM mn , is the subspace of n spanned by the rows of A. The following result may be a bit surprising in that the rows and col-
umns of the matrix AM mn need not even live in the same Euclid- ean spaces: The rows of A are in n while its columns are in m : 5.1 Matrix Multiplication 155
THEOREM 5.3 The dimension of the column space of any matrix AM mn equals the dimension of its row space. PROOF: We know, from Theorem 3.13, page 103, that the dimension
of the column space of AM mn equals the number of leading ones in rrefA . To see that the dimension of the row space of A is also equal to the number of leading ones in rrefA we reason as follows: Let B be a matrix obtained by performing any of the three ele- mentary row operations on A. Since each row in B is a linear combination of the rows in A, and vice version, the space spanned by the rows of A equals that spanned by the rows of B . Since rrefA is derived from A through a sequence of ele- mentary row operations, the row space of A equals that of rrefA , which is easily seen to be the non-zero rows of rrefA ; each of which contains a leading one.
DEFINITION 5.4 For AM mn , the rank of A, denoted by RANK OF A MATRIX rankA is the common dimension of the row and column space of AM mn .
EXAMPLE 5.2 Find a basis for both the row and the column 39–15–6 space of A = –32421 – . 26–2–6–4
39–15–6 13–02–2 rref SOLUTION: A = –32421 – 00110. 26–2–6–4 00000 A basis for the row space of A: 13– 0.– 2 2 00110 . A basis for the column space of A: 31– 2 12– 2 (see Theorem 3.13, page 103) While Theorem 3.13 assures us that he columns of A associated with the leading one columns of rrefA constitute a basis for the column space of A, the rows of A associated with the leading one rows of rrefA need not be a basis for the row space of A. A case in point: 156 Chapter 5 Matrices and Linear Maps
CHECK YOUR UNDERSTANDING 5.3
Find a basis for the column and the row space of: 25–4 3 A = –104 –68– . Answer: 240 – 5101 – 012– 4
SYSTEM OF EQUATIONS REVISITED As you know, a system of equations can take the form of an augmented matrix multiplication matrix 24– 4 x 6 matrix [see page 3, and Figure (a) and (b) below]. That system can also 26 4 y = 0 be represented as the product of its coefficient matrix with a (column) 11 2 z –2 variable matrix [see Figure 5.1(c) and margin]. 2x + 4y – 4z 6 2x +64y – 4z = 24–6 4 24– 4 x 6 2x ++6y 4z = 0 2x ++6y 4z = 0 26 4 0 26 4 y = 0 xy++2z –2 xy++2z = – 2 11 2– 2 11 2 z –2 2x +64y – 4z = System of Equations Augmented Matrix Matrix Product Form 2x ++6y 4z = 0 (a) (b) (c) xy++2z = – 2 Figure 5.1
The next result tells us that the product of AM mn with a column
variable matrix Xn 1 is a linear combination of the columns of A. Spe- cifically: C C C 1 2 n THEOREM 5.4 a11 a12 a1n x1 Let Amn = aij . For each 1 j n let a a a x 21 22 2n 2 = C = a (the jth “column matrix” of A). j ij m 1 Then, for any Xx= i : am1 am2 amn xn n 1 n a a a 11 12 1n AX= x C (see margin). a a a i i x 21 ++x 22 x 2n 1 2 n i = 1 n am1 am2 amn PROOF: For BAXb== , and Dx==C d : C1 C2 Cn i m 1 i i i m 1 i = 1 m m b ==a x = x a d i ij j j ij i j = 1 j = 1 5.1 Matrix Multiplication 157
Consider the following system of equations AX= B : A X B x 2134 b1 2x +++1y 3z 4w b1 y 183– 5 = b 1x ++8y 3z – 5w = b z 2 2 12–16 b 1x –12y ++z 6w b w 3 3 Invoking Theorem 5.2, we find that:
2 1 3 4 b1
x 1 +++y 8 z 3 w –5 = b2
1 –2 1 6 b3 It follows that the given system AX= B is consistent if and only if
b1
b2 is in the column space of A.
b3 In general: THEOREM 5.5 A system of linear equations AXB= is consis- tent if and only if B is in the column space of A.
CHECK YOUR UNDERSTANDING 5.4 Prove that the solution set of any homogeneous system of m Answer: See page B-18. equations in n unknowns is a subspace of n .
FROM MATRICES TO LINEAR TRANSFORMATIONS While matrix spaces only come equipped with vector addition and sca- Note that: lar multiplication, matrix multiplication can serve to define a linear map Amn Xnz Mmz between such spaces — providing their dimensions “match up:”
THEOREM 5.6 For AM mn and any positive integer z, the
map TA:Mnz Mmz given by TAB = AB is linear.
PROOF: For B1 B2 Mnz and r Theorem 5.1(iv):
TArB1 + B2 ==ArB1 + B2 ArB1 + AB2 Theorem 5.1(iv): ==rAB1 + AB2 rTAB1 + TAB2 158 Chapter 5 Matrices and Linear Maps
For notational convenience we will let the symbol n denote the space Mn 1 (“vertical n-tuples”). We then have:
Note that X is a is a THEOREM 5.7 n m n 1 For AM mn the map TA: given “vertical n-tuple,” and that by T X = AX is linear. TAX is a vertical m-tuple. A PROOF: Simply set z = 1 in Theorem 5
DEFINITION 5.5 For A Mmn , the null space of A, denoted NULL SPACE by nullA , is the set: nullA = X n AX = 0
THEOREM 5.8 n The null space of AM mn is a subspace of . PROOF: Follows from Theorem 4.8(a), page 124, and the fact that:
nullA = KerTA Note: The dimension of the null space of A is called the nullity of A Fair terminology, in that nullA = KerTA .
EXAMPLE 5.3 Find a basis for the null space of the matrix: 2134 A = 18310– 12–16 SOLUTION: By definition, the null space of A is the solution set of the homogeneous system of equations: 2134x 0 y 18310– = 0 z 12–16w 0 x y z w x y z w 1 0 --7------14- 2134 5 5 rref 1 8 From: 18310– 0 1 --- –--- 5 5 12–16 0000 5.1 Matrix Multiplication 159
vertical 3-tuple 7 14 1 8 nullA = –---a – ------b –---a + ---bab ab R We see that: 5 5 5 5 = –147a – ba – + 8b5a 5b ab R
Letting a = 1 and b = 0 we obtain the vector –71–50 . Letting a = 0 and b = 1 we arrive at the vector –14 8 0 5 . The above two vectors are clearly independent, and also span nullA : –7a – 14ba – + 8b5a 5b = a–71–50+ b–14805 It follows that –71–50 –14805 is a basis for nullA .
We arrived at the two vectors in the basis for nullA in the above example by setting each of the two free variables to 1 and the other free variable to 0. Generalizing:
THEOREM 5.9 The nullity of AM mn equals the number of free variables in rrefA .
CHECK YOUR UNDERSTANDING 5.5
Find a basis for: 21 3 0 null14–7 2– Answer: 11– 11 30 1– 2
You get to draw the final curtain of this section: CHECK YOUR UNDERSTANDING 5.6
n m Let AM mn and let TA: be the linear map given by TAX = AX. Show that: Answer: Seepage B-18. nullityA = nullityTA and rankA = rankTA 160 Chapter 5 Matrices and Linear Maps
EXERCISES
Exercises 1-5. Perform the given matrix operations.
12 12 23 230 1.01– 2. 01– –41 –401 13 13
010 100 100 010 3. 200 002 and 002 200 003 030 030 003
12 12 12 13031 211 13 031 13 211 4. + 13 and 13 + 13 22123 420 22 123 22 420 01 01 01 Exercises 5-8. Find a basis for the null space of A. Determine rankA along with a basis for the column and row space of A.
321 5 201 1101 11–55 5. A = 10 6 1 10 6. A = 31–0 7. A = –011 –4 8. A = 10 3 2 42–0 1 633 53–7 2 01–2–1– 142 52–2 1 14 7 2
9. (a) Show that each column of 25 371 (as a vertical two-tuple) is a linear combination 13 204 of the columns of 25 . 13 (b) Show that each row of 25 371 (as a horizontal three-tuple) is a linear combination 13 204 of the rows of 371 . 204
th 10. Let AM mn and let Xi Mn 1 be the column matrix whose i entry is 1 and all other th entries are 0. Show that AXi is the i column of A, for 1 in .
r 0 2 2 11. (a) (Dilation and Contraction) Let A = for r 0 . Show that TA: maps 0 r every point in the plane to a point r times a far from the origin.
10 2 2 (b) (Reflection about the x-axis) Let A = . Show that TA: reflects every 01– point in the plane about the x-axis. 5.1 Matrix Multiplication 161
2 2 (c) Find the matrix A for which TA: reflects the point xy to the point twice the length of xy and reflected about the y-axis.
TAv = x y (d) (Rotation about the Origin) Let A = cos –sin . Show that . r sin cos v = xy 2 2 r . T : rotates the vector xy by in a counterclock- y A wise direction. x Suggestion: Start with x = rcos+ and y = rsin+ . 12. Show that for any given linear transformation T: n m there exists a unique matrix AM mn such that TX= AX .
13. Let AB Mmn . Prove that if AC= BC for every CM n 1 , then AB= .
14. Determine all AM 22 such that AB= BA for every BM 22 . 15. Prove Theorem 5.1(ii). 16. Prove Theorem 5.1(iii). 17. Prove Theorem 5.1(iv).
18. A square matrix Aa= ij nn for which aij = 0 if ij is said to be a diagonal matrix. n Show that if Aa= ij nn is a diagonal matrix and if Xx= i is a column matrix, 200 9 18 n then AX= aiixi . For example: 050 6 = 30 . 007 –1 –7 T 19. The transpose of a matrix Aa= ij Mmn is the matrix A = aij Mnm , where aij = aji . In other words, the transpose of A is that matrix obtained by interchanging the rows and columns of A. (a) Show that for any matrix A, AT T = A . (b) Show that for any matrix A and any scalar r, rA T = rAT . T T T (c) Show that for any AB Mmn , AB+ = A + B . T T T (d) Show that for any AB Mmn , AB– = A – B . T T T (e) Show that for AM nn , AA = A A . T T T (f) Show that for AM mn and BM nr , AB = B A .
20. A square matrix A is symmetric if the transpose of A equals A: AT = A (see Exercise 19).
(a) Show that the set of symmetric matrices in the space Mnn is a subspace of Mnn .
(b) Let AB Mmn be symmetric matrices. Show that AB+ is symmetric. T (c) Let AM nn . Show that AA+ is symmetric.
T (d) Let AM nn . Show that AA is symmetric.
(e) Let AB Mnn be symmetric. Show that AB is symmetric if and only if AB= BA 162 Chapter 5 Matrices and Linear Maps
21. A square matrix A is said to be skew-symmetric if AT = –A (see Exercise 19).
(a) Show that if the square matrix Aa= ij is skew symmetric, then each diagonal entry aii must be zero. (b) Show that the square matrix Aa= ij is skew symmetric if and only if aij = –aji for all i, j. T (c) Let AM nn . Show that AA– is skew symmetric. (d) Show that every square matrix A can be uniquely written as ASK= + , where S is a symmetric matrix and K is a skew symmetric matrix.
2 22. A matrix AM nn is said to be idempotent if A = A . (a) Find an idempotent 22 matrix A containing no zero entry. (b) Let A and B be idempotent square matrices of the same dimension. Prove that if AB= BA , then AB is idempotent.
k 23. A matrix AM nn is said to be nilpotent if A = 0 for some integer k. (a) Find a nilpotent 22 matrix A distinct from the zero matrix. (b) Let A and B be two nilpotent matrices of the same dimension. Prove that if AB= BA , then AB is again nilpotent.
24. The sum of the diagonal entries in the matrix Aa= ij Mnn is called the trace of A and is n
denoted by traceA : traceA ==a11 ++a22 ann aii . i = 1 (a) Show that for any square matrix A: traceA = traceAT (see Exercise 19). (b) Show that for any two square matrices A and B of the same dimension: TraceAB+ = TraceA + TraceB (c) Show that for any two square matrices A and B of the same dimension: TraceAB = TraceBA Exercises 25-30. (PMI) Determine An for the given matrix A. Use the Principle of Mathematical Induction to substantiate your claim.
25.11 26.11 27. 10 01 11 02
28. a 0 29.10 30. 1 a 0 b 21 01
31. (PMI) Show that if AB= BA , then for any positive integer n, AB n = AnBn .
n 32. (PMI) Let AM mm and BM ms . Show that if AX= X , then A XX= for every positive integer n.
33. (PMI) Show that if the entries in each column of AM nn sum to 1, then the entries in each column of Am also sum to 1, for any positive integer m. 5.1 Matrix Multiplication 163
n 34. (PMI) Show that if AM nn is a diagonal matrix, then so is A . (See Exercise 18.)
n 35. (PMI) Show that if AM nn is an idempotent matrix, then A = A for all integers n 1 . (See Exercise 22.)
n T T n 36. (PMI) Show that for any AM nn , and for any positive integer n, A = A . (See Exercise 19.)
37. (PMI) Let Ai Mmn , for 1 in . Show that: T T T T A1 +++A2 An = A1 +++A2 An . (See Exercise 19.)
38. (PMI) Let Ai Mnn for 1 in . show that: T T T T A1 A2 An = An An – 1 A1 . (See Exercise 19.)
PROVE OR GIVE A COUNTEREXAMPLE
39. For AM mn and BM nr , if AB = 0 then either A = 0 or B = 0 . 40. Let A and B be two-by-two matrices with A 0 . If AB= AC , then BC= . 41. If A and B are square matrices of the same dimension, and if AB is idempotent, then AB= BA . (See Exercise 22.)
2 2 42. For all AB M22 , A – B = AB+ AB– .
2 43. For any given matrix AM 22 , all entries in the matrix A are nonnegative.
44. For AM mn and BM nr , if A has a column consisting entirely of 0’s, then so does AB.
45. For AM mn and BM nr , if A has a row consisting entirely of 0’s, then so does AB.
46. For AM mn and BM nr , if A has two identical columns, then so does AB.
47. For AM mn and BM nr , if A has two identical rows, then so does AB.
48. For AM nn , BM nn AB = 0 is a subspace of Mnn .
49. For AM nn , BM nn AB= BA is a subspace of Mnn .
50. For AM nn , BM nn TraceAB 0 is a subspace of Mnn . (See Exercise 24.)
51. If A is a nilpotent matrix, then so is A2 . (See Exercise 23.) 52. A is idempotnet if and only if AT is idempotent. (See Exercise 22 and 19.) 164 Chapter 5 Matrices and Linear Maps
5
§2. INVERTIBLE MATRICES
Since all identity matrices The identity matrix of dimension n, denoted by In , is the nn are square we can get away by specifying just one of matrix that has 1’s along its main diagonal (top-left corner to bottom- its dimensions, as with: right corner), and 0’s elsewhere. In the event that the dimension n of
1 0 0 the identity matrix is understood, we may simply write I rather than In . I = 3 0 1 0 Just as 1 a = a and a 1 = a , for any number a, it is not difficult 0 0 1 to show that for any AM : I AA== and AI A . In particu- instead of I33 . mn m n
lar, for any square matrix AM nn : InAAI==n A .
INVERTIBLE MATRICES
DEFINITION 5.6 A square matrix AM nn is said to be INVERTIBLE invertible if there exists a matrix BM nn MATRIX such that: AB== BA I We will soon show, that a The matrix B is then said to be the inverse of matrix A can have but one –1 inverse. A, and we write BA= . If no such B exists, then A is said to be non-invertible, or singular.
EXAMPLE 5.4 Determine if the matrix A = –25 is 94– invertible. If it is, find its inverse.
ab SOLUTION: Let us see if there exists a matrix B = such that cd AB= I2 :
–25 ab = 10 94– cd 01
– 5a + 2c –25b + d = 1 0 9a –94c b – 4d 01 Equating corresponding entries leads us to the following pair of sys- tems of two equations in two unknowns:
– 5a +12c = –25b +0d = 9a –04c = 9b –14d = 5.2 Invertible Matrices 165
If you take the time to solve the above systems, you will find that the one on the left has solution a = –2 , c = –--9- ; and the one on the 2
--5- right has solution b = –1 , and d = –2 ; which is to say: –12 – B = –--9- –--5- 2 2 Actually, you need not verify We leave it for you to verify that both AB and AB equal I and that there- that both AB and BA equal I, for if one does, then so must –12 – the other (Theorem 5.19). fore A–1 = (see margin). –--9- –--5- 2 2
EXAMPLE 5.5 Determine if the matrix A = 23 is –64 – invertible. If it is, find its inverse. SOLUTION: As we did in the previous example, we again try to find a ab matrix B = such that AB= I2 : cd
23 ab = 10 –64 – cd 01
2a +23c b + 3d = 10 –64a –4c –6b – d 01 Equating corresponding entries of the two matrices leads us to the following two systems of equations: 2a +13c = 2b +03d = –64a –0c = –64b –1d =
The system of equation on Multiplying the equation 2a +13c = in the system on the left by 2, the right also has no solution. and adding it to the bottom equation, leads to an absurdity: 4a +26c = –64a –0c = add: 02=
We have just observed that there does not exist a matrix ab such that cd
23 ab = 10, which tells us that 23 is a singular matrix. –64 – cd 01 –64 – 166 Chapter 5 Matrices and Linear Maps
GRAPHING CALCULATOR GLIMPSE 5.2 Example 5.4 Example 5.5
CHECK YOUR UNDERSTANDING 5.7 Answer: Invertible with 32– inverse –2515 . Determine if is invertible. If it is, find its inverse. –3545 41– While a square matrix need not have an inverse (Example 5.5), if it does, then it is unique: THEOREM 5.10 An invertible matrix A has a unique inverse.
PROOF: For B and C inverses of A, we have: BBIBAC=====BA CICC
since C is an since B is an inverse of A inverse of A Here are three additional results pertaining to invertible matrices:
THEOREM 5.11 (i) If A is invertible, then so is A–1 , and: A–1 –1 = A (ii) If A is invertible and r 0 , then rA is also invertible, and: 1 rA –1 = ---A–1 r This is sometimes refered (iii) If A and B are invertible matrices of the to as the same dimension, then AB is also invertible, shoe-sock theorem Can you guess why? and: AB –1 = B–1A–1
PROOF: (i) We must suppress the temptation to appeal to the familiar exponent rule a–1 –1 ==a–1 –1 a , for we are not dealing with matrices and not numbers. What we do, instead, is to turn to Defini- tion 5.6 which tells us that: 5.2 Invertible Matrices 167
If A–1 = A–1 for some matrix in the box, then the matrix A–1 is invertible, and the matrix in the box is its inverse: A–1 –1 = Now put A in the box:
A–1 AA== A–1 I
We have shown that A–1 –1 = A .
(ii) Exercise 25. (iii) Returning to our “box game,” we see that: From the given condi- AB B–1A–1 ====ABB–1 A–1 AIA–1 AA–1 I tions, we know that B–1 and A–1 exist. What we and: B–1A–1 AB ====B–1A–1A BB–1IB B–1BI do here is to show that the –1 –1 product B A is, in –1 –1 –1 fact, the inverse of AB. The above shows that the inverse of AB: AB = B A . In words: The inverse of a product of invertible matrices is the product of their inverses, in the reverse order.
CHECK YOUR UNDERSTANDING 5.8 Use the Principle of Mathematical Induction to show that if
A1A2 An are invertible, then so is their product, and: Answer: See page B-19. –1 –1 –1 –1 A1A2 An = An An – 1 A1 As you know, for any nonzero number a: 1 (i) a0 = 1 and (ii) a–n = ----- (for a 0 ) an While we can adopt the notation in (i) for invertible matrices, we can- 1 not do the same for (ii), since the expression --- is simply not defined A for a matrix A. We can however rewrite (ii) in the form a–n = a–1 n , and use that form to define A–n : DEFINITION 5.7 For A, an invertible matrix, and n a positive integer, we define: 0 –n A and A (i) A0 = I (ii) A–n = A–1 n 168 Chapter 5 Matrices and Linear Maps
ELEMENTARY MATRICES
Elementary row operations DEFINITION 5.8 An elementary matrix is a matrix that is were introduced on page 3. ELEMENTARY obtained by performing an elementary row MATRIX operation on an identity matrix.
Here are some examples of elementary matrices: 100 0 01 0 050 0 1 6 E1 = 10 0 E2 = E3 = 001 0 0 1 00 1 000 1 Interchange Add 6 times the I the first and Multiply the second second row of 2 second row of I row of by 5 to the first row of I 3 I4 2
CHECK YOUR UNDERSTANDING 5.9
Show that the above three elementary matrices, E1 , E2 and E3 are Answer: See page B-19 invertible, and that the inverse of each is itself an elementary matrix. The situation in the above Check Your Understanding box is not a fluke. You are invited to establish the following theorem in the exercises: THEOREM 5.12 Every elementary matrix is invertible, and its inverse is also an elementary matrix. Indeed: (a) If E is the elementary matrix obtained by interchanging rows i and j of the identity matrix I, then E–1 is the elementary matrix obtained by again interchanging rows i and j of I. (b) If E is obtained by multiplying row i of I by c 0 , then E–1 is obtained by multi- 1 plying row i of I by --- . c (c) If E is obtained by adding c times row i to row j of I, then E–1 is obtained by adding –c times row i to row j. Consider the row operation that is performed in the first and second line of Figure 5.2. As you can see in line 3, the matrix resulting from performing the elementary row operation on A coincides with the prod- uct matrix EA : 5.2 Invertible Matrices 169
multiply row 3 by -2 and add it to the first row
2103– 2R + R R –4 –5 –8 1 3 1 1 A = 1326 1326
3341 3341 SAME
10 0– 2R + R R 10 –2 3 1 1 I3 ==01 0 01 0 E 00 1 00 1
10 –22103 –4 –5 –8 1 EA ==01 01326 1326 00 13341 3341 Figure 5.2 The following theorem, a proof of which is relegated to the exercises, tells us that the situation depicted in Figure 5.2 holds in general: THEOREM 5.13 The matrix obtained by performing an elemen- tary row operation on a matrix AM mn equals that matrix EA , where E is the matrix obtained by performing the same elementary
row operation on the identity matrix Im .
001 CHECK YOUR UNDERSTANDING 5.10 Answer: (a) E = 010 (a) Switch the first and third rows of the matrix A of Figure 5.2 to 100 arrive at a matrix B, and then find the elementary matrix E such that EA= B . 001 (b) Multiply the second row of the matrix A of Figure 5.2 by 2 to (b) E = 0 2 0 arrive at a matrix B, and then find the elementary matrix E such 100 that EA= B . We remind you that two matrices are equivalent if one can be derived from the other via a sequence of elementary row operations (Definition 1.1, page 3). The following theorem asserts that if a matrix A is invert- ible, then every matrix equivalent to A is also invertible. THEOREM 5.14 For A and B are equivalent square matrices: A is invertible if and only if B is invertible.
PROOF: Since A and B are equivalent, there exist elementary matrices
E1E2 Es such that: BE= s E2E1A 170 Chapter 5 Matrices and Linear Maps
Theorem 5.11 tells us that each Ei is invertible. Consequently, if A is invertible then so is B, for it is a product of invertible matrices. By sym- metry we also have that if B is invertible then so is A. You can quickly determine whether or not a matrix is invertible by looking at its row-reduced-echelon form:
THEOREM 5.15 AM nn is invertible if and only if:
rrefA = In
PROOF: If rrefA = In , then A and I are equivalent. Since I is invertible (with inverse I), A is also invertible (Theorem 5.15).
We establish the converse by showing that if rrefA In , then A is not invertible:
Let rrefA = CI n . Since Cc= ij has less than n leading ones (otherwise it would be In ), its last row must consist entirely of zeros. This being the case, for any given nn matrix Dd= ij , the product matrix CD= eij cannot be the identity matrix, as its lower-right-corner entry is not 1: n n
enn === cndn 0 dn 01 = 1 = 1 Since there does not exist a matrix D such that CD= I , rrefA = C is not invertible. It follows, from Theorem 5.155, that A is not invertible. The following theorem provides a systematic method for finding the inverse of an invertible matrix: THEOREM 5.16 If a sequence of elementary row operations reduces the invertible matrix A to I, then applying the same sequence of elementary row operations on I will yield A–1 .
PROOF: Let E1E2 Es be the elementary matrices correspond- ing to elementary row operations which take A to I: Es E2E1AI= Since elementary matrices are invertible, their product is also invert- ible and we have (CYU 5.9):
–1 –1 –1 –1 –1 AE===sE2E1 IEsE2E1 E1 E2 Es Thus: 5.2 Invertible Matrices 171
–1 –1 –1 –1 –1 –1 –1 –1 –1 –1 –1 A ==E1 E2 Es Es E2 E1
==EsE2E1 EsE2E1I
We have shown that EsE2E1I is the inverse of the matrix A; to put it another way: The exact same sequence of row operations that trans- form A to I can also be used to transform I to A–1 . We now illustrate how the above theorem can effectively be used to find the inverse of an invertible matrix, by hand:
EXAMPLE 5.6 12 3 Determine if A = 25 3 is invertible. If it 10 8 is, find its inverse.
SOLUTION: We know that A is invertible if and only if rrefA = I . Rather than obtaining the row-reduced-echelon form of A, we will find the row-reduced echelon form of the matrix AI which is that matrix obtained by “adjoining” the 33 identity matrix to A: 12 3100 AI = 25 3010 10 8001 for that can feed two birds with one seed: First bird: A is invertible if and only if rrefAI = I (By Theorem 5.15) Second bird: If A is invertible, then rrefAI = IA–1 (By Theorem 5.17) Using the Gauss-Jordan Elimination Method (page 10), we have:
12 3100 12 3 100 10 9 5–0 2 – 2R1 + R2 R2 – 2R + R R A I = 25 3010 01 –23 –10 2 1 1 – 1R + R R 2R + R R 01 –23–10 1 3 3 2 3 3 10 8001 02– 51–01 00 –51–21
–1R R 10 9 5–0 2 10 0–169 40 3 3 – 9R + R R 01 –23–10 3 1 1 = I A–1 3R + R R 01 0 13–3 5– 3 2 2 00 1 5–1 2– 00 1 5–1 2–
Oh well: 172 Chapter 5 Matrices and Linear Maps
Answer: A is invertible CHECK YOUR UNDERSTANDING 5.11 with inverse: –557 – 6 101 2 –233 – 2 021 4 10–10 7 –8 Determine if A = is invertible. If it is, find its inverse. 111 0 –111 – 1 031 1 Theorem 5.16 says that a matrix A is invertible if and only rrefA = I . Here are some other ways of saying the same thing:
THEOREM 5.17 Let AM nn . The following are equivalent: (i) A is invertible. (ii)AX= B has a unique solution for
every BM n 1 . (iii)AX = 0 has only the trivial solution. (iv)rrefA = I . (v) A is a product of elementary matrices.
PROOF: We show i ii iii iv v i : i ii : Let A be invertible. For any given equation AX= B , we have: AX= B A–1AX = A–1B A–1A XA= –1B IX= A–1B XA= –1B unique solution
ii iii : If AX= B has a unique solution for any B, then AX = 0 has a unique solution. But AX = 0 has the trivial solution X = 0 . It follows that AX = 0 has only the trivial solution.
iii iv : Assume that AX = 0 has only the trivial solution. Since A and rrefA are equivalent matrices, rrefA X = 0 has only the trivial solutions. This tells us that rrefA does not have a free variable, and must therefore have n leading ones. As such: rrefA = I . 5.2 Invertible Matrices 173
iv v : If rrefA = I then: Es E2E1AI= for some elementary matrices E1E2 Es . Since elementary matrices are invertible (Theorem 5.12), their product is also invertible and we have: –1 –1 –1 –1 –1 AE===sE2E1 IEsE2E1 E1 E2 Es
a product of elementary matrices (see CYU 5.8)
v i : If A is a product of elementary matrices, then it is a product of invertible matrices, and is therefore invertible [CYU 5.9]. We defined a matrix A to be invertible if there exists a matrix B such that AB== BA I . As it turns out, B need only “work” on the left (or right side) of A:
THEOREM 5.18 Let AM nn . If there exists BM nn such that BA= I , then A is invertible and A–1 = B .
PROOF: Assume that BA= I , and consider the equation AX = 0 . Multiplying both side of that equation on the left by B we have: AX = 0 BAX= B0 IX = 0 X = 0 Since the equation AX = 0 has only the trivial solution, the matrix A is invertible (Theorem 5.18). Upon multiplying both sides of BA= I on the right by A–1 , we have: BA= I BAA–1 = IA–1 BI= A–1 BA= –1 In a similar fashion: Answer: See page B-20. CHECK YOUR UNDERSTANDING 5.12
Let AM nn . Show that if there exists BM nn such that AB= I , then A is invertible and A–1 = B . 174 Chapter 5 Matrices and Linear Maps
EXERCISES
Exercises 1-6. Find the inverse of the given invertible matrix A, and then check directly that AA–1 ==A–1AI.
1.20 2.12 3. 22 05 32 13
10 0 01 1 21 3 4.12 0 5.12 0 6. 12 3 12 3 01 2 20 4 Exercises 7-12. Determine if the given matrix A is invertible and if so, find its inverse.
7.23 8.32 9. –41 –64 – 31 520–
12 3 140 13 0 10. 23 2 11.22–3 12. 22– 3 33 3 13212 12 1 Exercises 12-15. For what values of a and b is the given matrix invertible?
13.a 3 14.a 2 15. ab 4 b b 1 1 ab
Exercises 16-18. (a) Find elementary matrices E1 and E2 such that E2E1IA= for the given matrix A. (b) Find two elementary row operations which will take A to I.
1 00 0 0 1 1 0 0 16.A = 03– 0 17.A = 01 0 18. A = 00 1 0 0 1 5 0 0 7 1 0 Exercises 19-24. Assume that A, B, and X are square matrices of the same dimension and that A and B are invertible. (a) Solve the given matrix equation for the matrix X.
(b) Challenge your answer in (a) using A = 12 and B = 21 . 01 31 19.2XA= AB 20. 2XA= BA 21. AXB–1 = BA 22.BXA= B2 23.BXB= BAB 2 24. BXA= AB BA –1
25. Prove Theorem 5.11(ii). 5.2 Invertible Matrices 175
26. Prove Theorem 5.12. 27. Prove Theorem 5.13. 28. Prove that if A is invertible, then A–2A–3 = A–5 .
29. Let AM mn . Prove that there exists an invertible matrix CM mm such that CA = rrefA .
30. Let A1A2 As be a linearly independent set of vectors in Mnn , and let AM nn
be invertible. Show that AA1AA2 AAs is linearly independent.
31. Let A1A2 An2 be a basis for Mnn , and let AM nn be invertible. Show that
AA1AA2 AAn2 is also a basis. 32. Show that a (square) matrix that has a row consisting entirely of zeros cannot be invertible. 33. Show that a (square) matrix that has a column consisting entirely of zeros cannot be invertible. 34. Show that if a row of a (square) matrix is a multiple of one of its other rows, then it is not invertible. 35. State necessary and sufficient conditions for a diagonal matrix to be invertible. (See Exercise 18, page 161.)
n 36. Prove that AM nn is invertible if and only if the rows of A constitute a basis for .
n 37. Prove that AM nn is invertible if and only if the columns of A constitute a basis for .
38. Prove that the transpose AT of an invertible matrix A is invertible, and that AT –1 = A–1 T . (See Exercise 19, page 161.) 39. Prove that if a symmetric matrix is invertible, then its inverse is also symmetric. (See Exer- cise 20, page 161.)
40. Prove that if AM nn is an idempotent invertible matrix, then AI= n . (See Exercise 22, page 162.) 41. Prove that every nilpotent matrix is singular. (See Exercise 23, page 162.)
42. (a) Prove that A = ab is invertible if and only if ad–0 bc . cd
(b) Assuming that A = ab is invertible, find A–1 (in terms of a, b, c, and d). cd
2 43. Let AM nn be such that A – 2A +0I = . Show that A is invertible.
2 44. Let AM nn be such that A ++sA tI = 0 , with t 0 . Show that A is invertible. 176 Chapter 5 Matrices and Linear Maps
3 45. Let AM nn be such that A –23A +0I = . Show that A is invertible. 46. (PMI) Show that if A is invertible, then so is An for every positive integer n. 47. (PMI) Let A and B be invertible matrices of the same dimension with AB= BA . Sow that: (a)B–1 nA–1 = A–1B–1 n for every positive integer n. (b)AB –n = A–nB–n for every positive integer n.
PROVE OR GIVE A COUNTEREXAMPLE
48. If A is invertible, then so is –A , and –A –1 = –A–1 .
49. If A1A2 As is a linearly independent set in the vector space Mnn , and if AM nn is not the zero vector, then AA1AA2 AAs is linearly independent.
50. Let A be an nn invertible matrix, and BM nm . If AB = 0 , then B = 0 .
51. Let AM nn be invertible, and BM mn . If BA = 0 , then B = 0 52. If A and B are nn invertible matrices, then AB+ is also invertible, and AB+ –1 = A–1 + B–1 .
2 53. If AM nn and A = 0 , then A is not invertible. 54. If a square matrix A is singular, then A2 = 0 .
55. If A and B are nn matrices, and if AB is invertible, then both A and B are invertible.
56. If A and B are nn matrices, and if AB is singular, then both A and B are singular. 5.3 Matrix Representation of Linear Maps 177
5
§3. MATRIX REPRESENTATION OF LINEAR MAPS
The main objective of this section is to associate to a general linear transformation T: VW , where dimV = n and dimW = m , a matrix AM mn which can be used to find the value of Tv , for every v V . The importance of all of this is that, in a way, the linear transfor- mation T is “linked to a finite object:” an mn matrix A. The notion of an ordered basis for a vector space plays a role in the current development. The only difference between a basis and an ordered basis is that, in the latter, the listing of the vectors is of conse- quence. For example, while 12 35 and 35 12 are one-and-the-same bases, they are not the same ordered bases (different ordering of the elements). Choosing an ordered basis for a vector space V of dimension n enables us to associate vertical n-tuple to each vector of V:
DEFINITION 5.9 Let = v1v2 vn be an ordered COORDINATE basis for a vector space V, and let VECTOR v = c1v1 +++c2v2 cnvn We define the coordinate vector of v rela- tive to be the column vector:
c1 c2 v =
cn
EXAMPLE 5.7 In Example 3.10, page 94, we showed that = 130 204 012 is a basis for 3 . Find the coordinate vector of 1118 with respect to .
SOLUTION: As is often the case, the problem boils down to that of solving a system of linear equations: If: 1118 = a130 ++b204 c012 , then: a b c a b c 100 2 a = 2 a ++2b 0c = 1 120 1 2 augS rref 1 1 S: 010–--- b = –--- 1118 = –--1- 3a ++0bc= 11 30111 2 2 2 0a ++4b 2c = 8 042 8 001 5 c = 5 5 178 Chapter 5 Matrices and Linear Maps
CHECK YOUR UNDERSTANDING 5.13
–14 Find the coordinate vector of –112 3 with respect to: Answer: –38 74 = 130 204 012
Throughout this sec- tion the term “basis” THEOREM 5.19 If = v1v2 vn is a basis for a vector will be understood to n mean “ordered basis.” space V, then the function L: V given
We remind you that we are by Lv = v (a vertical n-tuple) is linear. n using to denote Mn 1 [Indeed, it is an isomorphism (Exercise 32).] n n
PROOF: For v = aivi and v = bivi in V, and r , we have: i = 1 i = 1 n n n L rvv+ L rv+ v ==L rav + b v L ra + b v i i i i i i i i = 1 i = 1 i = 1
ra1 + b1 ra1 b1 a1 b1 == + =r +
ran + bn ran bn an bn
n n = rL a v + L b v = rL v + L v = rLv + Lv i i i i i = 1 i = 1
Consider a linear transformation T: VW , with dimV = n and
dimW = m . If we fix a basis = v1v2 vn for V and a basis
(gamma) is the Greek = w1w2 wm for W, then for every v V we can determine letter c. the coefficient matrix v of v with respect to , as well as the coeffi-
cient matrix Tv of Tv with respect to . The following defini-
tion provides an important relation between v and Tv : 5.3 Matrix Representation of Linear Maps 179
DEFINITION 5.10 Let T: VW be a linear map from a vector Note the order of the two MATRIX REPRESEN- space V of dimension n to a vector space W subscripts in T . It kind TATION OF A LIN- of dimension m. Let = v1v2 vn of “looks backward,” EAR MAP since T “goes from to and = w1w2 wm be bases for V .” As you will soon see, and W, respectively. We define the matrix however, the chosen order is the more suitable representation of T with respect to and for the task at hand, that th of representing a linear to be the matrix T Mmn whose i map in matrix form. column is Tvi .
The above definition looks intimidating, but appearances can be mis- leading. Consider the following example.
EXAMPLE 5.8 3 Let T: P2 be the linear map given by: Tabc = 2ax2 ++ab+ xc
Determine the matrix representation T of T with respect to the bases: = 130 204 012 and = 2x2 + x 3x2 – 1 x
SOLUTION: Definition 5.10 tells us that the first column of T is the coefficient matrix of T130 with respect to [entries are the values of a, b, c stemming from (1) below], the second column is T204 [values of a, b, c stemming from (2)], and the third column is T012 [values stemming from (3)]: (1): T130 ==2x2 ++4x 0 a2x2 + x ++b3x2 – 1 cx (2): T204 ==4x2 ++2x 4 a2x2 + x ++b3x2 – 1 cx (3): T012 ==0x2 ++1x 2 a2x2 + x ++b3x2 – 1 cx Equating coefficients in each of the above we come to the following three systems of equations:
0x2 ++1x 0 3x2 + 0x – 1 2 2x ++1x 0 a b c a b c 2302 2304 2300 (1): 1014 (2): 1012 (3): 1011 01–00 01–04 01–02 180 Chapter 5 Matrices and Linear Maps
We want the solutions of the above systems of three equations in the three unknowns a, b, c. Noting that the coefficient matrix of all three systems are one and the same, we can “compress” all three systems into a single matrix form: system (3) solutions of system (3) system (2) solutions of system (2) system (1) solutions of system (1) a b c 2302 4 0 100183 rref 18 3 1014 2 1 01004– –2 T = 04– –2 T 01–00 4 2 00136– –2 36– –2 Figure 5.2
CHECK YOUR UNDERSTANDING 5.14
3 Let T: M22 be the linear map given by:
T ==ab bac + d cd Determine T for: Answer: 33 72 11– 26 12–14 3 4 = 10–238 1 8 22 11 51 59 0 1 13 4 54 and = 130 204 012 . To know the coordinates of a vector v in a finite dimensional vector space V with respect to a fixed ordered basis for that space is the same as knowing the vector v itself. This being the case, the following theo- rem tells us that the action of a linear transformation T: VW can, in effect, be realized by means of a matrix multiplication. THEOREM 5.20 Let T: VW be a linear map from a vector Note that the dimensions space V of dimension n to a vector space W of match up: dimension m. Let = v1v2 vn and Tv = T v = w1w2 wm be bases for V and W, M M N n mn m respectively. Then, for every v V : 1 1 Tv = T v m m PROOF: Noting that T v : V and Tv : V are Since both linear, we need only show that T vi = Tvi for each vi = 0v1 ++1vi ++ 0vn, 0 1 in (Theorem 4.6, page 115):
0 th v = th T v is the i column of T , namely: Tv (see i 1 i position i i margin and Exercise 10, page 160). By Definition 5.10, 0 th Tvi is also the i column of the matrix T . 5.3 Matrix Representation of Linear Maps 181
EXAMPLE 5.9 Let T: 2 3 be the linear map given by: Tab = ab+ b 2a
Determine T of T with respect to the bases: = 12 30 and = 100 110 111 and then show directly that:
T57 = T 57 SOLUTION: Proceeding as in Figure 5.2 of Example 5.8 we find
T . Since T12 ==322 and T30 306 :
T a b c 11133 1001 3 (*) 01120 rref 0100– 6 00126 0012 6 INCIDENTALLY, noting that the coefficient matrix The three vectors in Solving the two systems of equations: of system (*) is identical 322 a100 ++b110 c111 = to that of (**) we could 306 save a bit of time by doing this Finding T57 . Since T57 = 12710 :
T57 11112 100 5 rref (**) 011 7 010– 3 00110 00110 T Finding 57 : T57
57
7 10--- 135 rref2 2 207 01--1- 2 The two vectors in We leave it for you to verifying that 7 5 13 --- 2 T57 = T 57 ; that is: –3 = 06– . --1- 10 26 2 182 Chapter 5 Matrices and Linear Maps
CHECK YOUR UNDERSTANDING 5.15 Referring to the situation described in CYU 5.15, verify directly that:
13 13 T= T Answer: See page B-20. 20 20 The next result is particularly nice — it tells us that a matrix of a com- posite of linear transformations is a product of the matrices of those transformations:
THEOREM 5.21 Let T: VW and L: WZ be linear V T W L Z THE COMPOSITION maps, and let , and be bases for the LT THEOREM finite dimensional spaces V, W, and Z, respec- tively. Then: LT = L T LT = L T PROOF: For any v V we have: Theorem 5.20
LT v = LT v
==LTv L Tv =L T v The result now follows from Exercise 34 which asserts that if A is a matrix such that LT v = Av for every v V , then the matrices LT and A must be one and the same.
EXAMPLE 5.10 2 Let T: M22 and L: M22 P2 be the linear maps given by:
Tab = aa+ b 02b and ab L= ax2 ++bx c+ d cd Show directly that:
LT = L T For bases: = 12 04
11 02 00 00 = 00 10 11 01 = x2x2 + x x2 ++x 1 5.3 Matrix Representation of Linear Maps 183
SOLUTION: Determining T :
T04 = 04 08
T12 = 13 04
11 02 00 00 00 10 11 01
1 00 0 1 0 1 00 0 1 0 10 120034 010012 12 T = 011000 0010–2 1– –21 – 001148 0001510 510
Determining L : L L L 01 00 L 11 00 10 02 00 11 = = = 0 0 =1 x x 0 2 1 1 1 2 1 x ++ x x x x 2 ++ 2 2 2 2 0 ++ 0 ++ ++ ++ ++ x x 2 0 1 1 x x x x x 1 2 1 0 0 0 1
1 11 1 0 0 0 1 000–00 2 02–00 rref 0111200 01011–1 2– L = 11–1 2– 0010121 00101 2 1 01 2 1
Determining LT :
LT04 ===LT04 L04 0x2 ++4x 8 1 1 1 08 x x x 2 2 2 ++ ++ ++ 13 2
0 1 LT12 ===LT12 L x ++3x 4 1
x x
x 04 0 0 1
1 11 1 0 100–4 2– –42 – 01134 rref 010–4 1– LT = –41 – 00148 001 4 8 48 And it does all fit nicely together, as you can easily check:
LT L T
1 0 –42 – 02–00 1 2 –41 – = 11–1 2– –1 –2 48 01 2 1 5 10 184 Chapter 5 Matrices and Linear Maps
CHECK YOUR UNDERSTANDING 5.16
3 2 Let T: P2 and L: P2 be the linear maps given by: Tabc = bx2 ++ax c and Lax2 ++bx c = aa ++ b c Show directly that:
LT = L T for bases: = 111 110 100 = x2x 2 Answer: See page B-21. = 01 11
A proof of the following result is relegated to the exercises: We recall that I denotes n THEOREM 5.22 Let be a basis for a vector space V of the identity matrix of dimension n. Then: dimension n, and that IV denotes the identity map I = I from V to V. V n As might be expected: THEOREM 5.23 Let T: VW be an isomorphism. Let and be bases for V and W, respectively. Then: –1 –1 T = T
PROOF: We have:
V T W T – 1 V
–1 T TI= V –1 T T = IV –1 Theorems 5.21 and 22: T T = I –1 –1 T = T
CHECK YOUR UNDERSTANDING 5.17
Show that if the matrix representation of T: VW with respect to any chosen basis for V and for W is invertible, then the linear Answer: See page B-21. map T is itself invertible (an isomorphism). 5.3 Matrix Representation of Linear Maps 185
EXERCISES
Exercises 1-3. Let T: 2 2 be the linear operator given by Tab = ab+2 b . Find 23 and T23 for the given (ordered) basis . 1. = 10 01 2. = 01 10 3. = 12 –2 1
Exercises 4-5. Let T: 3 3 be the linear operator given by Tabc = ab+ bac– . Find 231 and T231 for the given basis . 4. = 100 010 001 5. = –101010 –211
2 2 Exercises 6-7. Let T: R P2 be the linear map given by T ab = ax – bx + ab– . Find 12 and T12 for the given bases and . 6. ==21 10 2x2 –x 2
7. ==31 – 12 2x2 + 12 xx 2 ++x 1
3 2 ad– Exercises 8-9. Let T: P3 M22 be the linear map given by Tax+++bx cx d = . bc 2 2 Find x ++x 1 and Tx++x 1 for the given bases and .
11 01 00 00 8. ==x32x2 x + 1 x – 1 11 11 11 01
11 01 00 00 9. ==x3 + x2 x2 + xx + 11 11 11 11 01
3 2 Exercises 10-11. Let T: be the linear map given by Tabc = ab+2 c . Find T with respect to the given bases and , and show directly that T121 = T 121 . 10. ==110 101 111 22 01 11. ==011 101 110 01 22
2 2 Exercises 12-13. Let T: P2 be the linear map given by T ab = ax – bx + ab– . Find T with respect to the given bases and , and show directly that T12 = T 12 . 12. ==22 01 xx2 x + 1 13. ==12 21 xx2 + x x2 ++x 1 186 Chapter 5 Matrices and Linear Maps
Exercises 14-15. Let T: P1 P2 be the linear map given by Tpx = xp x + p0 . Show directly that T2x + 1 = T 2x + 1 for the given bases and . 14. ==42 x 42 x 4x2 15. ==x + 12 x + 3 xx2 + x x2 ++x 1
3 3 3 Exercises 16-17. Let I3: be the identity map on . Find I3 with respect to the given bases and , and show directly that I3121 = I3 121 16. ==110 101 111 110 101 111 17. ==110 101 111 101 111 110
ab 12 ab Exercises 18-19. Let T: M22 M22 be the linear map given by T = . Find cd 01 cd
T for the given bases and , and show directly that T 12 = T 12 . 21– 21–
11 01 00 00 01 00 11 00 18. == 11 11 11 01 11 01 11 11
01 00 11 00 10 00 01 01 19. == 11 01 11 11 01 11 00 01
Exercises 20-22. Let T: 2 2 be the linear operator given by Tab = ab+2 b . Find T with respect to the given basis , and show directly that T13 = T 13 . 20. = 10 01 21. = 01 10 22. = 12 –2 1
23. (Calculus Dependent) Let T: P2 P3 be the linear map given by Tpx = xp x and let D: P3 P2 be the differentiation linear function: Dpx = px . Determine the given 2 2 3 matrices for the basis = 1xx of P2 , and the basis = 1xx x of P3 .
(a) T (b) D (c) DT
(d) TD (e) TDT (f) DTD
24. (Calculus Dependent) Let V be the subspace of F spanned by the three vectors 1, sinx , and cosx . Let D: VV be the differentiation operator. Determine D for = 1 sinxx cos , and show directly that D52+ sinx = D 52+ sinx .
25. (Calculus Dependent) Let D: P3 P3 be the differentiation operator. Determine D for 3 2 3 2 3 2 = 4x 3x 2x 1 , and show directly that D5x + 3x = D 5x + 3x . 5.3 Matrix Representation of Linear Maps 187
2 2 11 26. Find the linear function T: , if T = for = 12 20 . 02
2 2 11 27. Find the linear function T: if T = for = 12 20 and 02 = 11 12 . 10 2 28. Find the linear function T: P2 if T = 01 for = 12 20 and 11 = x2 2x + 12 . 1111 0111 29. Find the linear function T: M22 M22 if T = for 0011 0001
10 11 11 11 10 01 00 00 = and = . 00 00 10 11 00 00 10 01
2 3 ab– 30. Let T: M22 and L: M22 be the linear maps given by: Tab = –0a
ab and L= bac + d . Show directly that LT = L T for bases: cd
11 02 00 00 = 12 01 = = 111 110 100 00 10 11 01
2 3 3 31. Let T: and L: P2 be the linear maps given by: Tab = –a0 ab+
2 and Labc = bx – cx + a . Show directly that LT = L T for bases: = 12 01 = 111 110 100 = x2x +31 32. Prove that the linear function of Theorem 5.21 is an isomorphism. 33. Prove Theorem 5.22. 34. Let T: VW be a linear map from a vector space V of dimension n to a vector space W of dimension m. Let and be bases for V and W, respectively. Show that if AM mn is such that Av = Tv for every v V , then AT= .
2 Exercises 35-36. Prove that the function T : P1 given by T ab = ax + 2b is an isomor- –1 2 –1 phism. Find the linear map T : P1 . Determine T and T for the given bases –1 –1 and , and show directly that T = T . 35. ==22 01 xx + 1 36. ==12 21 x 2 188 Chapter 5 Matrices and Linear Maps
4 aa+ b Exercises 37-38. Prove that the function T : M22 given by T abcd = is an dc– d –1 4 –1 isomorphism. Find the linear map T : M22 R . Determine T and T for the –1 –1 given bases and , and show directly that T = T .
11 01 00 00 37. = 1100 0110 0011 1000 = 11 11 11 01
01 00 00 11 38. = 1111 1110 1100 1000 = 11 11 01 11
T 39. Let L: M32 M23 be given by LA= A (See Exercise 19, page 161). Let:
11 11 11 11 11 10 100 110 111 111 111 111 = 1111 11 10 00 00 , = . 000 000 000 100 110 111 11 10 00 00 00 00
(a) Determine L . –1 (b) Show that L is an isomorphism, and use Theorem 5.25 to find L .
40. Let T: VV be a linear operator. A nontrivial subspace S of V is said to be invariant under T if TS= Tv v S S . Assume that dimV = n and dimS = m . Show AB that there exists a basis for V such that T = , where 0 is the zero nm– m 0 C matrix.
41. Let T: VW be a linear function and let and be bases for the finite dimensional vec- tor spaces V and W, respectively. Let AT= . Show that:
(a)v KerT if and only if v nullA .
(b)w ImT if and only if w is in the column space of A.
42. Let V and W be vector spaces of dimensions n and m, respectively. Prove that the vector space LVW of Exercise 35, page 122, is isomorphic to Mmn . Suggestion: Let and be bases for V and W, respectively. Show that the function : LVW Mmn given by T = T is an isomorphism.
43. (PMI) Let V1V2 Vn be vector spaces and let i be a basis for Vi , 1 in . Let Ti: Vi Vi + 1 be a linear map, 1 in– 1 . Use the Principle of Mathematical Induction to show that T T T = T T T T . n – 1 n – 2 1 n1 n – 1 nn – 1 n – 2 n – 1n – 2 2 32 1 21 5.3 Matrix Representation of Linear Maps 189
PROVE OR GIVE A COUNTEREXAMPLE
44. Let T: VW be an isomorphism, and let = v1v2 vn be a basis for V. Then, for
every v V , v = Tv , where = Tv1 Tv2 Tvn .
45. Let T: VW be linear, and let and be bases for V and W, respectively. Let
rT: VW be defined by rT v = rTv . Then: rT = rT .
46. Let = v1v2 vn be a basis for V, and let = 2v12v2 2vn . If T: VV is
a linear operator on V, then T = 2T .
47. If Z: VW is the zero transformation from the n-dimensional vector space V to the m-
dimensional vector space W, then Z is the mn zero matrix for every pair of bases and for V and W, respectively.
48. Let IV : VV be the identity map on a space V of dimension n, and let and be
(ordered) basis for V. Then IV = In if and only if = .
49. Let T: 2 2 be given by Tab = a + 3b 2a + 2b . There exists a basis such that
T is a diagonal matrix (See Exercise 18, page 161).
50. Let T: 2 2 be given by Tab = a + 3b –2a + 2b . There exists a basis such
that T is a diagonal matrix (See Exercise 18, page 161).
n n n n n 51. For T: and T: and any basis for : TL+ = T + L . 190 Chapter 5 Matrices and Linear Maps
5
§4. CHANGE OF BASIS
In the previous section we observed that by choosing a basis in an n-dimensional vector space V one can associate to each vector its coor- dinate vector relative to (Definition 5.9, page 177). The following result tells us how the coordinate vector of v V changes when switching from one basis to another:
THEOREM 5.24 For and bases for the finite dimensional vector space V, and for v V we have:
v = IV v
where IV denotes the identity map from V to V.
PROOF: Consider the identity map IV: VV along with accompa- nying chosen bases and : V V v . IV .v
Applying Theorem 5.20, page 180 we have our result:
v ==IVv IV IVv =IV v
CHANGE OF BASE The above matrix IV is called the MATRIX change of base matrix from to .
EXAMPLE 5.11 Find the change-of-base matrix I for the P2 basis = x2x 1 , = 1 + xx + x2 x of P2 , and then verify directly that v = I v for v = 2x2 + 3 . P2
SOLUTION: Definition 5.10, page 179, tells us that the first column of I is the coefficient matrix of I x2 with respect to , the P2 P2 second column is I x , and the third column is I x ; P2 P2 bringing us to the following three vector equations:] I x2 ==x2 a1 + x ++bx+ x2 cx P2 I x ==xa1 + x ++bx+ x2 cx P2 I 1 ==1 a1 + x ++bx+ x2 cx P2 5.4 Change of Basis 191
2 Noting that we also have to find 2x + 3 we might as well throw in the fourth vector equation: 2x2 + 3 = a1 + x ++bx+ x2 cx Equating coefficients in each of the above four vector equations we come to the following four systems of equations in the three unknowns a, b, and c: xx 1 +1 +0 x
2 2 2 2 x I x ==x 1x ++0x 0 P2 = = = 2 0 I x ==x 0x ++1x 0
x P x
x 2 2 2 2 ++ ++ ++ I 1 ==10x2 ++0x 1
1 P 1
1 2 x x x 2x2 +23 = x2 ++0x 3 0 0 1 a b c 0 10 1 0 0 2 1 00 0 0 1 3 rref 1110100 0101002
1000013 001–11 1} –5
IP 2 2 2x + 3 At this point we have two of the three matrices in the equation 2x2 + 3 = I 2x2 + 3 P2 2 2 2 As for 2x + 3 , it is simply 0 , since = x x 1 . 3 We leave it for you to verify that:
3 001 2 2 = 100 0 5 –111 – 3
CHECK YOUR UNDERSTANDING 5.18 Let V = 2, = 12 21 , and = 03 21 – . Determine the change-of-base matrix I and verify directly 2
Answer: See page B-21. that 23 = I 2 23 .
EXAMPLE 5.12 Find the coordinates of the point P = 13 with respect to the coordinate axes obtained by rotating the standard axes by 45 in a counterclockwise direction. 192 Chapter 5 Matrices and Linear Maps
SOLUTION: We are to find the coordinate vector of the point 13 with respect to vectors of length 1 (unit vectors) in the direction of the x - and y-axis depicted in Figure 5.3. y P. ==13 x y y x 2 1 45 1 1 1 1 1 1 cos45 sin 45 = ------ ------–45cos45 sin = –------ ------. . 2 2 2 2 1 45 x 1
Figure 5.3
We begin by finding the change-of-base matrix I 2 , for 1 1 1 1 ------ ------ –------ ------ = 10 01 and = : 2 2 2 2 I 10 = 10 2 I 201 = 01
1 1 2 2 ------–10------10------2 2 rref 2 2 1 1 2 ------01 01–------2------2 2 2 2
Applying Theorem 5.20, page 180, we have:
------2------2- 2 2 1 22 13 ===I 13 2 3 –------2------2 2 2 Conclusion: In the x y -coordinate system of Figure 5.3:
P = 22 2 Check: 1 1 1 1 22------ ------+212–------ ------==–21 + 23 2 2 2 2
CHECK YOUR UNDERSTANDING 5.19 Find the coordinates x y of the point P = 13 with respect to Answer: 13– 31+ the coordinate axes obtained by rotating the standard axes by 60 in a clockwise direction. 5.4 Change of Basis 193
The adjacent identity map is pointing in two directions. The left-to-right direction gives rise to the change-of- V V base matrix I , while the right-to- v I V . V .v left directions brings us to IV . Are IV and IV related? Yes:
I In other words: V THEOREM 5.25 Let and be two bases for a finite I and V are invert- dimensional vector space V. Then: ible, with each being the –1 IV = IV inverse of the other. PROOF: A direct consequence of Theorem 5.23, page 184, with –1 IV: V V playing the role of T: V W (note that IV = IV ).
We now turn our attention to the matrix representations of a linear operator T: VV .
A generalization of this THEOREM 5.26 Let T: VV be a linear operator on a finite result appears in Exer- CHANGE OF BASIS dimensional space V. Let and be two cise 24. bases for V. Then: T = IV T IV
PROOF: Consider the following figure: V T V
IV IV In reading the composition T of functions, you kind of V V have to read from right to left: the right-most func- T = IV TIV tion being performed first. Top path
Dotted path: IV at left of figure, then T, then IV at right of figure Figure 5.4
Since the identity map IV does not move anything, we have:
T = IVTIV
and therefore: T = IVTIV Applying Theorem 5.21, page 182, to the above equation, we have:
top path in Figure 5.4 dotted path in Figure 5.4 } T = IV T IV this function first then this function and finally this function 194 Chapter 5 Matrices and Linear Maps
EXAMPLE 5.13 Verify, directly, that Theorem 5.26 holds for the linear operator T: 3 3 given by: Tabc = 2ab + c 0 and bases: = 111 110 100 = 101 010 001
SOLUTION: We determine the four matrices:
T T I3 I3
For T
T101 = 210 T010 = 010 T001 = 010
1 0 0 20 0 100 200 rref 010111 010111 101000 001–00 2
For T
T111 = 220 T110 = 210 T100 = 200
1 1 1 22 2 100 000 rref 1 1 0 2 10 010 210 1 0 0 00 0 001 012
For I3
I111 = 111 I 110 = 110 I100 = 100 1 0 0 11 1 100 111 rref 010 1 10 010 110 1 0 1 10 0 001 01–1– 5.4 Change of Basis 195
For I3
I101 = 101 I 010 = 010 I001 = 001 1 1 1 10 0 100 101 rref 110 0 10 010–11 1 – 1 0 0 10 1 001 11–0 As advertised:
111 00 0 101 200 I T I ===110 21 0 –111 – 111 T 01–1– 01 2 11–0 –002
CHECK YOUR UNDERSTANDING 5.20 Verify, directly, that Theorem 5.26 holds for the linear operator 2 2 T: P2 P2 given by Tax++bx c = bx –2ax + c , and bases: Answer: See page B-22. = x2x2 + x x2 ++x 1 = x2 + 1x2 –1x
In Exercise 21 you are asked DEFINITION 5.11 AB M are similar if there exists an to show that “similar” is an nn equivalence relation on SIMILAR MATRICES invertible matrix PM nn such that –1 Mnn . (See Exercises 37-39, BP= AP. page 147 for the definition of an equivalence relation). Theorem 5.26 tells us that if T: V V is a linear operator on a vector
space of dimension n, and if and are basis for V, thenT and
T are similar matrices. The following result kind of goes in the oppo- site direction:
THEOREM 5.27 Let = v1v2 vn be a basis for V, and let T be a linear operator on V. If A is similar to T , then there exists a basis for V, such that AT= .
PROOF: Since A is similar to T , there exist a matrix Pp= ij such –1 that AP= T P . In Exercise 22 you are asked to verify that
= v1v2vn , where vi = p1iv1 +++p2iv2 pnivn , is a basis for V. Applying Theorem 5.26 we have:
T = IV T IV . 196 Chapter 5 Matrices and Linear Maps
ith I By its very construction: I = P (see margin). Moreover: The column of V , V namely: –1 I v = v I = P (Theorem 5.23, page 184). Hence: V i i V th –1 equals the i column of P, T ===IV T IV P T P A since: vi = p1iv1 +++p2iv2 pnivi EXAMPLE 5.14 Let T: 2 2 be the linear map given by Tab = 3a +66b a .
(a) Find T for = 12 21 . –812 (b) Show that is similar to T . –1518 (c) Find a basis for 2 such that –812 T = –1518
T12 = 15 6 T21 = 12 12 T 1 2 15 12 10–4 1 SOLUTION: (a): rref 21612 0184
(b) We determine P = ab such that cd –1 –812 = ab –1 4ab –1518 cd 8 4 cd
ab –812 = –1 4ab cd –1518 8 4 cd The above leads us to a homogeneous system of four equations in four unknowns: –1812a – b = –4a + c –1811a –4b –0c +0d = 8a + 15b = –4b + d 8a ++16b 0c – 4d = 0 –1812c –8d = a + 4c –08a +0b –1816c – d = 8c +815d = b + 4d 0a –88b ++c 11d = 0
a b c d a b c d 102--9- –1811 –4–0 4 81604– 01– 1–-----11- rref 8 (*) –0168 –18– 0000 08–811 0000 5.4 Change of Basis 197
Matrix (*) has two free variables, telling us that there are infinitely –1 many ab for which –812 = ab –1 4ab . Letting cd –1518 cd 8 4 cd d = 0 and c = 1 , we arrive at a solution, namely: a ====–12 b c 1 d 0 . And so we have: –1 –812 = –12 –1 4–12 –1518 10 8 4 10
(c) Following the procedure spelled out in the proof of Theorem 5.27,
with P = –12 , and = 12 21 , we determine a basis 10
–812 = v1 v2 such that T = : –1518
v1 ==– 212 + 21 03 –
v2 ==112 + 021 12 –812 Let’s verify that T = for = 03 – 12 : –1518 T03 – = –18 0 T12 = 15 6 T
01 –1518 rref 10 –812 –23 06 01 –1518
CHECK YOUR UNDERSTANDING 5.21 Referring to Example 5.14, determine a basis, distinct from
–812 = 03 – 12 , for which T = . Answer: See page B-22. –1518 198 Chapter 5 Matrices and Linear Maps
EXERCISES
Exercises 1-7. Verify directly that v = IV v holds for the given vector space V, the vector v V , and the bases and : 1.V = 2 , v = 25 , = 10 01 , and = 12 –2 1 . 2.V = 2 , v = 31 – , = 10 01 , and = 12 –2 1 . 3.V = 3 , v = 23– 1, = 110 101 111 , and = 100 010 001 . 4.V = 3 , v = 30– 2, = 102 001 112 , and = 101 011 110 , 2 2 2 2 5.VP= 2 , v = 2x ++x 1 , = x xx+ 1 , and = xx+ x x ++x 1 . 2 2 2 6.VP= 2 , v = x + 1 , = 22 x 2x , and = 1x + 1 x + x .
20 11 11 11 10 7.VM= 22 , v = , = , and 11 11 10 00 00
00 00 01 11 = . 01 11 11 11
8. Find the coordinates of the point Pxy= in the xy-plane with respect to the coordinate axes obtained by rotating the standard axes in a counterclockwise direction. (See Example 5.12.)
Exercises 9-13. Verify directly that T = IV T IV holds for the given vector space V, the linear operator T, and the bases and : 9.V = 2 , T: VV given by Tab = –b a , = 10 01 , and = 12 –2 1 .
10.V = 3 , T: VV given by Tabc = –bac , = 010 101 111 , and = 120 02– 1 101 .
2 2 2 11.VP= 2 , T: VV given by Tax++bx c = cx + b , = x x +11 , and = 2x 1 + x2 .
ab –b c 12.VM= 22 , T: VV given by T= , cd da–
11 11 11 10 00 00 01 11 = , and = . 11 10 00 00 01 11 11 11 5.4 Change of Basis and Similar Matrices 199
13. (Calculus Dependent) VP= 3 , T: VV given by Tpx = px , = 1xx2 x3 , and = 1 x 2x2 3x3 .
14. Let T: 2 2 be the linear operator given by Tab = 2ab – . Find a basis for 2
–1 10 such that T = P T P , where = 10 01 and P = . 11 15. Let T: 3 3 be a linear operator. Find the basis for 3 such that –021 – –1 T = P T P , where: = 100 110 111 and P = 011 . 101
16. Let T: P2 P2 be a linear operator. Find the basis for P2 such that 110 –1 2 T = P T P , where = 1x + 1 x + x and P = 011 . 001
17. Show that 20 and 20 are similar. –12 – 11–
18. Show that 21 and 20 are not similar. –12 – 11–
19. Find all matrices that are similar to the identity matrix In . 20. Let T: 2 2 be the linear map given by Tab = ab+ b .
(a) Find T for = 12 21 . 20 8 (b) Show that is similar to T . –1738 –
2 20 8 (c) Find a basis for such that T = . –1738 –
21. Show that “similar” is an equivalence relation on Mnn . (See Exercises 37-39, page 147 for the definition of an equivalence relation).
22. Show that = v1v2vn in the proof of Theorem 5.27 is a basis for V.
n n 23. Let AB Mnn be similar. Show that there exists a linear operator T: and bases n and for such that AT= and BT= . 24. (A generalization of Theorem 5.26) Let T: VW be a linear map from the finite dimen- sional vector space V to the finite dimensional vector space W. Let and be bases for V,
and let and be bases for W. Prove that: T = IW T IV . 200 Chapter 5 Matrices and Linear Maps
Exercises 25-29. Referring to Exercise 24, show directly that T = IW T IV holds for the given linear transformation T: VW , the bases and for V, and the bases and for W.
25.V ==2W 3 , Tab = –baa+ b, = 10 01 , = 12 01 , = 110 –201 111 , and = 111 001 110
26.V ==3W 2 , Tabc = abc+ – , = 110 –201 111 , = 111 001 110 , = 10 01 , and = 12 01 .
3 2 27.V == W P2 , Tabc = bx + cx– a, = 110 –201 111 , = 111 001 110 , = 21 + x 2 – x2 , and = 1xx2 .
3 2 2 28.VP==2W , Tax++bx c = ab+ 0 –c , = 1xx , = 21 + x 2 – x2 , = 21 + x 2 – x2 , = 101 001 110 and = 110 –201 111 .
2 2 29.VP= 2 W = P1 , Tax++bx c = bx+ a+ c , = x xx+ 1 , = 21 + x 2 – x2 , = xx – 1 and = x 2x + 1 .
30. Let T: VW be linear. Let be bases for the n-dimensional space V, and let be bases for the m-dimensional space W. Prove that there exists an invertible matrix QM mm and an invertible matrix PM nn such that T = QTP . Suggestion: Consider Exercise 24. 31. Let T: VW and L: WZ be linear maps. Let be bases for V, be bases for W,
and be bases for Z. Show that LT = IZ L T IV .
PROVE OR GIVE A COUNTEREXAMPLE
32. Let T: VV be a linear operator, and let = v1v2 vn and be a bases for V. If
T = 2T , then = 2v12v2 2vn . 33. If A and B are similar matrices, then A2 and B2 are also similar.
34. If A and B are similar invertible matrices, then A–1 and B–1 are also similar. 35. If A and B are similar matrices, then at least one of them must be invertible. 36. If A and B are similar matrices, then so are their transpose. (See Exercise 19, page 161.) 37. If A and B are similar matrices, and if A is symmetric, then so is B. (See Exercises 20, page 161.) 38. If A and B are similar matrices, and if A is idempotent, then so is B. (See Exercises 22, page 162.) 39. If A and B are similar matrices, then TraceA = TraceB . (See Exercises 24, page 162.) Chapter Summary 201
CHAPTER SUMMARY
MULTIPLYING MATRICES You can perform the product Cmn = Amr Brn of two matrices, if the number of columns of A equals the number of rows of B, and you get cij of the product matrix C by running across the ith row of A and down the jth column of B: cij = ai1b1j ++++ai2b2j ai3b3j airbrj Properties Assuming that the matrix dimensions are such that the given operations are defined, we have: (i) AB+ C = AB+ AC (ii) AB+ CACBC= + (iii) ABC= AB C (iv) rAB==rA BArB A connection between matrix For AM mn and any positive integer z, the map multiplication and linear T :M M given by T X = AX is linear. transformations. A nz mz A In particular:
n m For AM mn the map TA: given by
TAX = AX is linear, where X is a vertical n-tuple and TAX is a vertical m-tuple.
NULL SPACE OF A MATRIX, n For AM mn , the null space of A is the subspace of R con- AND NULLITY sisting of the solutions of the homogeneous linear system of equations AX = 0 . It is denoted by nullA .
INVERTIBLE MATRIX A square matrix A is said to be invertible if there exists a matrix B (necessarily of the same dimension as A), such that: AB== BA I The matrix B is then said to be the inverse of A, and we write BA= –1 . If no such B exists, then A is said to be non-invert- ible, or singular. Need only “work” on one Let A be a square matrix. If B is a square matrix such that side either AB= I or BA= I, then A is invertible and A–1 = B . Uniqueness An invertible matrix has a unique inverse. 202 Chapter 5 Matrices and Linear Maps
Properties If A is invertible and r 0 , then rA is also invertible, and: A–1 –1 = A 1 rA –1 = ---A–1 r
–1 –1 –1 –1 A1A2 An = An An – 1 A1
ELEMENTARY MATRIX A matrix that is obtained by performing an elementary row operation of an identity matrix
Invertibility Every elementary matrix is invertible. Inverses by means of The matrix obtained by performing an elementary row opera- multiplication tion on a matrix A Mmn equals that matrix obtained by multiplying A on the left by the elementary matrix obtained by performing the same elementary row operation on the identity
matrix Im . Inverses by row-reduction If a sequence of elementary row operations reduces the invert- ible matrix A to I, then applying the same sequence of elemen- tary row operations on I will yield A–1 .
EQUIVALENCES OF Let AM nn . The following are equivalent: INVERTIBILITY (i) A is invertible.
(ii)AXB= has a unique solution for every B Mn 1 . (iii)AX = 0 has only the trivial solution. (iv)rrefA = I . (v) A is a product of elementary matrices.
COORDINATE c where v = c v +++c v c v VECTOR 1 1 1 2 2 n n v = and
cn = v1v2 vn is a basis for V
MATRIX Let T: VW be linear, = v1v2 vn be a basis for V, and REPRESENTATION OF and = w w w a basis for W. A LINEAR MAP 1 2 m The matrix representation of T with respect to and is that th matrix T Mmn whose i column is Tvi Chapter Summary 203
The matrix representa- Let T: VW be linear. If = v1v2 vn is a basis for V, and tion of a linear map T = w w w is a basis for W, then: describes the “action” 1 2 m of T. Tv = T v
The matrix of a com- Let T: VW and L: WZ be linear maps, and let and be position function is the bases for the finite dimensional spaces V, W, and Z, respectively. product of matrices of Then: those functions. LT = L T
Relating coordinate Let and be bases for V. Then: vectors with respect to different bases. v = I v
The matrix of the Let T: VW be an invertible transformation and let and be inverse of a transforma- bases for V and W, respectively. Then: tion is the inverse of the T –1 = T –1 matrix of that transfor- mation. Relating matrix repre- Let T: VV be a linear operator, and let and be two bases for sentations of a linear V. Then: operator with respect to different bases. T = IV T I
SIMILAR The matrices AB Mnn are similar if there exists an invertible MATRICES –1 matrix PM nn such that BP= AP . Similar matrices repre- Let = v1v2 vn be a basis for V, and let T be a linear opera- sent linear maps with tor on V. If A is similar to T , then there exists a basis for V, respect to different basis. such that AT= . 204 Chapter 5 Matrices and Linear Maps 6.1 Determinants 205
6 CHAPTER 6 DETERMINANTS AND EIGENVECTORS As you know, a linear operator T: VV has a matrix representation
T , which depends on the chosen basis for V. A main goal of this
chapter is to determine if there exists a basis for which T turns out to be a diagonal matrix. Determinants, as you will see, play an essential role in that endeavor. §1. DETERMINANTS
We define a function that assigns to each square matrix a (real) num- jthcolumn ber: 1
st out row DEFINITION 6.1 For a 22 matrix:
out DETERMINANT det ab = ad– bc cd For AM , with n 2 , let A denote the nn nn 1j n – 1 n – 1 matrix obtained by deleting the first row and jth column of the matrix A resulting in a (see margin). Then: asquare matrix n of dimension 1 + j n – 1 detA = –1 a1j detA1j j = 1
IN WORDS: Multiply each entry a1j in the first row of A by the determinant of the (smaller) matrix obtained by discarding the first row and the jth column of A, and then sum those n products with alternating signs (starting with a + sign). It’s not as bad as you may think:L
EXAMPLE 6.1 29– 3 Evaluate: det 32–4 57– 6
SOLUTION:
2 93– 2 9 –3 29–3 3 –2 4 3 –2 4 32–4 5 76– 5 7 –6 57–6
29– 3 –2 4 34 32– det 32–4= 2det – 9det – 3det 76– 56– 57 57– 6 ==22– –6 – 47 –3375293– 6 – 45 –217 – – 206 Chapter 6 Determinants and Eigenvectors
GRAPHING CALCULATOR GLIMPSE 6.1
Definition 6.1 defines the determinant of a matrix by an expansion process involving the first row of the given matrix. The next theorem, known as the Laplace Expansion Theorem, enables one to expand along any row or column of the matrix. A proof of this important result is offered at the end of the section.
THEOREM 6.1 For given AM nn , Aij will denote the n – 1 n – 1 submatrix of A obtained th th Note that the sign of the by deleting the i row and j column of A. ij+ We then have: –1 has an alternat- n ing checkerboard pattern EXPANDING ALONG i + j _ _ _ _ detA = –1 a detA + + + + THE th ROW ij ij _ _ _ _ i + + + + + _ + _ + _ + _ j = 1 _ _ _ _ + + + + + _ + _ + _ + _ and: _ _ _ _ + + + + EXPANDING ALONG n _ _ _ _ + + + + th _ _ _ _ THE j COLUMN i + j + + + + detA = –1 aij detAij i = 1
Note: detAij is called the minor of aij , and ij+ th Cij = –1 aij detAij is called the ij cofactor of A
EXAMPLE 6.2 Evaluate: 29– 3 det 32–4 57– 6 by expanding about its second row.
SOLUTION:
_ 2 93– 2 9 –3 29–3 as in Example 6.1 same +_ _+ + 3 –42 3 –2 4 32– 4 + _ + 5 76– 5 7 –6 57–6
29– 3 93– 23– 2 9 det 32– 4 = –det3 –2det –4det 76– 56– 5 7 57– 6 ==–22639– 6 – 73– –– – 53– –217 427 – 59 6.1 Determinants 207
CHECK YOUR UNDERSTANDING 6.1
29– 3 Evaluate det 32–4 expanding along: 57– 6 Answer: See page B-23. (a) The third row. (b) The second column.
While it was not so bad to calculate the 33 determinant of Example 6.1, the task gets increasingly more tedious as the dimension of the matrix increases. If we were only interested in calculating determinants An upper triangular matrix of matrices with numerical entries, then we could avoid the whole mess is a square matrix with zero entries below its main diag- entirely and simply use a calculator. But this will not always be the case. onal. For example: In any event, one can easily calculate the determinant of any upper triangular matrix (see margin): 2501– 2 017–0 2 The determinant of an upper diagonal matrix 00034 THEOREM 6.2 00091 equals the product of the entries along its diagonal. 00005 PROOF: By induction on the dimension, n, of Mnn . A lower triangular matrix is a square matrix with zero ab entries above its main diago- I. Claim holds for n = 2 : det ==ad– b 0 ad . nal. For example: 0 d 2000 II. Assume claim holds for nk= . 5100 –3402 III. We establish validity at nk= + 1 : 4025 Let Aa= ij Mk + 1 k + 1 . Since all entries in the first
column below its first entry a11 is zero, expanding about the
first column of A we have detA = a11detA11 (where A11 is the k by k upper triangular matrix obtained by removing the first row and first column from the matrix A. As such, by II: detA11 = a22a33 ak + 1 k + 1 . Consequently: detA ==a11detA11 a11a22a33 ak + 1 k + 1
COROLLARY For any n: detIn = 1 .
CHECK YOUR UNDERSTANDING 6.2
Prove that the determinant of a lower diagonal matrix equals the prod- Answer: See page B-23. uct of the entries along its diagonal. 208 Chapter 6 Determinants and Eigenvectors
ROW OPERATIONS AND DETERMINANTS Since it is easy to find the determinant of an upper triangular matrix, ad since any square matrix can be transformed into an upper triangular matrix by means of elementary row operations, it would be nice to have relations between the determinant of a matrix and that obtained by per- forming an elementary row operations on that matrix. Niceness is at hand:
THEOREM 6.3 (a) If two rows of AM nn are inter- changed, then the determinant of the resulting matrix is –detA . (b) If one row of A is multiplied by a con- stant c, then the determinant of the result- ing matrix is cdetA . (c) If a multiple of one row of A is added to another row of A, then the determinant of the resulting matrix is detA .
PROOF: (a) By on the dimension of the matrix A. For n = 2 :
det ab ==ad– bc and det cd cb– da cd ab negative of each other Assume the claim holds for matrices of dimension k 2 (the hypothesis).
Let Aa= ij be a matrix of dimension k + 1 , and let Bb= ij denote the matrix obtained by interchanging rows p and q of A. Let i be the index of a row other than p and q. Expanding about row i we have: k + 1 k + 1 ij+ ij+ detA =det –1 aij Aij and det B =det –1 bij Bij j = 1 j = 1 Since rows p and q were switched to go from A to B, row i of B still equals that of A, and therefore: bij = aij . Since Bij is the matrix Aij with two of its rows interchanged, and since those matrices are of dimension k, we have: detBij = –detAij (the hypothesis). Consequently: k + 1 k + 1 detB = –1 ij+ b detB = –1 ij+ a –detA ij ij ij ij j = 1 j = 1
k + 1 ij+ ==– –1 aijdetAij –detA j = 1 6.1 Determinants 209
(b) Let B denote the matrix obtained by multiplying row i of matrix A by c. Expanding both matrices about the ith row, we have: n n ij+ ij+ detA =det –1 aij Aij and detB =det –1 bij Bij j = 1 j = 1
Matrix A and B differ only Since bij = caij , and since Bij = Aij (margin), we have: in the ith row, and that row n n has been removed from ij+ ij+ both A and B to arrive at detB =det –1 bij Bij = –1 caij detAij the matrices A and B . ij ij j = 1 j = 1
n pull out that common factor, c: ij+ ==c –1 aijdetAij cdetA j = 1
(c) Let B be the matrix obtained by multiplying row r of A by c and adding it to row i. If ir= , then the result follows from (b). Assume ir . Expanding B about its ith row, we have: n ij+ detB =det –1 aij + carj Aij j = 1 n n ij+ ij+ = –1 aijdetAij + c –1 arjdetAij j = 1 j = 1 n ===detA + c –1 ij+ a detA detA + c detA rj ij 0 j = 1
n ij+ –1 arjdetAij is the determinant of a matrix with th th j = 1 two equal rows: the i and r row The result now follows from CYU 6.3 below
CHECK YOUR UNDERSTANDING 6.3
Answer: See page B-23. Show that if two rows of a matrix A are identical, then detA = 0 . 210 Chapter 6 Determinants and Eigenvectors
The following example illustrates how Theorem 6.3 can effectively be used to calculate determinants.
EXAMPLE 6.3 1268 Evaluate: det 1002 2135 –1111
SOLUTION: Theorem 6.3(a) Theorem 6.3(c) (applied three times)
1268 1002 100 2 det 1002 ==–det 1268 –det 026 6 2135 2135 013 1 –1111 –1111 011 3
Theorem 6.3(b) (see margin) a11 a1n 100 2 100 2 100 2 det ca ca i1 in 013 3 013 3 013 3 ==–det2 –det2 =2det =8 013 1 000 –2 00– 2 0 an1 ann 011 3 00– 2 0 000– 2 a a 11 1n Theorem 6.3(c) Theorem 6.3(a) Theorem 6.2 (applied two times) = c det ai1 ain
an1 ann CHECK YOUR UNDERSTANDING 6.4
Evaluate: 2 101 det 012 2 101 4 Answer: 15 411 3
To establish any of the following three elementary matrix results, you
need but substitute the identity matrix In for A in Theorem 6.3:
Note that det In = 1 THEOREM 6.4 (a) If EM nn is obtained by interchang- ing two rows of In , then detE = –1 .
(b) If EM is obtained by multiplying The restriction c 0 is nn imposed in (b) since we a row of In by c 0 , then detE = c . are concerned with ele- mentary row operations (see page 3). (c) If EM nn is obtained by adding a multiple of one row of In to another row, then detE = 1 . 6.1 Determinants 211
Soon, we will be in a position to show that the determinant of a prod- uct of any two nn matrices is equal to the product of their determi- nants. For now:
THEOREM 6.5 For any BM nn and any nn elementary matrix E: detEB = detE detB
PROOF: Let EM nn be an elementary matrix obtained by inter- changing two rows of In . By Theorem 5.13, page 169: (*) EB is the matrix obtained by interchanging the same two rows in the matrix B. Consequently: detEB ==–detdetB E detB (*) and Theorem 6.4(a) detE = –1 [Theorem 6.4(a)] As for the rest of the proof: CHECK YOUR UNDERSTANDING 6.5
(a) Establish Theorem 6.5, for the elementary matrix EM nn : (i) Obtained by multiplying a row of In by c 0 . (ii) Obtained by adding a multiple of one row of In to another row of In .
(b) Let BM nn and E1E2 Es Mnn be elementary matri- ces. Use the Principle of Mathematical Induction to show that: detEs E2E1B = detEs E2E1 detB Answer: See page B-24. = detEs detE2 detE1 detB
We now come to one of the most important results of this section: You can add this result to the list of equivalences for THEOREM 6.6 A matrix AM is invertible if and only invertibility appearing in nn Theorem 5.17, page 172: if detA 0 . (vi) detA 0 PROOF: Let E1E2 Es be a sequence of elementary matrices such that Es E2E1A = rrefA (Theorem 5.13, page 169). Appeal- ing to CYU 6.5(b), we have: detEs detE2 detE1 detA = det rrefA
By Theorem 6.4, detEs detE2 detE1 0 . Consequently: Ifrref AI= , then: det rrefA 0 if and only if detA 0 det rrefA = 10 If rrefA I , then its last margin: if and only if rrefA = I row consists entirely of zeros, and Theorem 5.17(iv), page 172: if and only if A is invertible det rrefA = 0 212 Chapter 6 Determinants and Eigenvectors
We are now in a position to establish another powerful result, Austin Cauchy, a prolific attributed to Cauchy: French mathematician (1789-1857). THEOREM 6.7 For AB Mnn : detAB = detA detB
PROOF: Case 1: A is invertible. By Theorem 5.17, page 172, A can be expressed as a product of elementary matrices:
AE= sE2E1
Then: detAB = detEsE2E1B CYU 6.5(b): ==detEsE2E1 detB detA detB
Case 2: A is not invertible. AB is not invertible; for: AB invertible CAB C = I ABC= IA invertible -- a contradiction. It follows, from Theorem 6.6, that both detAB and detA are zero, and we again have equality: detAB ==0 0 detB =detA detB
CHECK YOUR UNDERSTANDING 6.6 Prove that if A is invertible, then: 1 detA–1 = ------Answer: See page B-24. detA
For the brave at heart: PROOF OF THE LAPLACE EXPANSION THEOREM
The column-expansion part We use on n to show that the determinant of AM can be eval- of the theorem is relegated nn to the exercises. uated by expanding about any row of A: The claim is easily seen to hold when n = 2 :
a11 a12 det ==a11a22 – a12a21 –a21a12 + a11a22 a21 a22 Assume that the claim holds for nk= (the hypothesis).
Let AM k + 1 k + 1 . We show that for any 1 t k + 1 : k + 1 k + 1 This will show that the 1 + j ts+ expansion about any row –1 a1jdetA1j = –1 atsdetAts equals that of expanding j = 1 s = 1 } about the first row. } (*) (**) expanding about first row expanding about row t 6.1 Determinants 213
Working with (*), we employ the hypothesis and evaluate the deter- minant of each kk matrix A1j along its t – 1 row, which is row t of A (see Figure 6.1). In so doing, we need to keep in mind that just as st th each A1j has a row and a column removed from A (1 row and j column), so then will each submatrix in the expansion of detAij have two rows and two columns of A removed. column numbers for A } 1 2 j – 1 j j + 1 n
row numbers for a11 a12 a1j – 1 a1j a1j + 1 a1n
1 row numbers for 2 a21 a22 a2j – 1 a2j a2j + 1 a2n 1
t as1 as2 asj– 1 asj asj+ 1 asn t – 1 A A { a a a a a a } n1 n2 nj– 1 nj nj+ 1 nn n n – 1 ij 1 2 j – 1 j n – 1 }
column numbers for Aij Figure 6.1
Let A1tjs denote the submatrix of A with rows 1 and t, and columns j and s of A removed. Using the hypothesis we can obtain the determi- th nant of A1j by expanding about its t – 1 row (which is the t row of A), breaking that sum into two pieces the “before-j” piece, and the “after-j” piece we have: j – 1 k + 1 t – 1 + s t – 1 + s – 1 detA1j =1 –1 ats detA1tjs + – ats detA1tjs s = 1 sj= + 1 Bringing us to: k + 1 1 + j –1 a1jdetA1j j = 1
k + 1 j – 1 k + 1 1 + j t – 1 + s t – 1 + s – 1 = –1 a1j –1 ats detA1tjs + –1 ats detA1tjs j = 1 s = 1 sj= + 1
jts++ jts++– 1 =1 –1 a1jatsdetA1tjs + – a1jatsdetA1tjs (A) sj sj 1 + J + t – 1 + s 1 ++jt– 1 +s – 1 214 Chapter 6 Determinants and Eigenvectors
Turning to (**), we again appeal to the hypothesis, and expand about the first row to calculate each detAts : k + 1 ts+ –1 atsdetAts s = 1 k + 1 s – 1 k + 1 ts+ 1 + j 1 + j – 1 = –1 ats–1 a detA1tjs + –1 a detA1tjs 1j 1j s = 1 j = 1 js= + 1
= –1 tsj+++1 a a detA + –1 tsj++ a a detA ts 1j 1tjs ts 1j 1tjs (B) js js To complete the proof, we observe that the left summation in (A) is equal to the right summation in (B), and that the right summation in (A) equals the left in (B):
–1 jts++a a detA = –1 tsj++ a a detA 1j ts 1tjs ts 1j 1tjs sj js –1 jts++– 1a a detA = –1 tsj+++1 a a detA 1j ts 1tjs ts 1j 1tjs sj js
note that – 1 jts++– 1 = –1 tsj+++1 6.1 Determinants 215
EXERCISES
Exercises 1-8. Compute the determinant of the given matrix.
210 153 224 124 1.–211 2.–011 3.5711 4. –199 101 429 369 461
2101 0301 1304 6339 5.–2121 6.–2421 7.–2461 8. 33–36 1010 4040 0042 30312 0202 1235 1035 9036
ab c Exercises 9-14. Given that det def = 9 , find: ghi
gh i a b c a +2d b +2e c + 2f 9.det ab c 10.det 2d 2e 2f 11. det de f def –3g – 3h – 3i gh i
2 ab c ab c adg+– beh+– cfi+ – 12. det def 13. det ghi 14. det de f ghi def gh i
Exercises 15-18. Find all values of k for which the given matrix is invertible.
15. k 1 k –01 kk2 0 k 00 1 16. 2 k k 0 k –1 17. 0 k k 18. 0 k 1 0 241 k2 k 0 10k 0 011 k Exercises 19-22. Verify:
x –01 111 2 19. det 0 x –1 = ax ++bx c 20. det abc = ab– bc– ca– cba a2 b2 c2
111 1 +11a 21. det abc = ab– bc– ca– abc++ 22. det 11+1b = abc+++ bc ac bc a3 b3 c3 111+ c
23. While one can certainly find matrices AB Mnn such that AB BA , prove that one can not find matrices AB Mnn such that detAB detBA . 216 Chapter 6 Determinants and Eigenvectors
111 24. Show that the matrix abc is invertible if and only if the numbers a, b, and c, are all distinct. a2 b2 c2 25. Prove that if a matrix A contains a row (or column) consisting entirely of zeros, then detA = 0 .
26. If Dd= ij Mnn is a diagonal matrix and if Xi Mn 1 is the column matrix whose th i entry is 1 and all other entries are 0, then DXi = diiXi . T T 27. Let AM nn . Prove that detA = detA , where A denotes the transpose of A (see Exercise 19, page 161). n 28. Prove that if AM nn is skew-symmetric, then detA = –1 detA (see Exercise 21, page 162). What conclusion can you draw from this result?
29. For AM nn , let B be obtained from A by interchanging pairs of rows of A m times. Express detB as a function of m and detA . 30. Let A be similar to B (see Definition 5.11, page 195). Prove that: (a) detA = detB (b) detAcI– = detBcI– for every c . Suggestion: Consider Theorem 5.1, page 153.
x y 1 x y 1 31. Show that det 1 1 = 0 is an equation of the line passing through the points x1 y1 x2 y2 1 2 and x2 y2 in .
x y z 1 x1 y1 z1 1 32. Show that det = 0 is an equation of the plane passing through the points x2 y2 z2 1 x3 y3 z3 1 3 x1y1 z1 , x2y2 z2 , and x3y3 z3 in .
33. Show that the area of the triangle with vertices at x1 y1 , x2 y2 , and x3 y3 is given by
x1 y1 1 1 ---det x y 1 , where the sign ( ) is chosen to yield a positive number. 2 2 2 x3 y3 1 34. (Cramer’s Rule) If AX= B is a system of n equations in n unknowns, with A invertible, then the system has a unique solution x1x2 xn [Theorem 5.17(ii), page 172]. Cramer’s rule asserts that: detA detA detA x = ------1 x = ------2 x = ------n 1 detA 2 detA n detA th where Ai is the matrix obtained by replacing the i column of A with B. Use Cramer’s rule to solve the system of: (a) Example 1.3, page 9. (b) CYU 1.3, page 10. 6.1 Determinants 217
35. Prove the “column-expansion-part” of Theorem 6.3 (Laplace Expansion Theorem).
Exercises 36-39. Use the Principle of Mathematical Induction to show that: 36. For any m and Ai Mnn , detA1 A2 Am = detA1 detA2 detAm .
n 37. For any AM nn and cR : detcA = c detA .
m m 38. Prove that for AM nn and any positive integer m: detA = detA .
cI X 39. If AM nn is of the form A = , where I is the rr identity matrix, 0 is 0 Y the nr– r zero matrix, and X and Y are rnr – and nr– nr– matrices, respectively, then: detA = crdetY .
X Y 40. If AM nn is of the form A = , where X and Z are square matrices and 0 Z
0 is a zero matrix, then: detA = det X det Z .
PROVE OR GIVE A COUNTEREXAMPLE
41. For AM nn , if detA = 1 , then AI= .
42. For AM nn , if detA = 0 , then A is the zero matrix.
–1 43. For AB Mnn , if detAB = 1 , then both A and B are invertible and AB= .
44. For any AB Mnn , detAB+ = detA + detB .
45. For any AM nn , det–A = –detA .
46. If AM nn is nilpotent, then detA = 0 (see Exercise 23, page 162).
X Y 47. If AM 44 and XYZW M22 , and if A = , then: Z W detA = detX detW – detY detZ . 218 Chapter 6 Determinants and Eigenvectors
6
The German word eigen §2. EIGENSPACES translates to: characteristic. At one time, eigenvalues were called latent values, We begin by defining eigenvalues and eigenvectors for matrices: and it is for this reason that (lamba), the Greek letter DEFINITION 6.2 An eigenvalue of a matrix AM nn is a for “l” is used. EIGENVALUES AND scalar (which may be zero) for which , EIGENVECTORS there exists a nonzero vector X n such We remind you that we use (FOR MATRICES) that: n to denote M , and n 1 AX = X that v n is the vector Any such vector X is said to be an eigenvec- v n in “column form.” tor corresponding to the eigenvalue .
EXAMPLE 6.4 Show that 21 and –1 3 are eigenvectors of
the matrix A = 32 . 32–
32 2 8 2 SOLUTION: Since ==4 , 21 is an eigenvec- 32– 1 4 1 tor of A corresponding to the eigenvalue 4. By the same token, –13 is an eigenvector corresponding to the eigenvalue –3 :
32 –1 ==3 –3 –1 32– 3 –9 3
Since the set of eigenvectors corresponding to an eigenvalue of a matrix AM nn does not contain the zero vector, it cannot be a sub- space of n . If you throw in the zero vector, however, you do end up with a subspace: eigenvectors, along with the zero vector n-by-n identity matrix Recall that null(A) denotes the solution set of the XAX= X ==XAX– X = 0 XAX– InX = 0 homogeneous system of equations AX = 0. ==A – In X = 0 nullA – In a subspace of n (Theorem 5.4, page 159)
Bringing us to: DEFINITION 6.3 The eigenspace associated with an eigen- value of a matrix AM nn , denoted by E , is given by:
E = nullA – In 6.2 Eigenspaces 219
EXAMPLE 6.5 Find a basis for the eigenspace E4 of the
matrix A = 32 of Example 6.4. 32–
32 10 –21 SOLUTION: E4 ==null– 4 null 32– 01 36– –21 null is the solu- 36– –21 rref 1 –2 tion set of the homoge- From we see that E4 = 2rr rR neous system: 36– 00 –2x +0y = with basis 21 . 3x –06y = CHECK YOUR UNDERSTANDING 6.7
Find a basis for the eigenspace E–3 of the matrix A = 32 of Answer: 13 – 32– Example 6.4.
CHARACTERISTIC POLYNOMIALS At this point, there is a gap in our eigenvector development; namely: How does one go about finding the eigenvalues of a matrix? The answer hinges on the following objects:
DEFINITION 6.4 For AM nn , the n-degree polynomial CHARACTERISTIC detA – I is said to be the characteristic POLYNOMIAL n (FOR MATRICES) polynomial of A, and detA – In = 0 is said to be the characteristic equation of A. We are now in the position to state the main theorem of this section:
THEOREM 6.8 The eigenvalues of AM nn are the solutions of the characteristic equation detA – In = 0 .
PROOF: To say that is an eigenvalue of A is to say that there exists a nonzero vector X n such that: AX = X AX – X = 0
AX – In X = 0
A – In X = 0 220 Chapter 6 Determinants and Eigenvectors
But, to say that A – In X = 0 has a nontrivial solution is to say that detA – In = 0 (Theorem 6.6, page 211, and Theorem 5.17(iii), page 172).
EXAMPLE 6.6 Find the eigenvalues and corresponding eigens- paces of the matrix: 1 0 1 A = 2 2 1 1 0 1
SOLUTION: The eigenvalues are the solutions of the equation: 1 – 0 1 detA – I3 ==det 2 2– 1 0 1 0 1– Expanding about the third row, we have: A better choice is to expand 01 1 –1 1 – 0 about the second column. If 1 det – 0det + 1 – det = 0 you do, pay particular atten- 2 – 1 21 22– tion to the checkerboard sign pattern of page 206. –12 – + – 1 – 2 – = 0 2 – –11 + – 2 = 0 2 – –121 ++– 2 = 0 2 – – 2 – = 0 –02 – 2 = ==0 2 We now determine the eigenspaces for the two eigenvalues, 0 and 2. Finding E0 : 1 0 1 1 00 1 0 1 E0 ==null2 2 1– 0 0 10 null2 2 1 1 0 1 0 01 1 0 1 x y z x y z 1 0 1 1 0 1 rref 1 2 2 1 0 1 –--- 2 1 0 1 0 0 0 Setting the free variable z equal to r, we have: r E0 ==–r --- r r –2rr2r rR 2 with basis: –212 . 6.2 Eigenspaces 221
Finding E2 : 1 0 1 1 00 –1 0 1 E2 ==null2 2 1– 2 0 10 null2 0 1 1 0 1 0 01 1 0 – 1
x y z x y z –1 0 1 1 00 rref 2 0 1 001 1 0 – 1 000 Setting the free variable y equal to r, we have: E2 = 0r 0 r with basis 010 .
EXAMPLE 6.7 Find the eigenvalues and corresponding eigenspaces of the matrix: 02–0 2 A = 110– 1 –121 –1 –121 –1
SOLUTION: To find the eigenvalues of A we need to solve the charac- teristic equation: A TI-92 teaser: –22 –0 11–0 – 1 detA – I4 ==det 0 –121 – –1 –121 –1–
In this endeavor, we first use Theorem 6.3, page 208, to express
detA – I4 as the determinant of a matrix in upper triangular form, and then take advantage of Theorem 6.2, page 207, to finish the job: 222 Chapter 6 Determinants and Eigenvectors
switch the first two rows R1 + R2 R2 ;R1 + R3 R3 ;R1 + R4 R4
–22 –0 11–0 – 1 11–01 – 2 det 11–0 – 1==–detdet –22 –0– 0 2+2– – – –121 – –1 –121 – –1 02– – 2– 0 –121 –1– –121 –1– 02–2 – – switch rows 2 and 4 (introduces another minus sign) – 1R + R R 2 3 3 – 1 – R2 + R4 R4
11–01 – 11–01 – 11–0 – 1 02–2 – – 02–2 – – 02–2 – – ===det det det 02– – 2– 0 0 0 – 00– 0 2+2– 2 – – 0 2+2– 2 – – 0022
2R3 + R4 R4
11–0 – 1 02–2 – – ===det 1 2 – – 2 + 2 2 – 2 + 2 00– 2 000 + 2 Theorem 6.2, page 207
2 Setting detA – I4 = – 2 + 2 to 0, we see that there are three distinct eigenvalues: = 0 = 2 , and = –2 . As for their corresponding eigenspaces:
02–0 2 1000 02–0 2 E0 ==null110– 1– 0 0100 null110– 1 –121 –1 0010 –121 –1 –121 –1 0001 –121 –1 x y z w x y z w 02–0 2 1 01– 1 rref 110– 1 0 1 –01 –121 –1 0000 –121 –1 0000 Setting the free variables z and w equal to r and s respectively, we find that E0 = – r + srrs rs . By letting r ==1 s 0 , and then r ==0 s 1 we arrive at the basis –1110 1001 . 6.2 Eigenspaces 223
02–0 2 1000 –222 –0 E 2 ==null110– 1– 2 0100 null11–01– –121 –1 0010 –141 –1 –121 –1 0001 –121 –1–
x y z w x y z w –222 –0 1 –001 rref 11–01– 001 0 –141 –1 0001 –121 –1– 0000 Setting the free variable y equal to r, we have E2 = rr00 rR with basis 1100 .
02–0 2 1000 22–0 2 E –2 ==null110– 1+ 2 0100 null130– 1 –121 –1 0010 –1011 –121 –1 0001 –121 –3
x y z w x y z w 22–0 2 1 00– 1 130– 1 rref 0 1 00 –1011 001 –1 –121 –3 0000
Setting the free variable w equal to r, we find that E–2 = r0 rr r with basis 1011 .
CHECK YOUR UNDERSTANDING 6.8 Find the eigenvalues and corresponding eigenspaces of the matrix: 16 3 2 A = –384 – Answer: See page B-24. –62 –11 224 Chapter 6 Determinants and Eigenvectors
TURNING TO LINEAR OPERATORS Shifting our attention from matrices to linear operators we have: DEFINITION 6.5 Let T: VV be a linear operator. An eigen- Compare with Definition 6.3, EIGENVALUES AND value of T is a scalar (which may be page 218. EIGENVECTORS zero) for which there exists a nonzero vector (FOR LINEAR OPERATORS) v V such that: Tv = v Any such v is then said to be an eigenvector corresponding to the eigenvalue .
EXAMPLE 6.8 Show that 21 and –1 3 are eigenvectors of the linear operator T: 2 2 given by: Note that the linear map T stretches the eigenvector Tab = 3a +32b a – 2b 21 by its eigenvalue 4, SOLUTION: and –31 by –3 : (a) T21 ===6262+ – 84 421 by d” che ret f 4 “st r o 421 and T–13 ===– 3 + 63 – – 6 39 – –3–13 . acto –13 a f We see that 21 is an eigenvector corresponding to the eigenvalue 4, 21 and that –13 is an eigenvector corresponding to the eigenvalue –3 . Since the set of eigenvectors corresponding to an eigenvalue of a
“s linear operator T: VV does not contain the zero vector, it cannot be tre f tc ac h to ed r ” a subspace of V. As it was with matrices, however, if you throw in the of by -3 a zero vector, then you do end up with a subspace, for: –3–13 eigenvectors, along with the zero vector vTv= v = v Tv – v = 0
==v T – IV v = 0 KerT – IV
a subspace of V (Theorem 4.8, page 126) Bringing us to: Compare with Definition 6.4, DEFINITION 6.6 The eigenspace associated with an eigen- page 219. value of a linear operator T: VV , denoted by E , is given by:
E = KerT – IV
EXAMPLE 6.9 Find a basis for the eigenspace E4 of the lin- Compare with Example 6.5. ear operator: Tab = 3a +32b a – 2b of Example 6.8. 6.2 Eigenspaces 225
SOLUTION: The kernel of the linear operator:
T – 4I 2 ab ==Tab – 4ab 3a + 2b – 4a 3a – 2b – 4b = – a +32b a – 6b is, by definition, the set: ab – a +32b a – 6b = 00 Equating coefficients, we have: a b a b –2a +0b = –21 rref 1 –2 3a –06b = 36– 00 Setting the free variable b equal to r, we have E4 = 2rr r with basis 21 .
CHECK YOUR UNDERSTANDING 6.9 Find a basis for the eigenspace E–3 of the linear operator Answer: –31 Tab = 3a +32b a – 2b of Example 6.8. How does one go about finding the eigenvalues of a linear operator? Like this: THEOREM 6.9 The eigenvalues of a linear operator T: VV on a vector space of dimension n are the eigen- values of the matrix T Mnn , where is any basis for V.
PROOF: We show that v is an eigenvector for the linear operator T corresponding to the eigenvalue , if and only if v is an eigenvec-
tor for the matrix T corresponding to the eigenvalue : Tv = v with v 0
Note that Tv = v
v ==0 v 0 Theorem 5.22, page 180: T v = v with v 0 (Why?)
The above theorem leads us to the following definition: DEFINITION 6.7 Let T: VV be a linear operator on a vector Theorem 5.26, page 193, and Exercise 30(b), page CHARACTERISTIC space V of dimension n. The characteristic 216, tell us that POLYNOMIAL polynomial of T is the n-degree polynomial detT – I (FOR LINEAR OPERATORS) n detT – In where is any basis for V, = detT – In and detT – In = 0 is said to be the for any bases and characteristic equation of T. 226 Chapter 6 Determinants and Eigenvectors
Embedding the above terminology in the statement of Theorem 6.9, we come to: THEOREM 6.10 Let V be a vector space of dimension n. The Compare with Theorem 6.8 eigenvalues of the linear operator T: VV are the solutions of the characteristic equation detT – In = 0 , where is any basis for V.
EXAMPLE 6.10 Find the eigenvalues and corresponding eigenspaces of the linear operator T: P2 P2 given by: Tax2 ++bx c = ac+ x2 ++2a ++2bcxac+ 2 PROOF: With respect to the basis = x x 1 in P2 , we have:
Tx2 = 1x2 ++2x 1 1 01 Tx= 0x2 ++2x 0 T = 2 21 2 T1 = x ++x 1 1 01
Theorem 6.10 tells us that the eigenvalues are the solutions of the equation: 1 – 0 1 see Example 6.6 2 detT – I3 ===det 2 2– 1 –02 – 1 0 1– We now determine the eigenspaces associated with the two eigenval- ues, = 0 and = 2 . Finding E0 : E0 =kerkerT – 0I ==T ax2 ++bx c Tax2 ++bx c = 0 P2
ac+ x2 ++02a ++2bcxac+ = x2 ++0x 0 Equating coefficients brings us to the following homogeneous system of equations: a b c a b c 1 0 1 ac+0= 101 1 rref 0 1 –--- 2a ++2bc= 0 221 2 ac+0= 101 0 0 0 6.2 Eigenspaces 227
As shown in Example 6.6, –2rr2r r is the solution set of the above system of equations. Thus: E0 = – 2rx2 ++rx 2rr with basis – 2x2 ++x 2 Finding E2 : E2 = kerT – 2I = ax2 ++bx c T – 2I ax2 ++bx c = 0 P2 P2
ac+ x2 ++2a ++2bcxac+ – 2ax2 ++bx c = 0 ==– a + c x2 ++2ac+ xac– 0x2 ++0x 0 Equating coefficients: a b c a b c – a +0c = –1 0 1 100 rref 2ac+0= 2 0 1 001 ac–0= 1 0 – 1 000 As shown in Example 6.6, 0r 0 r is the solution set of the above system of equations. Thus: E2 = rx r with basis x
CHECK YOUR UNDERSTANDING 6.10 Find the eigenspaces of the linear operator of Example 6.10 using Answer: See page B-25. = x2 ++x 1 x + 11 instead of = x2x 1 . 228 Chapter 6 Determinants and Eigenvectors
EXERCISES
Exercises 1-14. Determine the eigenvalues and corresponding eigenspaces of the given matrix.
1.11 2.14 3. 71– 234 62 21– 62 4. 230 005
32– 1 1111 1000 2 1 000 5. 26– 2 6.1111 7. 0200 01– 000 –21 –3 1111 31–0 1 8. 0 0 401 1111 –0022 0 0 530 0 0 008 A factorization for the characteristic polynomial in the next six exercises can be obtained with the help of the following result: A zero of a polynomial px is a Since –1 is a zero px= x3 –67x – , number which when substituted for x – –1 = x + 1 must be a factor, and we the variable x yields zero. For have: example, –1 is a zero of the poly- x2 –6x – 3 nomial px= x3 –23x – , since x + 1 x –67x – 3 2 p–1 = 0 . One can show that: c x + x –7x2 –6x – is a zero of a polynomial if and only – x2 – x if xc– is a factor of the polyno- –66x – mial. –66x – The adjacent example illustrates So: x3 –67x – = x + 1 x2 –6x – how the above result can be used to factor certain polynomials. = x + 1 x + 2 x – 3
20 1 401 12– 2 9.A = 03 4 10.A = 232 11. A = –522 – 00 1 –021 –636 –
32– 2 –1101 42–2 2 12. A = –13 –3 13.A = 000– 1 14. A = 131– 1 120 0100 0020 011– 1 11–5 3 6.2 Eigenspaces 229
Exercises 15-31. Determine the eigenvalues and corresponding eigenspaces of the given linear operator. 15.T: given by Tx= –5x . 16.T: 2 2 where T10 = 20 and T01 = 11 . 17.T: 2 2 given by Tab = 8a – 6b 12a – 19b . 18.T: 3 3 given by Tabc = 0 ac+ 3bc– . 19.T: 3 3 given by Tabc = a –99b + ca –35b + c 2a –46b + c . 20.T: 4 4 , where Tabcd = adbc . 21.T: 4 4 , where Tabcd = 2aa – b 3a + 2cdab– – + c – 2d .
22.T: P1 P1 , where Tax+ b = ab+ xb– .
23.T: P1 P1 , where T1 = x and Tx= 1 . 2 24.T: P2 P2 given by Pax++bx c = –2bx – c . 2 2 25.T: P2 P2 given by Pax++bx c = – cx + bx– a . 2 2 26.T: P2 P2 , if Tx= 3x –42x + , Tx= 7x – 8 , and T1 = 1 . 3 2 3 2 27.T: P3 P3 , if Tax+++bx cx d = ad+ x +++– 2a – c + d x 2c – 2d xb– d .
ab –0c 28.T: M22 M22 given by T= . cd ad–
10 00 01 –29 – 00 00 29.T: M22 M22 , where T = , T = , T = , and 01 11– 00 00 10 –12
T 00 = 00. 01 –27
30.I: VV , where I is the identity map: Iv = v . 31.Z: VV , where Z is the zero map: Zv = 0 .
32. (Calculus Dependent) Let V be the vector space of differentiable functions, and let D: VV be the derivative operator. Show that v = e2x is an eigenvector for D. 33. Prove that a square matrix A is invertible if and only if 0 is not an eigenvalue of A. 34. Let A be an invertible matrix with eigenvalue 0 and corresponding eigenvector v. Prove 1 that --- is an eigenvalue of A–1 with corresponding eigenvector v.
35. Let 1 and 2 be distinct eigenvalues of AM nn . Prove that E1 E2 = 0 230 Chapter 6 Determinants and Eigenvectors
36. (a) Show that similar matrices have equal characteristic polynomials (see Definition 5.11, page 195). (b) Let T: VV be a linear operator on a finite dimensional space V. Show that if and are bases for V then: T and T have equal characteristic polynomials.
–1 37.Let AP Mnn , with P invertible. Prove that if v is an eigenvector of A, then P v is an eigenvector of P–1AP .
38. Let A = ab . Prove that a, and c are eigenvalues of A. 0 c
39. For A = ab , find necessary and sufficient conditions for A to have: cd (a) Two eigenvectors. (b) One eigenvector. (c) No eigenvector.
a11 a12 a13 40. Let A = 0 a22 a23 . Prove that a11 a22 , and a33 are eigenvalues of A. 00a33
41. (a) Let T: VV be a linear operator with eigenvalue . Prove that: E = vv is an eigenvector of T 0
(b) Let AM nn with eigenvalue . Prove that: E = vv is an eigenvector of A 0
42. For AM nn , show that nullA – I = kerTA – I . 43. Prove that 0 is an eigenvalue for a linear operator T: VV if and only if kerT 0 . 44. Show that if v is an eigenvector for the linear operator T: VV be a linear operator, then so is rv for any r 0 . 45. Let T: VW be an isomorphism. Show that v is an eigenvector in V if and only if Tv is an eigenvector in W. 46. Let v be an eigenvector for the linear operators T: VV and L: VV . Show that v is also an eigenvector for the linear operator LT: VV . Find a relation between the eigenvalues corresponding to v for T, L, and LT .
47. Show that if 1 and 2 are distinct eigenvalues of a linear operator T: VV , then E1 E2 = 0 .
48. Let v1 and v2 be eigenvectors corresponding to distinct eigenvalues 1 and 2 of a linear operator T: VV . Show that v1 v2 is a linearly independent set. 6.2 Eigenspaces 231
49. Let T: VV be a linear operator on a vector space V of dimension n, and let L: V n be an isomorphisms. Prove that is an eigenvalue of T if and only if is an eigenvalue of –1 n the matrix AL= TL S , where S is the standard basis of , and that E = L–1v v nullA – I . 50. Let T: VW be an isomorphism. Show that if v is an eigenvector of the linear operator L: VV , then Tv is an eigenvector of the linear operator TLT–1 : WW . 51. Let be a basis for a space V of dimension n, and L: VV a linear operator. Prove that if v V is an eigenvector of T with eigenvalue , then v is an eigenvector of T : Rn Rn with eigenvalue . L
52. Show that if is an eigenvalue of AM n n then is also an eigenvalue of the transpose AT . (See Exercise 19, page 162)
53. Show that if AM nn is nilpotent, then 0 is the only eigenvalue of A. (See Exercise 23, page 163.)
54. Show that the characteristic polynomial of AM 22 can be expressed in the form 2 –detTraceA + A , where Trace(A) denotes the trace of A (see Exercise 24, page 163).
55. Let AM 22 . Prove that the characteristic polynomial of A is of the form p = 2 ++b detA , and that A2 ++bA detA I = 0 . (This is the Cayley-Hamilton Theorem for square matrices of dimension 2.)
56. (PMI) Let AM nn . Use the Principle of Mathematical Induction to show that the coeffi- cient of the leading term of the characteristic polynomial of A is 1 .
57. (PMI) Let AM nn . Show that the constant term of the characteristic polynomial of A is detA .
58. (PMI) Let 12 k be the distinct eigenvalues of A for AM mm . Prove that n n n n 12 k are the distinct eigenvalues of A . 59. (PMI) Let A be a square matrix with eigenvalue and corresponding eigenvector v. Show that for any positive integer n, n is an eigenvalue of An with corresponding eigenvalue v.
60. (PMI) Let A be a square matrix with eigenvalue and corresponding eigenvector v. Show that for any integer n, n is an eigenvalue of An with corresponding eigenvalue v. 232 Chapter 6 Determinants and Eigenvectors
61. (PMI) Let be an eigenvalue for a linear operator T: VV . Use the Principle of Mathe- matical Induction to show that n is an eigenvalue for T n: VV , where Tn is defined 1 k + 1 k inductively as follows: T = T , and T = TT .
PROVE OR GIVE A COUNTEREXAMPLE
62. If is an eigenvalue for T: VV then it is also an eigenvalue for kT : VV , where kT v = kTv .
63. If is an eigenvalue for the two operators T: VV and L: VV , then it is also an eigenvalue for the operator TL+ : VV , where TL+ v = Tv + Lv .
64. For AB M22 , if A and B are eigenvalues of A and B, respectively, then A + B is
an eigenvalue of TAB+ .
65. For AB M22 , if A and B are eigenvalues for A and B, respectively, then AB is an eigenvalue for AB.
66. If is an eigenvalue of the linear operator T: VV , then 2 is an eigenvalue of TT : VV .
67. If T and L are eigenvalues for the linear operators T: VV and L: VV , respectively,
then AB is an eigenvalue for TL : VV .
68. If T and L are eigenvalues for the linear operators T: VV and L: VV , respectively,
then A + B is an eigenvalue for TL+ : VV .
69. If v is an eigenvector for T: VV and L: VV , then v is also an eigenvector for TL+ : VV .
70. If T: VV is a linear operator with eigenvector v, then v + rv is also an eigenvector of T for every r .
2 71. For AM 22 , A = 0 if and only if 0 is the only eigenvalue of TA .
72. Let T be a linear operator on a vector space V of dimension n. Let be an eigenvalue for T
and let v1v2 vm be a basis for E . Then, for any r , + r is an eigenvalue for
TrI+ n , and v1v2 vm is a basis for E + r . 6.3 Diagonalization 233
6
§3. DIAGONALIZATION
We begin with: DEFINITION 6.8 A square matrix Aa= ij nn for which a 0 0 11 DIAGONAL MATRIX aij = 0 if ij is said to be a diagonal 0 a22 0 matrix (see margin). 00 a DIAGONALIZABLE nn A linear operator T: VV on a finite OPERATOR dimensional vector space V is said to be diagonalizable if there exists a basis for
which T is a diagonal matrix. There is an intimate connection between diagonalizable matrices and eigenvectors, namely: THEOREM 6.11 Let T: VV be a linear operator on a finite dimensional vector space. Then: T is diagonalizable if and only if there exists a basis for V consisting of eigenvectors of T.
PROOF: Assume that T is diagonalizable. Let = v1v2 vn be such that T is a diagonal matrix: T = aij with aij = 0 th for ij . Since the i column of T consists of the coefficients of the vector Tvi with respect to the basis , we have: Tvi ==0v1 ++ 0vi – 1 +aiivi +0vi + 1 ++ 0vn aiivi
From the above we see that vi is an eigenvector for T corresponding to the eigenvalue aii .
Conversely, let = v1v2 vn be a basis for V consisting of eigenvectors, and let 12 n be the eigenvalues corresponding to v1v2 vn . From: Tvi ==ivi 0v1 +++++0v2 ivi 0vn we have (see Definition 5.10, page 179):
1 0 0 The i ‘s can be zero and need not be distinct (sev- 0 2 0 T = eral of the eigenvectors in may share a com- 00 n – 1 0 mon eigenvalue). 00n 234 Chapter 6 Determinants and Eigenvectors
CHECK YOUR UNDERSTANDING 6.11
Let T: 3 3 be the linear map given by: Tabc = 3ab– – c 2a – 2c 2ab– – c Show that = 120 122 211 is a basis for 3 con- sisting of eigenvectors of T. Determine T and show that its diag- Answer: See page B-25. onal elements are eigenvalues of T.
In our quest for bases consisting of eigenvectors, we note that:
THEOREM 6.12 If 12 m are distinct eigenvalues of a
linear operator T: VV , and if vi is any
eigenvector corresponding to i , for
1 im, then v1v2 vm is a linearly independent set.
PROOF: By induction on m:
If m = 1 , then v1 consists of a single nonzero vector and is there- fore linearly independent (Exercise 33, page 92). Assume the assertion holds for mk= (the induction hypothesis).
Let v1v2 vk + 1 be a set of eigenvector corresponding to distinct
eigenvalues 12 k + 1 . We are to show that v1v2 vk + 1 is a linearly independent set. With this in mind, we consider the linear combination:
a1v1 +++a2v2 akvk + ak + 1vk + 1 = 0 (*) Applying T to both sides, we have:
Ta1v1 +++a2v2 akvk + ak + 1vk + 1 = T0 Since vi is an eigenvector a Tv +++a Tv a Tv + a Tv = 0 corresponding to i : 1 1 2 2 k k k + 1 k + 1 a v +++a v a v + a v = 0 (**) Tvi = ivi 1 1 1 2 2 2 k k k k + 1 k + 1 k + 1
Multiply both sides of (*) by k + 1 :
a1k + 1v1 +++a2k + 1v2 akk + 1vk + ak + 1k + 1vk + 1 = 0 (***) Subtract (***) from (**):
a11 – k + 1 v1 +++a22 – k + 1 v2 akk – k + 1 vk = 0
By the induction hypothesis, the k eigenvectors v1v2 vk corre- sponding to the distinct eigenvalues 12 k are linearly inde- pendent. Consequently:
a11 – k + 1 ====a22 – k + 1 akk – k + 1 0 6.3 Diagonalization 235
Since the eigenvalues 12 k + 1 are distinct, none of the above i – k + 1 is equal to 0. Hence:
a1 ====a2 ak 0 Returning to (*), we then have:
ak + 1vk + 1 = 0
Being an eigenvector, vk + 1 0 , and therefore ak + 1 = 0 (Theorem 2.8, page 54). We have just observed that if you take eigenvectors corresponding to different eigenvalues you will end up with a linearly independent set of vectors. More can be said:
THEOREM 6.13 Let 12 m be distinct eigenvalues of a linear operator T: VV , and let S = v v v be any linearly inde- i i1 i2 iri pendent subset of Ei . Then:
SS= 1 S2 Sm is a linearly independent set.
PROOF: Consider the vector equation:
The r1 vectors in S1 The rm vectors in Sm linear combination = 0 a v ++ a v ++a v ++ a v = 0 (*) 11 11 1r1 1r1 m1 m1 mrm mrm (we will show that every coefficient must be zero) For 1 im , let v = a v ++ a v . Assume, without loss of i i1 i1 iri iri generality, that vi 0 for 1 it and that the rest are zero vectors. As for any of the zero vectors, v ==a v ++ a v 0 , its i i1 i1 iri iri coefficients must be zero, as v v v is given to be a i1 i2 iri linearly independent set. As for the nonzero vectors, we begin by rewriting (*) in the form: v1 ++v2 +vt = 0 (**)
Since the nonzero vectors v1v1 vt are eigenvectors associated
with distinct eigenvalues 12 t , they are linearly independent
(Theorem 6.12). It follows, from (**), that each vi must be 0: v ==a u ++ a u 0 i i1 i1 iri iri Using, again, the fact that each S = u u u is a linearly each coefficient is 0 i i1 i1 iri independent set, we conclude that each scalar aij must be zero CHECK YOUR UNDERSTANDING 6.12
Let T: VV be a linear operator on a space of dimension n. Prove Answer: See page B-26. that if T has n distinct eigenvalues, then T is diagonalizable. 236 Chapter 6 Determinants and Eigenvectors
RETURNING TO MATRICES
To say that T: VV is diagonalizable is to say that V contains a basis consisting of eigenvectors of T (Theorem 6.11). Let’s modify this characterization to accommodate matrices:
DEFINITION 6.9 A matrix AM nn is diagonalizable if DIAGONALIZABLE n MATRIX there exists a basis for consisting of eigenvectors of A. Here is a link between diagonalizable matrices and diagonalizable linear operators:
THEOREM 6.14 AM nn is diagonalizable if and only if the n n linear map TA: given by TAXAX= is diagonalizable. (X is a vertical n-tuple: a column matrix.)
n n PROOF: The linear map TA: is diagonalizable if and only if: n there exists a basis X1X2 Xn of , and scalars
12 n , such that TAXi = iXi if and only if:
AXi = iXi (Definition of TA ) if and only if A is diagonalizable (Definition 6.9). Definition 6.9 is okay, but just where do diagonal matrices come into play? Here:
This theorem asserts that THEOREM 6.15 Let AM nn be diagonalizable. Let any diagonalizable matrix n is similar to a diagonal = X1X2 Xn be any basis for matrix. The converse also consisting of eigenvectors of A, with associated holds (Exercise 37). And eigenvalues . If PM is the so we have: 1 2 n nn th AM nn is diagonal- matrix whose i column is Xi , then: izable if and only if it is DP= –1AP where Dd= is the diag- similar to a diagonal ij matrix. onal matrix, with diagonal entry dii = i .
PROOF: Let Sn = e1e2 en denote the standard basis for n (see page 94). Employing Theorem 5.26 (page 193), and The- n n orem 5.23 (page 184) to the linear map TA: given by TAX = AX, we have: 6.3 Diagonalization 237
T = P–1T P (*) A A SnSn th n T n where PI= (see margin). Since IX= X , the i col- A Sn i i umn of P is simply the vector Xi (recall that Sn is the standard IIbasis). Now: th n TA n The i column of T is: A SnSn Sn Sn T e ==Ae the ith column of A (See page 193) A i Sn i Sn Hence: T = A . A SnSn Since is a basis of eigenvectors:
TAXi == iXi 0X1 +++++0X2 iXi 0Xn
It follows that TA is the diagonal matrix Dd= ij with dii = i . Putting all of this together we have (see *): DP==–1AP, or: APDP–1
EXAMPLE 6.11 Show that the matrix: 02–0 2 A = 110– 1 –121 –1 –121 –1 is diagonalizable, and find a matrix P such that P–1AP is a diagonal matrix.
SOLUTION: In Example 6.7, page 221, we found that 0, 2, and –2 are the eigenvalues of A. We also observed that –1110 1001 is a basis for E0 , 1100 is a basis for E2 , and that 1011 is basis for E–2 . It is easy to see that the four eigen- vectors –10101101 1100 1011 are lin- early independent, and therefore constitute a basis for 4 . Taking P to be the matrix with columns the above four eigenvectors: –1111 P = 1010 1001 0101 we have: 238 Chapter 6 Determinants and Eigenvectors
02–0 2 000 0 P–1 110– 1P = 000 0 –121 –1 002 0 –121 –1 000– 2 Q CHECK YOUR UNDERSTANDING 6.13 Q–1AQ Determine if the given matrix is diagonalizable. If it is, use Theorem 6.15 to find a matrix P such that P–1AP is a diagonal matrix. –011 32– 1 (a) A = –301 (b) A = 26– 2 Answer: See page B-26. –1314 – –21 –3
The next result plays an important role in many eigenvector applica- tions:
THEOREM 6.16 If AM mm is diagonalizable with APDP= –1 , then An = PDnP–1 .
PROOF: (By induction on n) For n = 1 we have:
P–1AP= D1 APDP= –1
Assume that Ak = PDkP–1 (the induction hypothesis). Then:
P–1AP= D A = PDP–1
Ak + 1 ==AAk APDkP–1 =PDP–1 PDkP–1
induction hypothesis ==PDDk P–1 PDk + 1P–1 Answer: 0 0 0 0 CHECK YOUR UNDERSTANDING 6.14 0 0 0 0 0 01024 0 10 0 0 0 1024 Calculate A for the diagonalizable matrix A of Example 6.11.
ALGEBRAIC AND GEOMETRIC MULTIPLICITY OF EIGENVALUES
An eigenvalue 0 of a matrix AM nn (or of a linear operator T on a vector space of dimension n) has algebraic multiplicity k if k 0 – is a factor of A’s (or T ’s) characteristic polynomial, and k + 1 0 – is not. We also define the geometric multiplicity of 0 to be the dimension of E0 (the eigenspace corresponding to 0 ). 6.3 Diagonalization 239
EXAMPLE 6.12 Find the algebraic and geometric multiplicity of the eigenvalues of the matrix: 1 0 1 A = 2 2 1 1 0 1
SOLUTION: In Example 6.6, page 220, we showed that –2 – 2 is the characteristic polynomial of A. It follows that the eigenvalue 0 has algebraic multiplicity 1, and that the eigenvalue 2 has algebraic multiplicity 2. Since both of the eigenspaces E0 and E2 were seen to have dimension 1, the geometric multiplicity of both eigen- values is 1. The above example illustrates the fact that the geometric multiplicity of an eigenvalue can be less than its algebraic multiplicity; it cannot go the other way around:
THEOREM 6.17 If 0 is an eigenvalue of AM nn with algebraic multiplicity ma and geometrical multiplicity mg , then mg ma .
PROOF: (By contradiction) Assume that ma mg and let v v v be a basis for E . Expand v v v to 1 2 mg 0 1 2 mg a basis = v v v v v for n . Since, for 1 2 mg mg + 1 n Recall that T : Rn Rn is A 1 img : the linear operator given by: TAvi ==Avi 0vi TAv = Av = 0v1 ++ 0vi – 1 +0vi +0vi + 1 ++ 0vn
the matrix T is of the following form: Recall that the ith column A m of T consists of the g A coefficients of the vector 0 0 0 T v A i with respect to the 0 0 0 mg X basis = v v v . 1 2 n TA = 00 0 0 0 0 0 0 0 Y
0 0 0 nn In the proof of Theorem 6.15 we observed that T = A (where A SnSn n Sn is the standard basis in ). It follows, from Exercise 36(a), page 230, that the characteristic polynomial of A equals that of TA , namely, detTA – I : 240 Chapter 6 Determinants and Eigenvectors
mg cI X det = crdetY –0 0 0 Exercise 40, page 217 0 Y 0 – 0 (see margin) 0 m X
g m det 00 0 – = – gdetY 0 0 0 0 0 0 0 Y 0 0 0 nn
This leads to a contradiction, for the factor 0 – cannot appear with exponent greater than ma in the characteristic polynomial of A (remember that ma is the algebraic multiplicity of 0 ). In certain cases, the algebraic and geometric multiplicities of a linear operator can be used to determine if the operator is diagonalizable: THEOREM 6.18 Assume that the characteristic polynomial of a linear operator T: VV (or of a matrix), can be factored into a product of linear fac- tors (with real coefficients). Then T is diago- nalizable if and only if the algebraic multiplicity of each eigenvalue of T is equal to its geometric multiplicity.
PROOF: Assume that T is diagonalizable. By Theorem 6.11, there exists a basis = v1v2 vn for V consisting of eigenvectors of T. Let 12 k be the set of T’s (several of the vi ’s may cor- respond to the same eigenvalue). If necessary, reorder so that = S1 S2 Sk , where Sj consists of the eigenvectors in corresponding to j . Since the vectors in Sj are linearly independent, and since their com- bined sum equals the dimension of V, it follows, from Theorem 6.13, that the number of vectors in Sj must equal gj = dimE j , the geo- metric dimension of j . Hence: ng= 1 +++g2 gk . We are given that the characteristic polynomial of T can be factored into a product of linear factors:
a1 a2 ak (*) detT – I = – 1 – 2 – k It follows that: ng==1 +++g2 gk a1 +++a2 ak n
Theorem 6.17 degree of the characteristic polynomial (*) 6.3 Diagonalization 241
Consequently: a1 +++a2 ak = g1 +++g2 gk Or: a1 – g1 +++0a2 – g2 ak – gk =
Knowing that ai gi (Theorem 6.17), we conclude that ai = gi for each 1 ik.
Conversely, assume that the multiplicity of each eigenvalue j , 1 jk, is equal to its degree: aj ==dimEj gj . Let Sj be a basis for Ej . By Theorem 6.13, the set S1 S2 Sk , which contains g1 +++g2 gk vectors, is linearly independent. It is in fact a basis, since it contains n vectors (Theorem 3.11, page 99): degree of the characteristic polynomial g1 +++g2 gk ==a1 +++a2 ak n Possessing a basis of eigenvectors, T is diagonalizable (Theorem 6.11).
EXAMPLE 6.13 Appeal to the previous theorem to show that the linear operator T: 4 4 given by: Tabcd = 2b – 2ca + b– d– a ++b – 2c d – a ++b – 2c d
is diagonalizable. Find a basis for which T is a diag- onal matrix.
SOLUTION: For S the standard basis of 4 , we have: T1000 = 01– 1 – 1 02–0 2 110– 1 T SS = –121 –1 –121 –1 The above matrix was encountered in Example 6.7, page 221, where we found its characteristic polynomial to be 2 – 2 + 2 . We also showed that: The eigenspace corresponding to the eigenvalue 0, of multiplicity 2, has dimension 2 (with basis –1110 1001 ). The eigenspace corresponding to the eigenvalue 2, of multiplicity 1, has dimension 1 (with basis 1100 .) The eigenspace corresponding to the eigenvalue –2 , of multiplicity 1, has dimension 1 (with basis 1011 ). Since the algebraic multiplicity of each eigenvalue equals its geometric multiplicity, the linear operator is diagonalizable. Moreover, since = –1110 1001 1100 1011 242 Chapter 6 Determinants and Eigenvectors
is a basis for V of eigenvectors, we know that T will be a diagonal matrix with diagonal entries equal to the eigenvalues corresponding to the eigenvalues of ; namely: 000 0 000 0 T = 002 0 000–2
CHECK YOUR UNDERSTANDING 6.15 Verify that the algebraic multiplicity of each eigenvalue of the diago- nalizable matrix of CYU 6.13(b) equals that of its geometric multi- Answer: See page B-27. plicity. 6.3 Diagonalization 243
EXERCISES
Exercises 1-19. Determine if the given linear operator T: VV is diagonalizable. If it is, find a basis for V such that T is a diagonal matrix. (Note: To factor the characteristic polynomial of the given operator, you may need to use the division process discussed above Exercise 9 on page 228.)
1.T: 2 2 given by Tab = 2aa + 2b . 2.T: 2 2 given by Tab = 7ab– 6a + 2b . 3.T: 2 2 given by Tab = – 4a +8b – a + 2b . 4.T: 2 2 where T10 = 41 – and T01 = 12 . 5.T: 2 2 where T11 = 12 and T01 = 20 . 6.T: 3 3 given by Tabc = –a 2c 2b + 3c . 7.T: 3 3 given by Tabc = 13a – 4b 8b – 2c 5c . 8.T: 3 3 where T100 = 011 – – , T010 = 145 , and T001 = 112 – – . 9.T: 3 3 where T111 = 101 , T011 = 110 , and T002 = 110 . 10.T: 3 3 given by Tabc = 2a +34b – 4c – b + 5c – 6b + 8c . 11.T: 4 4 given by Tabcd = badc . 12.T: 4 4 given by Tabcd = –a0 a + 2c –2b . 13.T: 4 4 where T1000 = 1000 , T0100 = 32– 00 , T0010 = 2 – 140 , and T0001 = 605– 3 . 14.T: 4 4 where T1100 = 1001 , T0011 = 0010 , T1001 = 0111 , and T0001 = 1111 . 15.T: 5 5 given by Tabcde = 2aa – b 4c + 5d 3dc + 8e .
16.T: P2 P2 given by Tpx = px+ 1 .
2 2 17.T: P2 P2 given by Tax++bx c = 3a – 2b + c x ++2bc– xb .
2 2 2 18.T: P2 P2 where Tx= x , Tx= 2x + 1 and T1 = x – 1 .
ab ab– 19.T: M22 M22 given by T= . cd 3c 0 244 Chapter 6 Eigenvectors and Diagonalization
Exercises 20-34. Determine if the given matrix A is diagonalizable. If it is, find a matrix P such that P–1AP is a diagonal matrix.
20.A = 11 21.A = 16– 22. A = 24– 11 –21 33–
135 500 2 34 23.A = 02–6 24.A = –501 25. A = 2 30 004 232 0 05
203 000 12– 1 26.A = –031 27.A = 123 28. A = 24– 2 102 –231 – –21 –1
1234 5000 02–0 2 29.A = 02–3 1 30.A = 2200 31. A = 110– 1 0051 –9101 –121 –1 0001 3357 –121 –1
253– 6 312–4 2 21000 32. A = –4022 01442 01– 000 –104 –6–12 33.A = 00211 34. A = 00401 173– 5 00030 00530 00003 00008
2 35. Let AM nn be such that A = I . Show that: (a) If is an eigenvalue of A, then = 1 or = 0 . (b) A is diagonalizable.
36. Let AM nn be diagonalizable. Prove that the rank of A is equal to the number of nonzero eigenvalues of A.
37. Prove that if AM nn is similar to a diagonal matrix, then A is diagonalizable.
T 38. Let AM nn . Prove that A and its transpose A have the same eigenvalues, and that they occur with equal algebraic multiplicity (see Exercise 19, page 161).
39. Let AM nn . Prove that if is an eigenvalue of A with geometric multiplicity d, then is an eigenvalue of its transpose AT with geometric multiplicity d (see Exercise 19, page 161). 40. Let L: VW be an isomorphism on a finite dimensional vector space. Prove that: (a) The linear operator T: VV and LTL–1: WW have equal characteristic polyno- mials. (b) The eigenspace corresponding to an eigenvalue of LTL–1 is isomorphic to the eigenspace corresponding to that eigenvalue of T. (c) T is diagonalizable if and only if LTL–1 is diagonalizable. 6.3 Diagonalization 245
PROVE OR GIVE A COUNTEREXAMPLE
41. Let T: VV be a linear operator on a space of dimension n. If 12 m are distinct eigenvalues of T, and if there exists a basis for V such that T is a diagonal matrix, then mn= .
42. Let 12 k be the distinct eigenvalues of a linear operator T: VV on a vector space V of dimension n. The operator T is diagonalizable if and only if kn= .
43. If AB Mnn are both diagonalizable, then so is AB .
44. If AB Mnn are such that AB is diagonalizable, then both A and B are diagonalizable. 246 Chapter 6 Determinants and Eigenvectors
6
§4. APPLICATIONS
Applications of eigenvectors surface in numerous fields. In this sec- tion we focus on recurrence relations, and on differential equations.
FIBONACCI NUMBERS AND BEYOND
The Fibonacci sequence is that sequence whose first two terms are 1, Leonardo Fibonacci (Ital- and with each term after the second being obtained by summing its ian; circa 1170 - 1250), is two immediate predecessors: considered by many to be 11+ the best mathematician of 813+ the Middle Ages. The 1123581321 34 55 sequence bearing his name 12+ 34+ 55 evolved from the follow- ing question he posed and What is the 100th Fibonacci number? Given enough time, we could resolve in 1220: keep generating the numbers of the sequence, eventually arriving at its Assume that pairs of rab- th bits do no produce off- 100 element. There is a better way: spring during their first th Letting sk denote the k Fibonacci number we have s1 ==s2 1 , month of life, but will produce a new pair of and, for k 3 : offspring each month sk = sk – 1 + sk – 2 thereafter. Assuming that We observe that s is the top entry of the matrix product: no rabbit dies, how many k pairs of rabbits will there s s + s s be after k months? 11 k – 1 ==k – 1 k – 2 k (*) 10 sk – 2 sk – 1 sk – 1
11 sk Letting F = and Sk = we can express (*) in the form: 10 sk – 1
Sk = FSk – 1 In particular 1 1 2 1 k – 2 1 S3 ==FS2 F S4 ==FS3 FF =F Sk =F 1 1 1 1 th Note that the k Fibonacci number sk in (*) is the top entry in the
sk k – 2 1 matrix Sk ==F , which is simply the sum of the s 1 ab 1 ab+ k – 1 = k – 2 cd 1 cd+ entries in the first row of F (see margin). But this is of little benefit
unless we can readily find the powers of the matrix F = 11 . We can: 10 6.4 Applications 247
Since the characteristic polynomial of F is 2 –1 – (margin), the 11 10 15– 15+ det– = 0 matrix F has eigenvalues 1 = ------and 2 = ------. Let’s find 10 01 2 2 eigenvectors associated with those eigenvectors: 1 –1 det = 0 15+ 15– 1 – ------a ------ab+0= 15+ 2 11 a = ------a ab+ = 2 2 –1 – = 0 2 10 b b a 15+ 15+ 114 + ------b a–------b = 0 = ------2 2 2 homogeneous system of equations a b a b solution 15– ------1 rref 15+ 15+ 2 1 –------ a = ------b 2 2 1 –------15+ 00 2
15+ Setting the free variable b to 1 we find that a = ------. It follows that 2 a ------15+ = 2 is an eigenvector associated with the eigenvalue 1 . b 1
------15– ------15– -1+ In the same manner one can show that is an eigenvector 11 15------– - 2 2 2 = 10 1 1 ------15– - 2 associated with the eigenvalue 2 — a fact that is verified in the margin. Theorem 6.15, page 236, tell us that: ------35– - = 2 a diagonal matrix 15– ------–1 15+ 2 15+ 15– 15+ 15– ------0 ------11 ------2 D ==2 2 2 2 15– 15– 15– = ------11 10 11 0 ------2 2 2 1 P–1FP Leading us to:
15+ –1 15+ 15– ------0 15+ 15– –1 ------2 ------F ==PDP 2 2 2 2 15– 11 0 ------11 2 Applying Theorem 6.16, page 238, we have:
k 15+ –1 15+ 15– ------0 15+ 15– k k –1 ------2 ------F ==PD P 2 2 2 2 15– k 11 0 ------11 2 248 Chapter 6 Determinants and Eigenvectors
You are invited to show that the first row of the above matrix product can be expressed in the following form:
15+ k + 1 15– k + 1 15+ k 15– k ------–------------– ------1 Fk = ------2 2 2 2 5 ********************** ****************** Recalling that the kth Fibonacci number is the sum of the entries in the first row of Fk – 2 , we have: k – 1 k – 1 k – 2 k – 2 k – 1 k – 2 15+ 15+ 1 15+ 15– 15+ 15– ------– ------sk = ------– ------+ ------– ------2 2 5 2 2 2 2 15+ k – 2 15+ = ------------– 1 2 2 1 15+ k – 1 15– k – 2 15– k – 1 15– k – 1 = ------------– ------– ------– ------15+ k – 2 15+ 2 = ------------5 2 2 2 2 2 2 k k 15+ k – 2 = ------1 15+ 15– 2 = ------– ------(see margin) 5 2 2 Looking at the above “5 -expression” from a strictly algebraic point of view, one would not expect to find that each sk is an integer. Being a Fibonacci number, that must be the case. In particular, the 100th Fibo- nacci number is: 100 100 1 15+ 15– s100 ==------– ------354,224,848,179,261,915,075 5 2 2
Using the TI-92: 15+ The number = ------has an interesting history dating back to the 2 time of Pythagoras (c. 500 B.C.). It is called the golden ratio ( is the first letter in the Greek spelling of Phydias, a sculptor who used the golden ratio in his work). Basically, and for whatever aesthetic rea- son, it is generally maintained that the most “visually appealing” partition of a line seg- L l ment into two pieces is that for which the ratio of the length of the longer piece L to the length of the sorter piece l equals the ratio of the entire line segment to that of the longer piece, leading us to: L --L- = Ll------+ - l L L2 – lL –0l2 = l ll+ 5 L 15+ L ==------ ------2 l 2 6.4 Applications 249
th The formula sk = sk – 1 + sk – 2 for the k element of the Fibonacci RECURSIVE RELATION sequence describes each element of the sequence in terms of previous elements. It is an example of a recurrence relation. You are invited to consider addition recurrence relations in the exercises, and in the fol- lowing Check Your Understanding box.
CHECK YOUR UNDERSTANDING 6.16 Answer: th 5 1 Find a formula for the k term of the sequence s1s2 s3 , if s = --- 2k – 1 + ---–1 k – 1 k 3 3 s1 = 2 , s2 = 3 , and sk = sk – 1 + 2sk – 2 for k 3 .
SYSTEMS OF DIFFERENTIAL EQUATIONS (CALCULUS DEPENDENT)
We begin by extending the concept of a matrix to allow for function entries; as with: 2x Ax==3xe and Bx 0 lnx 4 sinx x –25 x The arithmetic of such matrices mimics that of numerical matrices. For example:
2x 2x 3xe+ 0 lnx = 3xe+ lnx 4 sinx x –25 x x – 1 sinx + 2x and:
2x 2x 2x 3xe 0 lnx = e x – 5 3xxln+ 2xe 4 sinx x –25 x sinxx– 5 4lnx + 2xxsin We also define the derivative of a function-matrix to be that matrix obtained by differentiating each of its entry. For example:
2x 2x 2x If Ax= 3xe then Ax ==3x e x 2e 4 sinx 4 sinx 0 cosx In the exercises, you are invited to show that the following familiar derivative properties: fx+ gx = f x + gx fxgx = fxgx + gxf x cf x = cfx extend to matrices: 250 Chapter 6 Determinants and Eigenvectors
THEOREM 6.19 Let the entries of the matrices Ax and Bx be differentiable function, and let C be a matrix with scalar entries (real numbers). Then: (i) Ax+ Bx = A x + Bx (ii) AxBx = AxBx + BxA x (iii) CA x = CAx (assuming, of course, that the matrix dimen- sions are such that the operations are defined) Differential equations of the form: f x = af x or y = ay play an important role in the development of this subsection. As you may recall:
THEOREM 6.20 The solution set of f x = af x , consists of those functions of the form fx= ceax for some constant c.
PROOF: If fx= ceax , then: f x ==ceax aceax =af x At this point, we know that every function of the form yce= ax is a solution of the differential equation f x = af x . Moreover, if fx is any solution of f x = af x , then the derivative of the function fx gx= ------ is zero: eax since f x = af x fx gxf x – fxgx ------= ------gx 2 eaxf x – fxaeax f x – af x eax 0 gx gx ====------0 e2ax e2ax eax fx If the derivative of a func- It follows that gx==------c for some constant c, or that eax tion is zero, then the func- fx= ceax . tion must be constant.
We now turn our attention to systems of linear differential equation of the form: f1x = a11f1x +++a12 f2x a1n fnx f2x = a21f1x +++a22f2x a2n fnx fnx = an1f1x +++an2f2x ann fnx where the coefficients aij are real numbers. As it is with systems of lin- ear equations, the above system can be expressed in the form: Fx = AF x were Fx= fix Mn 1 and Aa= ij Mnn . 6.4 Applications 251
In the event that A is a diagonal matrix, the system Fx = AF x is easily solved: THEOREM 6.21 f x 0 0 f x c e1x f1x = 1f1x 1 1 1 f1x 1 f x 0 0 f x f x = f x 2 2 2 f x 2x 2 2 2 = 2 c e If then: = 2 fnx 0 0 n fnx f x fnx = nfnx n nx cne where c1c2 cn . PROOF: Simply apply Theorem 6.20 to each of the n differential
equations: fix = ifix . We now consider systems of differential equations of the form: Fx = AF x (*) where A is a diagonalizable matrix. In accordance with Theorem 6.15, page 236, we know that for any chosen basis = v1v2 vn of eigenvectors of A: APDP= –1 th where the i column of PM nn is the eigenvector vi a1ia2i ani with eigenvalue i , and Dd= ij is the diago- nal matrix with dii = i . Substituting in (*), we have: Fx = PDP–1 Fx –1 Multiply both sides by P : P–1Fx = DP–1Fx Theorem 6.19(iii): P–1Fx = DP–1Fx Letting Gx= P–1Fx , brings us to: Gx = DGx Applying Theorem 6.21, we have:
1x 1x 1x c1e c1e c1e x x x 2 –1 2 2 Gx===c2e P Fx c2e Fx P c2e
nx nx nx cne cne cne At this point we have: v1 v2 vn
a a a 1x 11 12 1n c1e
a a a 2x Fx= 21 22 2n c2e a a a nx n1 n2 nn cne Appealing to Theorem 5.3, page 154, we conclude that: 1x 2x nx Fx= c1e v1 +++c2e v2 cne vn 252 Chapter 6 Determinants and Eigenvectors
Summarizing, we have:
THEOREM 6.22 Let AM nn be diagonalizable, and let n v1v2 vn be a basis for consisting entirely of eigenvectors of A with corre- sponding eigenvalues i . Then, the general solution of: Fx = AF x is of the form: 1x 2x nx c1e v1 +++c2e v2 cne vn
for c1c2 cn .
In alternate notation form: EXAMPLE 6.14 Find the general solution for: f x = – 3 f x + f x 1 1 2 y1 = –3y1 + y2 f2 x = 6 f1x + 2f2x y2 = 6y1 + 2y2
SOLUTION: Our first order of business is to find (if possible) a basis
2 –13 v1 v2 of consisting of eigenvectors of A = . 62
– 3 –1 From:det = – 3 – 2 – – 6 62– ==2 + – 12 + 4 – 3
we see that the matrix A = –13 is diagonalizable, with eigenvalues 62 –4 and 3. Here are their corresponding eigenspaces: –13 –04 11 E–4 ==null– null 62 04– 66 x y x y homogeneous 11 rref 1 1 system of equations: 66 00 Setting the free variable b equal to r, we have: E–4 = –rr rR And: –13 30 –16 E3 ==null– null 62 03 61–
a b a b 1 –16 1 –--- rref 6 61– 00 Bringing us to: E3 = r 6r rR 6.4 Applications 253
Choosing the eigenvector v = –11 for the eigenvalue –4 and Any other two eigenvec- 1 tors corresponding to the the eigenvector v2 = 16 for the eigenvalue 3, we obtain a basis of two eigenvalues will do 2 consisting of eigenvectors for A. Applying Theorem 6.22, we just as well. conclude that the general solution of the given system of differential equations is give by: –4x 3x y1 –4x –1 3x 1 –c1e + c2e ==c1e + c2e –4x 3x y2 1 6 c1e + 6c2e Which is to say: –4x 3x –4x 3x y1 ==–c1e + c2e and y2 c1e + 6c2e
y1 = –3y1 + y2 Let’s check our result in the given system : y2 = 6y1 + 2y2
–4x 3x –4x 3x y1 ==–c1e + c2e 4c1e + 3c2e and: –4x 3x –4x 3x –4x 3x – 3y1 + y2 ==– 3–c1e + c2e + c1e + 6c2e 4c1e + 3c2e
–4x 3x –4x 3x Similarly: y2 ==c1e + 6c2e –4c1e + 18c2e –4x 3x –4x 3x ==6–c1e + c2e + 2c1e + 6c2e 6y1 + 2y2
CHECK YOUR UNDERSTANDING 6.17 Find the general solution for:
y1 = 2y2 – 2y3 y2 = y1 + y2 – y4 y3 = – y1 ++y2 – 2y3 y4 y4 = – y1 ++y2 – 2y3 y4 Answer: See page B-28. Suggestion: Consider Example 6.11, page 237. Let us return momentarily to the system of equations of Example 6.14: –4x 3x y1 = –3y1 + y2 y1 = –c1e + c2e with general solution: (*) y = 6y + 2y –4x 3x 2 1 2 y2 = c1e + 6c2e To arrive a particular or specific solution for the system, we need some additional information. Suppose, for example, that we are given the ini-
y 0 tial condition Y0 ==1 –2 . Substituting in (*), we then y20 3 have: 254 Chapter 6 Determinants and Eigenvectors
–4 0 15 3 0 –2 = –c + c c1 = ------–2 = – c1e + c2e 1 2 7 1 3 = c e–4 0 + 6c e3 0 3 = c1 + 6c2 c = --- 1 2 2 7 1 –4x 15 3x y = – ---e + ------e 1 7 7 Solution: y = --1-e–4x + -----90-e3x 2 7 7
CHECK YOUR UNDERSTANDING 6.18 Answer: 2x 2x y1 = 1 – e y2 = 2 – e Find the specific solution of the system in CYU 6.17, if y = 2 y = 3 3 4 y10 = 0y20 = 1 y30 = 2 and y40 = 3
EXAMPLE 6.15 In the forest of Illtrode lived a small peaceful community of 50 elves, when they were sud- denly invaded by 25 trolls. The wizard Callan- dale quickly determined that: d -----Tt= --1-Tt– --1-Et dt 2 9
d -----Et= –Tt+ --1-Et dt 2 where Tt and Et represents the troll and elves populations t years after the troll inva- sion. Analyze the nature of the two popula- tions as time progresses. SOLUTION: To find the general solution of the system: 1 1 --- –--- T t = 2 9 Tt Et –1 --1- Et 2 we first find the eigenvalues of the above 22 matrix:
1 1 = --5- --- – –--- 1 1 6 det 2 9 ==--1- – 2 –0--- --1- – = --- 1 2 9 2 3 1 –1 --- – = --- 2 6
1 1 5 1 1 5 --- –------0 –--- –--- Then: E --- ==null2 9 – 6 null3 9 6 5 1 –1 --1- 0 --- –1 –--- 2 6 3 x y x y
homogeneous 1 1 1 –--- –--- rref 1 --- system of equations: 3 9 3 –1 –--1- 0 0 3 6.4 Applications 255
Setting the free variable y to 3, we obtain the eigenvector –31 for the eigenvalue --5- . In a similar fashion, you can show that 13 is an 6 1 eigenvector for --- . This leads us to the general solution: 6
--5-t --1-t 5 1 ---t ---t –c e6 + c e6 Tt 6 –1 6 1 1 2 ==c1e + c2e Et 3 3 --5-t --1-t 6 6 3c1e + 3c2e Turning to the initial conditions, we have: T0 25 25 = – c1 + c2 25 125 === c1 –------and c2 ------E0 50 50= 3c1 + 3c2 6 6 Leading us to the specific solution:
5 1 25 ---t 125 ---t Tt= ------e6 + ------e6 6 6 25 --5-t 125 --1-t Et= –------e6 + ------e6 2 2 A consideration of the graphs of the two functions reveals that while the troll population will continue to flourish in the region, the poor elves vanish around two-and-a-half years following the invasion:
trolls
elves
CHECK YOUR UNDERSTANDING 6.19
Turtles and frogs are competing for food in a pond, which currently contains 120 turtles and 200 frogs. Assume that the turtles’ growth rate and the frogs’ growth rate are given by 5Tt Ft 5Ft T t ==------– ------and F t ------– Tt 2 4 2 respectively; where Tt and Ft denote the projected turtle and frog population t years from now. Find, to one decimal place, the number of years it will take for the turtle population to equal that of Answer: 1.3 years the frog population. 256 Chapter 6 Determinants and Eigenvectors
EXERCISES
th Exercises 1-8. Find a formula for the k term of the sequence s1s2 s3 , if:
1. s1 = 2 , s2 = 2 , and sk = sk – 1 + sk – 2 for k 3 .
2. s1 = a , s2 = a , and sk = sk – 1 + sk – 2 for k 3 .
3. s1 = 1 , s2 = 2 , and sk = sk – 1 + sk – 2 for k 3 .
4. s1 = 1 , s2 = 6 , and sk = 6sk – 1 – 9sk – 2 for k 3 .
5. s1 = 1 , s2 = 4 , and sk = 3sk – 1 – 2sk – 2 for k 3 .
2 6. s1 = 1 , s2 = 2 , and sk = ask – 1 – bsk – 2 for k 3 and a – 4b 0 .
7. s1 = 1 , s2 = 2 , s3 = 3 , and sk = 2sk – 1 + sk – 2 – 2sk – 3 for k 4 .
21– 2 3 Hint: Note that S4 = 10 0 2 . 01 0 1
8. s1 = 1 , s2 = 2 , s3 = 3 , and sk = –2sk – 1 ++sk – 2 2sk – 3 for k 4 .
th 2 k + 1 9. (PMI) Let sk denote the k Fibonacci number. Prove that sksk – 2 –1sk – 1 = – , for k 3 . Suggestion: Use the Principle of Mathematical Induction to show that for A = 11 and 10 s s k 3 , Ak = k k – 1 . sk – 1 sk – 2
10. Let s0 and s1 be the first two elements of a sequence and let sk = ask – 1 + bsk – 2 be a recur- rence relation which defines the remaining elements of the sequence. Prove that if the quadratic 2 n n equation – a –0b = has two distinct solutions, 1 and 2 , then sk = c11 + c22 for some c1 c2 .
Suggestion: Replace the matrix 11 in the development of the Fibonacci sequence with 10 the matrix ab . 10 6.4 Applications 257
11. Let the entries of the matrices Ax and Bx be differentiable function, and let C be a matrix with scalar entries (real numbers). Given that the dimensions of the matrices are such that the operations can be performed, prove that: (i) Ax+ Bx = A x + Bx (ii) AxBx = AxBx + BxA x (iii) CA x = CAx
Exercises 12-17. Find the general solution of the given system of differential equations, then check your answer by substitution.
f1 x = 2f1x y1 = y1 – y2 12. 13. f2x = 3f1x – f2x y2 = 2y1 + 4y2
y1 = 3y1 + 2y2 f x = fx+ 2gx– hx 14. y2 = 6y1 – y2 15. gx = 2fx+ 4gx– 2hx hx = –fx – 2gx+ hx
x = 4x ++3y 3z x –261 x 16. y = –3x ++y 8z 17. y = –351 y z = –86x ++y 6z z 025 z
Exercises 18-21. Solve the given initial-value problem.
y y y 0 18. 1 ==31– 1 1 1 y2 62– y2 y20 –1
f x f x f 0 19. 1 ==21– 1 1 2 f2x –21 f2x f20 0
x 02–1x x0 1 20. y ==003 y y0 0 z 100 z z0 2
f x –261 fx f0 1 21. gx ==–351 gy g0 –1 hx 025 hz h0 0 258 Chapter 6 Determinants and Eigenvectors
22. Given enough space and nourishment, the rate of growth of plants A and B are given by At = --3-At and Bt = --3-Bt , respectively, where t denotes the number of months 2 2 after planting. One year, 50 of A and 30 of B were planted, and in such a fashion that the rates of growth of each of the two plants were compromised by the presence of the other; in accordance with: At = --3-At– Bt and Bt ==--3-Bt --1-At . Analyze the nature of 2 2 4 the two plant populations as time progresses. 23. Assume that initially, tank A contains 20 gallons of a liq- uid solution that is 10% alcohol, and that tank B contains 30 gallons of a solution that is 20% alcohol. At time t = 0 , the mixture in A is pumped to B at a rate of 1 gal- lons/minute, while that of B is pumped to A at a rate of 1.5 gallons/minute. Find the percentage of alcohol con- centration in each tank t minutes later. A B 6.5 Markov Chains 259
6
§5. MARKOV CHAINS Certain systems can occupy a number of distinct states. The transmis- Stochos: Greek for “guess.” sion in a car, for example, may be in the neutral state, or reverse (state), Stochastices: Greek for “one or first gear (state), etc. When chance plays a role in determining the who predicts the future.” current state of a system, then the system is said to be a stochastic pro- Andrei Markov: Russian cess, and if the probabilities of moving from one state to another Mathematician (1856-1922). remain constant, then the stochastic process is said to be a Markov process, or Markov chain. Here is an example of a two-state Markov process: State Y: Person x was involved in an automobile accident within the previous 12 month period. State N: Person x was not involved in an automobile accident within the previous 12 month period. Let’s move things along a bit by citing the following study: Probability of x being involved .23 if x is in Y in an accident within the next = 12 month period } { .19 if x is in N The above information is reflected in Figure 6.2(a) (called a transi- tion diagram), wherein each arrow is annotated with the probability of taking that path. The same information is also conveyed in the transi- Transition matrices are also tion matrix, Aa= , of Figure 6.2(b), where t represents the called probability matrices. ij ij probability of moving from the jth state to the ith state in the next move. The 0.19 in the upper right-hand corner of the matrix, for exam- ple, gives the probability of moving from state N to state Y in the next Since the entries in the transi- move, while the entry 0.81 is the probability of remaining in state N. tion matrix are probabilities, current state they must lie between 0 and Y N 1 (inclusive). Moreover, since .23 .77 .81 next state the entries down either col- .23 .19 Y umn account for all possible YN A : outcomes (staying in Y, or leaving Y, for example), their .19 .77 .81 N sum must equal 1. In particu- lar, since there is a 0.23 prob- Transition Diagram Transition Matrix ability that a person in state Y (a) (b) returns to state Y, there has to Figure 6.2 be a 0.77 probability that the Assume that, initially, 25% of the population was involved in an auto- person will leave that state and, consequently, move to mobile accident within the previous 12 month period (and 75% was state N. not). This given condition brings us to the so called initial state matrix .25 of the system: S0 = . .75 260 Chapter 6 Determinants and Eigenvectors
Utilizing matrix multiplication we can arrive at the next state, S1 :
T S0 S1 .23 .19 .25 (.23)(.25) + (.19)(.75) .20 Y (*) == .77 .81 .75 (.77)(.25) + (.81)(.75) .80 N The above tells us that there is a 0.20 probability that a person will be involved in an accident in the first 12 month period.
To get to the next state matrix, we replace S0 with S1 in (*): T S1 S2 .23 .19 .20 ==(.23)(.20) + (.19)(.80) .198 Y .77 .81 .80 (.77)(.20) + (.81)(.80) .802 N The above tells us that there is a 0.198 probability that a person will be involved in an accident in the next (second) 12 month period.
T S2 S3 .23 .19 .198 .19792 Y Similarly: = .77 .81 .802 .80208 N
Working backwards, we find that we can also arrive at S3 by multi- 3 plying the initial state matrix S0 by T :
2 2 3 S3 ==TS2 TTS1 =T S1 =T TS0 =T S0 Generalizing, we have: THEOREM 6.23 If T is the transition matrix of a Markov pro- cess with initial-state matrix S0 , then the nth state matrix in the chain is given by: n Sn = T S0
CHECK YOUR UNDERSTANDING 6.20 Of the 1560 freshmen at Bright University, 858 live in the dorms. There is a 0.8 probability that a freshman, sophomore, or junior cur- rently living in the dorms will do so in the following year, and a 0.1 probability that a currently commuting student will live on campus Answer: 757, 686, and 636 next year. Assuming (big assumption) that all 1560 freshmen will of the current freshmen will live in the dorm in their soph- graduate, determine (to the nearest integer) the number of the current omore, junior, and senior freshman that will be living in the dorms in their sophomore, junior, year, respectively. and senior years. 6.5 Markov Chains 261
POWERS OF THE TRANSITION MATRIX Let us formally define the concept of a transition matrix:
DEFINITION 6.10 A transition matrix TM nn is a TRANSITION MATRIX matrix that satisfies the following two properties: (1) T contains no negative entry. (2) The sum of the entries in each column of T equals 1.
Consider the three-state Markov process with transition matrix: I II III .3 .1 .6 I T: .2 .1 .4 II .5 .8 0 III Assume that at the start of the process we are in state II:
0 I S0 = 1 II 0 III I II III .3 .1 .6 0 .1 I Observe that: S1 ==TS0 .2 .1 .4 1 =.1 II .5 .8 0 0 .8 III same 2 .3 .1 .6 0 .41 .52 .22 0 .52 I 2 And that: S2 ==T S0 .2 .1 .4 1 =.28 .35 .16 1 =.35 II .5 .8 0 0 .31 .13 .62 0 .13 III same In general: THEOREM 6.24 Let T denote the transition matrix of a Mar- kov chain. If the process starts in state j, then the element in the ith row of the jth col- umn of T m represents the probability of end- ing up at state i after m steps.
EXAMPLE 6.16 Analyze the nature of the second column of T 2 , T 4 , and T 8 for the transition matrix: I II III .2 .6 .4 I T = .3 .1 .3 II .5 .3 .3 III Given that the system is initially in state II. SOLUTION: 262 Chapter 6 Determinants and Eigenvectors
.2 .6 .4 .2 .6 .4 .42 .30 .38 I 2 T ==.3 .1 .3 .3 .1 .3 .24 .28 .24 II .5 .3 .3 .5 .3 .3 .34 .42 .38 III The second column of T 2 tells us that if you start in state II, then there is a 0.30, 0.28, and 0.42 probability that you will end up at states I, II, and III, respectively, after two steps. .42 .30 .38 .42 .30 .38 .3776 .3696 .3760 I 4 2 2 T ==T T .24 .28 .24 .24 .28 .24 =.2496 .2512 .2496 II .34 .42 .38 .34 .42 .38 .3728 .3792 .3744 III The second column of P 4 tells us that if you start in state II, then there is a 0.3696, 0.2512, and 0.3792 probability that you will end up at states A, B, and C, respectively, after four steps.
.3776 .3696 .3760 .3776 .3696 .3760 .3750 .3750 .3750 8 4 4 T ==T T .2496 .2512 .2496 .2496 .2512 .2496 =.2500 .2500 .2500 .3728 .3792 .3744 .3728 .3792 .3744 .3750 .3750 .3750 Whoa! The three columns of T 8 are identical (to four decimal places). Moreover, if you take higher powers if T, you will find that you will again end up at the above T 8 matrix. This suggests that eventually there is, to four decimal places, a 0.375, 0.250, and 0.375 probability, respectively, that you will end up at states I, II, and III, independently of whether you start at state I, or II, or III! Indeed, no matter what initial state you start with, say the state .2 S0 = .5 , it looks like you will still end up at the same situation: .3 .2 .3750 .3750 .3750 .2 .3750 .3750 .3750 8 A T .5 ==.2500 .2500 .2500 .5 .2500 .2500 .2500 B .3 .3750 .3750 .3750 .3 .3750 .3750 .3750 C It appears that for this Markov chain, there is a probability of 0.375 that you will eventually end up in state A, a probability of 0.250 that you will end up in state B, and a probability of 0.375 that you will end up in state C, independently of the initial state of the process! Even more can be said; but first, a definition:
DEFINITION 6.11 n SF is a fixed state for a transition
matrix TM nn if TSF = SF .
EXAMPLE 6.17 .2 .6 .4 Show that the transition matrix T = .3 .1 .3 .5 .3 .3 of Example 6.16 has a fixed state. 6.5 Markov Chains 263
x SOLUTION: We are to show that there exists a state SF = y such z that TSF = SF : .2 .6 .4 x x .2x ++.6y .4z x .3 .1 .3 y = y .3x ++.1y .3z = y .5 .3 .3 z z .5x ++.3y .3z z Equating entries brings us to a system of three equations in three unknowns: –.6.8x ++y .4z = x 0 .2x +.6y +.4z = x .3x +.1y +.3z = y .3x –.3.9y +0z = .5x +.3y +.3z = z .5x +0.3y – .7z = It can be shown, however, that for any transition matrix T, the system x x of equations stemming from T y = y will always have more than z z one solution (Exercise 22). By adding the equation xyz++= 1 (the sum of the entries in any state matrix of the system must equal 1) to the system, we do end up with a unique solution: –.6.8x ++y .4z = 0 –.6.40.8 1003 8 .3x –.3.9y +0z = .3–.30 .9 rref 0101 4 .5x +0.3y – .7z = .5 .3–0 .7 0013 8 xyz++= 1 1111 000 0 We see that the matrix T has a unique fixed state; namely: .3750 .3750 .3750 38 .3750 8 T = .2500 .2500 .2500 SF ==14 .2500 , which, to four decimal places, coincides with .3750 .3750 .3750 38 .3750 the columns of the matrix T 8 in Example 6.14 (margin). The above rather surprising result, as you will soon see in Theorem 6.26, actually holds for the following important class of Markov chains: DEFINITION 6.12 A Markov chain with transition matrix T is For example: REGULAR MARKOV said to be regular if T k consists solely of 0.3.2 CHAIN positive entries for some integer k. The T = .8 .4 .5 transition matrix of a regular Markov chain .2 .3 .3 is said to be a regular transition matrix. is regular, since: Note that it is possible to eventually go from any state to any .28 .18 .21 2 other state in a regular Markov chain (see Theorem 6.24). T = .42 .55 .51 .30 .27 .28 264 Chapter 6 Determinants and Eigenvectors
In other words, AT is that matrix obtained by inter- DEFINITION 6.13 The transpose of a matrix changing the rows and col- TRANSPOSE Aa= ij Mmn is the matrix umns of A. For example: T A = aij Mnm , where aij = aji . 12 103 T If A = A = 04 245 35 The following results will be called upon within the proof of Theo- rem 6.25 below:
T T T A-1: If AM mn and BM nr , then AB = B A . [Exercise 19(f), page 161.]
A-2: If is an eigenvalue of AM n n then is also an eigenvalue of AT . (Exercise 52, page 231.)
THEOREM 6.25 = 1 is an eigenvalue of every transitional
matrix TM nn .
PROOF: Let w n be the n-tuple with every entry equal to 1. Being
a transition matrix, the columns of Tt= ij sum to 1. Hence:
t11 t12 t1n
t21 t22 t2n wT ===111 111 w
tn1 tn2 tnn
t11 +++t21 tn1 = 1 Taking the transpose of wT = w , we have T TwT = wT (A-1), and T T Note that wT is an eigen- this tells us that w is an eigenvector of T corresponding to the vector of the transpose of eigenvalue = 1 . Applying A-2 we conclude that = 1 is also an T, and not necessarily of T. eigenvalue of T.
Every regular transition matrix TM has In a sense, independently of THEOREM 6.26 nn FUNDAMENTAL THEOREM r its initial state: 1 OF REGULAR MARKOV The fixed state of a reg- r 2 ular transition matrix is CHAINS a unique fixed state vector SF = , and: also the final state of the r matrix n
r1 r1 r1 lim Ts = r2 r2 r2 s
rn rn rn (each column of the matrix Ts approaches SF as s increases). 6.5 Markov Chains 265
PROOF: Assume that Tt= ij consists of positive entries. (You are invited, in Exercise 27, to establish the result under the assumption that T k consists solely of positive entries for some integer k 1 .)
s s Consider the matrix T = cij , which must also consist of positive s s s s s That “(s)” in Mi is not entries. Let M = c and m = c denote the largest and i ijM i ijm an exponent; it is there to th s indicate that we are con- smallest entry in the i row of T . We will show that s s s th sidering the matrix T limMi – mi = 0 . This will tell us that all entries in the i row s of lim Ts are equal, which is the same as saying that the columns of s lim Ts are all equal. s s + 1 s + 1 s s From T ===cij T Tcij tij we have: n n s + 1 s c ==c t cs t + cs t ij i j ijm jmj i j
= 1 jm n = ms t + cs t i jmj i j
jm n ms t + Ms t i jmj i j j The entries in the jth column of the m s s transition matrix T sum to 1: = m t + M 1 – t i jmj i jmj
s + 1 th s + 1 We have shown that for every entry cij in the i row of T : cs + 1 ms t + Ms 1 – t ij i jmj i jmj In particular, for the largest entry in that row we have: s + 1 M ms t + Ms 1 – t i i jmj i jmj A similar argument (Exercise 26) can be used to show that for the s + 1 th s + 1 smallest entry mi in the i row of T we have: ms + 1 Ms t + ms 1 – t i i jMj i jMj Consequently: s + 1 s + 1 Mi – mi s ms t + M 1 – t – Ms t + ms 1 – t i jmj i jmj i jMj i jMj = Ms – ms 1 – t – t i i jmj jMj 266 Chapter 6 Determinants and Eigenvectors
Let t be the smallest entry in T. Since T consists of positive entries, and since the entries in every column of T sums to 1, we have 1 0 t --- , and, in particular that 1 – t – t 12– t . Hence: n jmj jMj s + 1 s + 1 s s Mi – mi Mi – mi 12– t Leading us to:
s s s – 1 s – 1 Mi –12mi – t Mi – mi
2 s – 2 s – 2 12– t Mi – mi
s – 1 12– t Mi – mi Since 012 – t 1 , 12– t s – 1 0 as s , and this tells us that the elements in the ith row of the matrix Ts must get arbitrarily close to each other as s tends to infinity. In turn, the columns of Ts
r1 must all tend to a common vector r2 . We complete the proof by
rn
r1 r showing that SF = 2 is the unique fixed state of T:
rn
x1 Employing Theorem 6.25, we start with an eigenvector X = x2 of T,
xn of eigenvalue 1. Since TXX= , TsXX= for all k. Hence:
x x r1 r1 r1 1 1 x x lim TsX ==r2 r2 r2 2 2 s
rn rn rn xn xn From Theorem 5.4, page 156:
r r r x r1 r1 r1 r 1 1 1 1 n 1 r r r x r r r r 2 2 2 2 ==x 2 +++x 2 x 2 x 2 1 2 1 i i = 1 r r r x n n n n rn rn rn rn 6.5 Markov Chains 267
Since TX= X :
x1 r n 1 x X ==2 x r2 i (**) i = 1 xn rn n
Since X 0 (it is an eigenvector), cx=0 i , and dividing both sides of (**) by c brings us to: i = 1 x 1 r1 1 x S ==--- 2 r2 F c
xn rn We then have: x x x r1 1 1 1 r1 r 1 x 1 x 1 x r T 2 ====T--- 2 ---T 2 --- 2 2 c c c rn xn xn xn rn The above argument also establishes the uniqueness of the fixed state vector, for if X is to be a (fixed) state vector, then c must equal one. EXAMPLE 6.18 An automobile insurance company classifies its customers as Preferred, Satisfactory, or Risk. Each year, 10% of those in the Preferred category are downgraded to Satisfactory, while 12% of those in the Satisfactory cate- gory move to Preferred. Twenty percent of Satisfactory drop to Risk, while 15% of Risk goes to Satisfactory. No customer is moved more than one slot in either direction in a sin- gle year. Find the fixed state of the system. SOLUTION:
.85 R .15 Pr S R Pr .9 .2 .9 .12 0 .1 Pr S .68 T = .1 .68 .15 S .12 0 .2 .85 R 268 Chapter 6 Determinants and Eigenvectors
The easiest way to go, is to take a “large” power of the transition How large is large enough? If matrix, and let any of its rows represent an approximation for the the rows look different, then fixed state matrix of the regular Markov process: take a higher power.
This establishes the fact that we are in a regular Markov situation (how)?
We conclude that roughly 34% of the company’s clients will (eventu- ally) fall in its Preferred category; 28% in its Satisfactory category; and 38% in its Risk category. But that is but an approximation, for:
.9 .12 0 .34 .3396 .1 .68 .15 .28 = .2814 0 .2 .85 .38 .3790 You can, however, find the exact steady state by the method of Exam- ple 6.17: .9x ++.12y 0z = x .9 .12 0 x x .1 .68 .15 y = y .1x ++.68y .15z = y 0 .2 .85 z z 0x ++.2y .85z = z
1 00-----18- 53 –.1x ++.12y 0z = 0 15 0 1 0 ------.1x –.32 +0 .15z = 53 rref 0x +0.2y – .15 = 001 -----20- 53 xyz++= 1 0000 see solution of Example 6.15.
18 15 20 We found ------------to be the (exact) steady state of the given Mar- 53 53 53 kov chain; telling us that the longer the process, the closer it will be that -----18-% of the customers, for example, will be in the preferred cate- 53 gory. 6.5 Markov Chains 269
CHECK YOUR UNDERSTANDING 6.21 The transition matrix T below represents the probabilities that an individual that voted the Democratic, Republican, or Green party ticket in the last election will vote D, R, or G in the next election. D R G D .73 .32 .09 T: .21 .61 .04 R Answer: Approximately 41%, .06 .07 .87 G 26%, 33% of the population, will vote democratic, republi- Determine the eventual percentage of the population in each of can, green, respectively. the three category. 270 Chapter 6 Determinants and Eigenvectors
EXERCISES
Exercises 1-6. Indicate whether or not the given matrix represents a transition matrix for a Markov Process. If not, state why not. If so, indicate whether or not the given transition matrix is regular.
.2 .1 01 1. 2. 3. 0 .1 .8 .9 10 –1 .9
.2 .4 .1 1 .4 .1 0 .3 .4 4..7 0 .3 5.0 0 .3 6. .5 .3 .6 .1 .6 .6 0 .6 .6 .5 .1 0 Exercises 7-8. Determine the transition matrix associated with the given transition diagram. 8. 7. .3 .4 .6 .1 .1 .2 A .5 B A .9 B .5 .4 .7 C .3
Exercises 9-11. Determine a transition diagram associated with the given transition matrix.
A B A B C A B C D A A A 9. .3 .4 1 0 1 0 1 .2 .6 10. .7 .6 B 000B 11. .5 0 .5 .4 B 010C 0 0 0 0 C .5 0 .3 0 D
12. Determine the probability of ending up at states A and B after two steps of the Markov chain associated with the transition matrix in Exercise 9, given that you are initially in state: (a) A (b) B 13. Determine the probability of ending up at states A, B and C after two steps of the Markov chain associated with the transition matrix in Exercise 10, given that you are initially in state: (a) A (b) B (c) C 14. Determine the probability of ending up at states A, B, C and D after two steps of the Markov chain associated with the transition matrix in Exercise 11, given that you are initially in state: (a) A (b) B (c) C (d) D 6.5 Markov Chains 271
Exercises 15-20. (a) Proceed as in Example 6.15 to find the stationary state matrix of the given regular transition matrix. (b) Use Theorem 6.26 and a graphing utility to check your answer in (a).
15..7 .2 16..1 .6 17. .5 .3 .3 .8 .9 .4 .5 .7
.8 .5 0 .5 .3 0 .6 .2 .1 18..2 .1 .6 19..1 .7 .6 20. .3 .5 .5 .0 .4 .4 .4 0 .4 .1 .3 .4
001 21. Show that the matrix A = 100 is not a regular matrix, by: 010 (a) Demonstrating that for each k, Ak will contain a row that does not consist solely of posi- tive entries. (b) Showing that A does not have a fixed state vector.
x x 22. Show that for any transition matrix T, the system of equations stemming from T y = y has z z infinitely many solutions. Suggestion: Use the fact that the sum of the elements in each column of T sum to 1.
23. Let TM nn be a regular transition matrix. Prove that x – 1 is a factor of the character- istic polynomial of T.
24. Show that if the entries in each column of AM nn sum to k, then k is an eigenvalue of A. 25. Referring to the proof of Theorem 6.26, show that: mk + 1 Mk b + mk 1 – b i i jMj i jMj 26. Establish Theorem 6.26 for an arbitrary transitional matrix T. Suggestion: Let r be such that AT= r consists of positive entries, and consider the matrix TA.
27. Prove that if is any eigenvalue of a regular transition matrix, then 1 .
28. Show that if is an eigenvalue of a regular transition matrix, then –1 .
29. (Rapid Transit) A study has shown that in a certain city, if a daily (including Saturday and Sunday) commuter uses rapid transit on a given day, then he will do so again on his next com- mute with probability 0.85, and that a commuter who does not use rapid transit will do so with probability 0.3. Assume that on Monday 57% of the commuters use rapid transit. Deter- mine, to two decimal places, the probability that a commuter will use rapid transit on: (a) Tuesday (b) Wednesday (c) Sunday 272 Chapter 6 Determinants and Eigenvectors
30. (Dental Plans) A company offers its employees 3 different dental plans: A, B, and C. Last year, 550 employees were in plan A, 340 in plan B, and 260 were in plan C. This year, there are 500 employees in plan A, 360 in plan B, and 290 in plan C. Assuming that the number of employees in the company remains at 1150, and that the current trend continues, determine the number of employees in each of the three plans: (a) A year from now. (b) Two years from now. (c) In 4 years (Suggestion: use the square of the 2-year matrix). (d) In 8 years (Suggestion: use the square of the 4-year matrix). (e) In 12 years (Suggestion: use the product of 4-year and the 8-year matrix). 31. (Campus Life) The following transition matrix gives the probabilities that a student living in the Dorms, at Home, or Off-campus (but not at home), will be living in the Dorms, at Home, or Off-campus (but not at home) next year (assume that all freshmen will graduate from the college in four years). D H O .62 .23 .25 D .11 .64 .09 H .27 .13 .66 O Currently, 55%, 24%, and 21% of the freshman class are living in the Dorms, at Home, and Off-campus (but not at home), respectively. Determine (to two decimal places) the probability that a current freshman will, three years from now, be living in the: (a) Dorms (b) Home (c) Off-campus. 32. (Higher Learning) The transition matrix below represents the probabilities that a female child will receive a Doctorate, a Masters, or a Bachelors (terminal degree), or No degree; given that her mother received a D, M, B (terminal degree), or No degree. mother D M B N
.31 .24 .11 .06 D daughter .25 .26 .09 .05 M .37 .42 .52 .49 B .07 .08 .28 .40 N
Given the initial state matrix (in column form), determine the S0 = .05 .09 .39 .47 probability that: (a) A granddaughter will receive a Bachelors degree. (b) A great granddaughter will earn a Doctorate. (c) A fifth generation daughter will receive no degree. 6.5 Markov Chains 273
33. (HMO Plans) A company offers its employees 5 different HMO health plans: A, B, C, D, and E. An employee can switch plans in January of each year, resulting in the following tran- sition matrix: this year A B C D E .54 .13 .08 .10 .06 A next year .11 .61 .17 .12 .18 B .17 .10 .56 .17 .15 C .06 .05 .08 .49 .10 D .12 .11 .11 .12 .51 E
Given the initial state matrix S0 = .11 .20 .31 .14 .24 (in column form), deter- mine, to three decimal places, the probability that: (a) An employee will chose plan B in the next enrollment period. (b) An employee will chose plan B two enrollment periods from now. (c) An employee will chose plan B three enrollment periods from now. (d) Determine to 5 decimal places, the fixed state of the system.
(e) Repeat (a) through (d) with initial state matrix S0 = .24 .31 .0 .26 .19 34. (Mouse in Maze) On Monday, a mouse is placed in a maze consisting of paths A and B. At the end of path A is a cheese treat, and at the end of path B there is bread. Experience has shown that if the mouse takes path A, then there is a 0.9 probability that it will take path A again, on the following day. If it takes path B, then there is a 0.6 probability that it will take that path again, the next day. The mouse takes path B on Monday. Determine the probability that the mouse will take path A on:
(a) Tuesday (b) Wednesday (c) Sunday (d) Answer parts (a), (b), and (c), under the assumption that the mouse takes path A on Monday. (e) Show that the transition matrix is regular, and then proceed as in Example 6.14 to determine the exact stationary state of that matrix. (f) Indicate the long-term state of the system (the probability that the mouse will take path A and the probability that the mouse will take path B, at the nth step of the process, for n “large”). 35. (Cities, Suburbs, and Country) Within the period of a year, 2% of a population currently residing in cities will move to the suburbs, while 2% of them will move to the country. 4% of those living in the suburbs will move to the cities, while 3% of them will move to the country. One percent of the country folks will move to the cities, while 2% of them will go to the sub- urbs. Currently, 65% of the population are in cities, and 20% are in the suburbs. Determine, to two decimal places, the percentage of city dwellers:
(a) Next year. (b) Two years from now. (c) Four years from now. 274 Chapter 6 Determinants and Eigenvectors
(d) Answer parts (a), (b), and (c), under the assumption that 50% of the population are in cities, and 35% are in the suburbs. (e) Determine to 5 decimal places, the fixed state of the system. 36. (Crop Rotation) A farmer rotates a field between crops of beans, potatoes and carrots. If she grows beans this year, then next year she will grow potatoes or carrots, each with 0.5 proba- bility. If she grows carrots, then she will grow beans with probability 0.2, potatoes with prob- ability 0.5 (and carrots with probability 0.3). If she grows potatoes, then she will grow beans with probability 0.5, and potatoes with probability 0.25. If she grows beans this year, what is the probability that she will grow beans again:
(a) Next year? (b) Two years from now? (c) Three years from now? (d) Answer parts (a), (b), and (c) under the assumption that she grows potatoes this year. (e) Determine to 5 decimal places, the fixed state of the system.
37. (Wolf Pack) A wolf pack hunts on one of four regions: A, B, C, and D:
A
D B
C
If the pack hunts in any given region one day, then it is as likely to hunt there again the next day as it is for it to hunt in either of its neighboring regions. On Monday, it hunted in region A. (a) Determine, to two decimal places, the probability that the pack will hunt in Region B on Tuesday. (b) Determine, to two decimal places, the probability that the pack will hunt in Region B on Sunday. (c) Determine the fixed state of the system. Chapter Summary 275
CHAPTER SUMMARY
DETERMINANTS The determinant of an nn matrix A, denoted det (A), is defined inductively as follows:
For a 11 matrix Aa= 11 , detA = a11 .
For a nn matrix A, with n 1 , let Aij denote the n – 1 n – 1 matrix obtained by deleting the ith row and jth column of the matrix A; Then: n 1 + j detA = –1 a1j detA1j j = 1
Laplace’s Theorem For AM nn and any 1 in : n n + _ + _ + _ + ______+ _+ _+ +_ detA = –1 ij+ a detA and det A = –1 ij+ a detA + + + + ij ij ij ij _ + _ + _ + _ + _ _ _ _ + + + + j = 1 i = 1 _ + _ + _ + _ + th + _ + _ + _ + _ Expanding along the i row Expanding along the jth row _ + _ + _ + _ + Determinants of diago- The determinant of a diagonal matrix or of an upper triangular matrix nal and upper triangu- is the product of the entries in its diagonal. lar matrices Determinants and (a) If two rows of AM nn are interchanged, then the determinant row operations of the resulting matrix is –detA . (b) If one row of A is multiplied by a constant c, then the determi- nant of the resulting matrix is cdetA . (c) If a multiple of one row of A is added to another row of A, then the determinant of the resulting matrix is detA Invertibility A matrix AM nn is invertible if and only if detA 0 . Product Theorem For AB Mnn : detAB = detA detB 276 Chapter 6 Determinants and Eigenvectors
EIGENVALUE AND An eigenvalue of a linear operator T: VV is a scalar R for EIGENVECTOR which there exists a nonzero vector v V such that: Tv = v Any such v is then said to be an eigenvector corresponding to the eigenvalue .
An eigenvalue of a matrix AM nn is a scalar R for which there exists a nonzero vector XR n such that: AX = X Any such X is then said to be an eigenvector corresponding to the eigenvalue .
IGENSPACE E The eigenspace of an eigenvalue of a matrix AM nn is denoted by E and is given by: E = nullA – I The eigenspace of an eigenvalue of a linear operator T: VV is denoted by E and is given by: E = kerT – I
CHARACTERISTIC For AM nn , the n-degree polynomial detA – I is said to be the POLYNOMIAL AND characteristic polynomial of A, and detA – I = 0 is said to be CHARACTERISTIC the characteristic equation of A. EQUATION For T: VV a linear operator on a vector space V of dimension n,
the n-degree polynomial detT – I , where is a basis for V, is said to be the characteristic polynomial of T, and
detT – I = 0 is said to be the characteristic equation of T. Finding Eigenvalues The eigenvalues of AM nn are the solutions of the characteristic equation detA – I = 0 . The eigenvalues of a linear operator T: VV on a vector space of
dimension n are the eigenvalues of the matrix T Mnn , where is any basis for V.
IAGONAL ATRIX D M A diagonal matrix is a square matrix Aa= ij with aij = 0 for ij . Chapter Summary 277
IAGONALIZABLE D A matrix AM nn is diagonalizable if A is similar to a diagonal MATRICES AND matrix. LINEAR OPERATORS A linear operator T: VV on a finite dimensional vector space V is
said to be diagonalizable if there exists a basis for which T is a diagonal matrix. Diagonalization Let T: VV be a linear operator on a finite dimensional vector Theorem space. The following are equivalent: (i) T is diagonalizable.
(ii)T is a diagonalizable matrix, for any basis of V. (iii)There exists a basis for V consisting of eigenvectors of T. Eigenvectors corre- If 12 m are distinct eigenvalues of a linear operator sponding to different T: VV , and if v is an eigenvector corresponding to , for eigenvalues are lin- i i early independent. 1 im, then v1v2 vm is a linearly independent set. The union of linearly Let 12 m be distinct eigenvalues of a linear operator independent subsets of T: VV , and let S = v v v be a linearly independent different eigenspaces is i i1 i1 iri again a linearly inde- subset of Ei . Then: pendent set. SS= 1 S2 Sm is a linearly independent set.
LGEBRAIC AND A An eigenvalue 0 of a matrix AM nn (or of a linear operator T on GEOMETRIC a vector space of dimension n) has algebraic multiplicity k if MULTIPLICITY OF k – 0 is a factor of A’s (or T ’s) characteristic polynomial, and EIGENVALUES k + 1 – 0 is not. We also define the geometric multiplicity of 0
to be the dimension of E0 (the eigenspace corresponding to 0 ) The geometric multi- If 0 is an eigenvalue of AM nn with algebraic multiplicity ma plicity cannot exceed and geometrical multiplicity m , then m m . the algebraic multi- g g a plicity. Another Diagonaliza- Assume that the characteristic polynomial of a linear operator tion Theorem. T: VV (or of a matrix), can be factored into a product of linear factors. Then T is diagonalizable if and only if the algebraic multiplic- ity of each eigenvalue of T is equal to its geometric multiplicity. 278 Chapter 6 Determinants and Eigenvectors
Diagonalizing a Matrix Let AM nn be diagonalizable. Let the columns of PM nn be any basis for Rn consisting of eigenvectors of A. Then DP= –1AP
is a diagonal matrix. Moreover, the diagonal entry dii D is the th eigenvalue i corresponding to the i column of P.
Power Theorem for –1 Let AM mm be diagonalizable with P AP= D . Then for any n: diagonalizable matri- ces An = PDnP–1 .
FIBONACCI The Fibonacci sequence is that sequence whose first two terms are 1, SEQUENCE and with each term after the second being obtained by summing its two immediate predecessors. 1 15+ k 15– k The kth Fibonacci number is given by ------------– ------. 5 2 2
SYSTEMS OF DIFFER- The solution set of f x = af x , consists of those functions of the ENTIAL EQUATIONS form fx= ceax for some constant c.
Let AM nn be diagonalizable, and let v1v2 vn be a basis for n consisting entirely of eigenvectors of A with corresponding eigenvalues i . Then, the general solution of: Fx = AF x is of the form: 1x 2x nx c1e v1 +++c2e v2 cne vn
for c1c2 cn .
MARKOV CHAINS Markov chain: When the probabilities of moving from one state of a system to another remain constant.
TRANSITION MATRIX A transition matrix TM nn is a matrix that satisfies the follow- ing two properties: (1) T contains no negative entry. (2) The sum of the entries in each column of T equals 1.
n FIXED STATE SF is a fixed state for a transition matrix TM nn if TSF = SF . If T is the transition matrix of a Markov process with initial-state th matrix S0 , then the n state matrix in the chain is given by: n Sn = T S0
UNDAMENTAL HEO F T - Every regular transition matrix TM nn has a unique fixed state vec- REM OF REGULAR tor and each column of the matrix Ts approaches S as s increases. MARKOV CHAINS F 7.1 Dot Product 279
7 CHAPTER 7 INNER PRODUCT SPACES Basically, an inner product space is a vector space augmented with an additional structure, one that will enable us to generalize the familiar concepts of distance and angles in the plane to general vector spaces.
§1. DOT PRODUCT
We begin by introducing a function which assigns a real number to each pair of vectors in n :
DEFINITION 7.1 The dot product of uu= 1u2 un DOT PRODUCT and vv= 1v2 vn , denoted by uv , is the real number: uv = u1v1 +++u2v2 unvn For example: 24– 3 1 507– 1 ==25 + 40 + – 3 71+ – 1 –12
The following four properties will lead us to the definition of an inner product space in the next section, much in the same way that the eight properties of Theorem 2.1, page 36, lead us to the definition of an abstract vector pace on page 40.
THEOREM 7.1 Let uvw n , and r . Then: positive-definite property: (i)vv 0 , and vv = 0 only if v = 0 commutative property: (ii) uv = vu homogeneous property: (iii) ruv ==ruv u rv distributive property: (iv) uv+ w = uw + vw
PROOF: We establish (iii) and invite you to verify the remaining three properties in the exercises:
ruv = ru1u2 un v1v2 vn scalar multiplication: = ru1ru2 run v1v2 vn definition 7.1: = ru1 v1 +++ru2 v2 run vn associative property: = ru1v1 +++ru2v2 runvn distributive property: = ru1v1 +++u2v2 unvn definition 7.1: = ruv 280 Chapter 7 Inner Product Spaces
CHECK YOUR UNDERSTANDING 7.1
Let uv n , and r . Prove that: Answer: See page B-30. ruv = u rv
DEFINITION 7.2 The norm of a vector vv= 1v2 vn , n NORM IN denoted by v , is given by: vvv=
2 2 2 For vv= 1 v2 , vvv== v1 + v2 represents the length (magnitude) of v [Figure 7.1(a)], and the same can be said about 3 v for vv= 1v2 v3 [Figure 7.1(b)].
2 2 2 2 2 v = v1 + v2 v = v1 ++v2 v3 vv= v 1 2 vv= 1v2 v3
v2 v3
v1 v1 v2
(a) (b) Figure 7.1 In general, for v n : v is defined to be the length of v. Moreover:
uv– is defined to be the distance between uv n .
2 In particular, for uu= 1 u2 and vv= 1 v2 in :
vv= 1 v2
2 2 uv– = u1 – v1 + u2 – v2 v2 – u2 v1 – u1 uu= 1 u2
7.1 Dot Product 281
CHECK YOUR UNDERSTANDING 7.2
Prove that for uv n and c : (a) cv = c v (b) uv– 2 = u 2 – 2uv + v 2 2 2 2 Answer: See page B-30. [Reminiscent of: ab– = a – 2ab + b ]
ANGLE BETWEEN VECTORS Applying the law of cosines [Figure 7.2(a)] to the vectors uv 2 in Figure 7.2(b), we see that: uv– 2 = u 2 + v 2 – 2 uvcos
c uv– a u b v c2 = a2 + b2 – 2abcos Law of Cosines uv– 2 = u 2 + v 2 – 2 uvcos (a) (b) Figure 7.2 From CYU 7.2, we also have: uv– 2 = u 2 – 2uv + v 2 Thus: u 2 + v 2 – 2 uvcos = u 2 – 2uv + v 2 –22 uvcos = – uv For any –1 x 1 , uv cos = ------ –1 cos x is defined to be uv that angle 0 –1 uv see margin: = cos ------whose cosine is x. uv Leading us to: In Exercise 44 you are asked to verify that DEFINITION 7.3 The angle between two nonzero vectors n ------uv 1 ANGLE BETWEEN uv is given by: uv VECTORS –1 uv Assuring us that: = cos ------uv –1 uv cos ------exists. uv 282 Chapter 7 Inner Product Spaces
EXAMPLE 7.1 Determine the angle between the vectors u = 120– 2 and v = –1312 .
SOLUTION: –1 uv –1 120– 2 –1312 ==cos ------cos ------uv 144++ 1914+++
–1 1 = cos ------ 85 315
CHECK YOUR UNDERSTANDING 7.3 Answer: –1 5 Determine the angle between the vectors u = 120 and cos ------ 83 55 v = –131.
ORTHOGONAL VECTORS IN Rn
The angle between the vectors uv 2 depicted in the adjacent figure has a measure of u We remind you that, for any –1 90 (--- radians ), and we say that those vectors are –1 x 1 , cos x is that angle 2 = 90 0 such that cos = x . perpendicular or orthogonal. Appealing to Defini- tion 7.3 we see that: v –1 uv So, if cos ------= 90 , uv –1 uv cos ------= 90 or uv = 0 uv then------ = 0 , or:uv = 0 . uv uv (see margin) Bringing us to:
DEFINITION 7.4 Two vectors u and v in n are orthogonal ORTHOGONAL VECTORS if uv = 0 .
Note: The zero vector in n is orthogonal to every vector in n .
CHECK YOUR UNDERSTANDING 7.4
Let v n . Show that the set v of vectors perpendicular to v: v = u n uv = 0 Answer: See page B-31. is a subspace of n .
It is often useful to decompose a vector v n into a sum of two vec- tors: one parallel to a given nonzero vector u, and the other perpendicular to u. The parallel-vector must be of the form cu for some scalar c (see Figure 7.3). 7.1 Dot Product 283
v – cu
v
cu = projuv v – cu u
Figure 7.3 ORTHOGONAL PROJECTION The vector cu in Figure 7.3 is said to be the orthogonal projection of v onto u and is denoted by projuv . To determine the value of c, we note that for v – cu to be orthogonal to u, we must have:
Theorem 7.1(iv): v – cu u = 0 Theorem 7.1(iii): vu – cu u = 0 vu – cuu = 0 vu vu c ==------ ------ uu u 2 Summarizing, we have: vv= – projuv + projuv THEOREM 7.2 Let v n and let u be any nonzero vector in n . VECTOR Then: DECOMPOSITION vv= – proj v + proj v v – projuv u u where: u uv projuv = ------u and v – projuv projuv = 0 uv uu proj v = ------u u uu The vector projuv is said to be the vector component of v along u, and the vector v – projuv is said to be the vector component of v orthogonal to u.
EXAMPLE 7.2 Express the vector 21– 3 as a sum of a vector parallel to 140 and a vector orthog- onal to 140 .
SOLUTION: For v = 21– 3 and u = 140 we have: uv 140 21– 3 6 proj v ==------u ------140 =------140 u uu 140 140 17 6 24 28 7 and v – proj v ==21– 3 – ------------0 ------–------–3 u 17 17 17 17 28 7 6 24 Check: ------–------–3 + ------------0 ==21– 3 v 17 17 17 17 28 7 6 24 28 6 7 24 and ------–------–3 ------------0 ==------------+ –------------0 17 17 17 17 17 17 17 17 284 Chapter 7 Inner Product Spaces
Answer: CHECK YOUR UNDERSTANDING 7.5 6 9 24 3–------– ------21 21 21 Express the vector 301– 1 as the sum of a vector parallel to 6 12 3 0241 and a vector orthogonal to 0241 . + 0------21 21 21
EXAMPLE 7.3 Find the distance from the point P = 313 to the line L in 3 which passes through the points 102 and 316 .
SOLUTION: We first find a direction vector for the given line (see Theorem 2.17, page 70): P u ==316 – 102 214 . . want this distance Choosing the point A = 102 on L v – proj v we determine the vector v from A to P: u L
v ==313 – 102 211 v Applying Theorem 7.2, we have: uv proj v = ------u u uu u 214 211 = ------214 proj v 214 214 . u A 9 6 3 12 ==------214 ---------21 7 7 7 Thus: 6 3 12 v – proj v = 211 – ---------u 7 7 7 8 4 5 1 1 105 ==------ –------84– 5 =--- 64++ 16 25 =------7 7 7 7 7 7 CYU 7.2(a)
CHECK YOUR UNDERSTANDING 7.6 (a) Find the distance from the point P = 25 to the line L in 2 passing through the points 12 – and 24 .
1 (b) Find the distance from the point P = 1013 to the line L in Answer: (a) ------(b) 4 4 37 passing through the points 1201 and 1221 . 7.1 Dot Product 285
PLANES REVISITED We now offer an alternative representation for a plane in 3 than that given in Theorem 2.19, page 71. Just as a line in 2 is determined by a point on the line and its slope, so then is a plane in 3 determined by a point on the plane and a nonzero vector orthogonal to the plane (a normal vector to the plane). To be more specific, suppose we want the equation of the plane with normal P n . vector n = abc that contains the point A0 = x0y0 z0 . For any . point Pxyz= on the plane we have: A0 n A0P = 0 or: normal form abc xx– 0 yy– 0 zz– 0 = 0 scalar form or: ax– x0 ++by– y0 cz– z0 = 0
or: ax++ by cz =d where d = ax0 ++by0 cz0 general form
Note that a normal to the plane can easily be spotted from any of the above forms. For example, n = 25– 4 is a normal to the plane: 25– 4 x – 1 y – 3 z + 2 = 0 2x – 1 +05y – 3 – 4z + 2 = 2x +255y – 4z =
EXAMPLE 7.4 Find a normal, scalar, and general form equa- tion of the plane passing through the point 13– 2 with normal vector n = 41– 5 . SOLUTION: normal: 41– 5 x – 1y – 3 z + 2 = 0 scalar: 4x – 1 –51y – 3 +0z + 2 = general: 4xy–5+9z = –
CHECK YOUR UNDERSTANDING 7.7
Find an equation of the plane passing through the point 13– 2 with normal parallel to the line passing through the points Answer: See page B-31. 110 021 .
EXAMPLE 7.5 Express the plane 3xy+6– 2z = in the vec- tor form of Theorem 2.19, page 71. 286 Chapter 7 Inner Product Spaces
SOLUTION: In order to write the plane in vector form we need two direction vectors and a translation vector. We chose 200 as our translation vector (corresponding to the vector wx= 0y0 z0 in Figure 2.9, page 71). Any two linearly independent vectors orthogonal to n = 31– 2 31– 2 021 = 0 can serve as direction vectors, say 021 and 203 [corre- and 31– 2 203 = 0 sponding to u = AB and v = AC in Figure 2.9]. This leads us to the following vector form equation of the plane: 200 ++r021 s203 rs R
CHECK YOUR UNDERSTANDING 7.8 With reference to Example 7.5, show, directly that: Answer: See page B-31. xyz 3xy+6– 2z = = 200 ++r021 s203 rs R
EXAMPLE 7.6 Find the distance from the point P = 432 to the plane 2x – 3y +5z = .
SOLUTION: We begin by choosing n Any point xyz satis- the point A = 005 on the plane fying the equation (note that 20 –530 +5= ). We P proj v 2x – 3y +5z = would position the normal vector n v . do just as well. n = 231 – so that its initial A point is at A, and then determine the we want this distance . vector v from A to P: B v ==432 – 005 43– 3 Applying Theorem 7.2, we have: vn proj v = ------n n nn 43– 3 231 – = ------231 – 231 – 231 – –4 4 6 2 ==------231 – –------ –--- 14 7 7 7 Hence: 1 1 56 214 proj v ==--- –462– --- 16++ 36 4 ==------n 7 7 7 7
CYU 7.2(a)
Answer: See page B-32. CHECK YOUR UNDERSTANDING 7.9 Note that if you apply this formula in Example 7.6 Prove that the distance D from a point x0y0 z0 to the plane you obtain: ax++ by cz = d is given by the formula: 24– 33+ 12– 5 D = ------ 22 ++32 12 ax ++by cz – d D = ------0 ------0 ------0 ==------4 ------214 a2 ++b2 c2 14 7 7.1 Dot Product 287
CROSS PRODUCT Here is a handy result:
THEOREM 7.3 If v1 = a1a2 a3 v2 = b1b2 b3 are lin- early independent vectors in 3 , then: va= 2b3 – a3b2– a1b3 + a3b1 a1b2 – a2b1 is perpendicular to both v1 and v2
PROOF: Rolling up our sleeves, we simply show that vv 1 = 0 , and leave it for you to verify that vv 2 is also zero: vv 1 = a2b3 – a3b2– a1b3 + a3b1 a1b2 – a2b1 a1a2 a3
= a2b3 – a3b2 a1 ++– a1b3 + a3b1 a2 a1b2 – a2b1 a3
==a1a2b3 – a1a3b2 –0a1a2b3 ++a2a3b1 a1a3b2 – a2a3b1
Handy or not, how is one to remember that complicated three-tuple v = a2b3 – a3b2– a1b3 + a3b1 a1b2 – a2b1 of Theorem 7.3? By formally evaluating the following determinant about its first row:
e1 e2 e3 a2 a3 a1 a3 a1 a2 det a1 a2 a3 = det e1 – det e2 + det e3 b2 b3 b1 b3 b1 b2 b1 b2 b3
= a2b3 – a3b2 e1 – a1b3 – a3b1 e2 + a1b2 – a2b1 e3 and then replacing e1e2 and e3 with the standard basis vectors 100 010 and 001 , respectively, to arrive at the three- tuple of Theorem 7.3:
va= 2b3 – a3b2– a1b3 + a3b1 a1b2 – a2b1
The above vector is called the cross product of v1 = a1a2 a3 and v2 = b1b2 b3 , and is denote by: v1 v2 . Moreover, to conform with standard notation, we shall replace the standard basis vectors e1e2 and e3 with the symbols i, j, and k, respectively; bringing us to:
DEFINITION 7.5 The cross product of v1 = a1a2 a3 CROSS PRODUCT and v2 = b1b2 b3 is denoted by v1 v2 , and is expressed in the form: ijk
v1 v2 = det a1 a2 a3
b1 b2 b3 288 Chapter 7 Inner Product Spaces
For example: ij k 234 31– 2 = det 23 4 31– 2
= det 34i – det 24j + det 23k 12– 32– 31 ==–46 – i –29–124 – j + – k –10 16– 7
EXAMPLE 7.7 Find the general form equation of the plane that contains the points A = 12– 1 , B = 231 , C = 312 – .
SOLUTION: Noting that the vectors AB ==231 – 12– 1 112 and AC ==312 – – 12– 1 23– 3 are parallel to the plane, we employ Theorem 7.3 to find a normal to the plane:
ijk n ===det 112 9ij+ – 5k 91– 5 23–3
Choosing the point A = 12– 1 on the plane, we proceed as in Example 7.4 to arrive at the general form equation of the plane: 91– 5 x – 1y – 2 z + 1 = 0 9x – 1 +0y – 2 – 5z + 1 = 9xy+0–165z – =
CHECK YOUR UNDERSTANDING 7.10
(a) Find the general form equation of the plane that contains the points A = 32–2, B = 25– 3, C = 412 – 3 . (b) Verify that your answer in (a) coincides with that of Example Answer: See page B-33. 2.15, page 72. 7.1 Dot Product 289
EXERCISES
Exercises 1-2. Evaluate uv for the given n-tuples. 1.u ==53 v 61 2. u ==0357 v 2104 Exercises 3-5. Determine the norm v for the given vector.
3.v = 32 4.v = 10– 5 5. v = 31– 21–
6. Find all values of c such that c231 = 9 . 7. Find all values of a such that the vector a 3 is orthogonal to the vector 2a –5 . 8. Find all values of a such that the vector 3a 2a is orthogonal to the vector a2 a . 9. Find all values of a and b such that the vector a3 b is orthogonal to the vector b3 a . Exercises 10-11. Determine the angle between the vectors u and v. 10.u ==53 v 61 11. u ==0357 v 2104 Exercises 12-13. Express the given vector v as a sum of a vector parallel to the given vector u and a vector orthogonal to u. 12.v ==15 u 32 13. v ==23– 1 u –204 Exercises 14-15. Find a normal form, the general form, and a vector form representation (Theo- rem 2.20, page 72) of the plane passing through the point A0 with given normal vector n.
14.A0 ==213 n 12– 1 15. A0 ==12– 1 n 213 Exercises 16-18. Find both a normal form equation and a vector form representation (Theorem 2.20, page 72) for the given plane. 16.2x – 3y +2z = 17.4xz+1= 18. x –23y –2z = –
19. Find the distance from the point P = 14 and the line L in 2 passing through the points 12 and 21 . 20. Find the distance from the point P = 14– 1 and the line L in 3 passing through the points 121 and 210 . 21. Find the distance from the point P = 14– 11 and the line L in 4 passing through the points 121– 2 and 2102 . 22. Find the distance from the point P = 21– 2 to the plane x –24y +3z = . 23. Find the distance from the point P = 14– 2 to the plane 3xy++4z = 2 . 24. Determine the angle of intersection of the planes x –23y +1z = and 2xyz+2– = . Suggestions: Consider the normals to those planes. 290 Chapter 7 Inner Product Spaces
25. Find the set of vectors in 3 orthogonal to: (a) the vector 132 . (b) the vectors 132 and 22– 1 . (c) the vectors 132 , 20– 1 , and 25– 3 . Exercises 26-27. Find the general form equation of the plane that contains the given points. 26. 21– 210– 1 –301 27. 001 200 030
28. Find the angle between a main diagonal and an adjacent edge of a cube of volume 8 in.3 . 29. Prove Theorem 7.1(i). 30. Prove Theorem 7.1(ii). 31. Prove Theorem 7.1(iv). 32. Establish the following properties for uvw n and r : (a) 0 v ==v 0 0 (b) uv w= uv + uw (c) u rv = ruv (d) uv– w = uw – vw (e) uvw – = uv – uw (f) –v w ==vw – –vw 33. Show that two nonzero vectors v and u are normal to a given plane if and only if each is a scalar multiple of the other. 34. (Normal form equation of a line in R2 ) Express the line ax+ by = c in the form nv = np , where vxy= , p is a point on the line, and n 0 is a normal to the line
n T 35. Let AM nn , and let uv . Show that Auv = u A v . (See Exercise 19, page 161).
n n 36.u . Show that the function pu: given by puv = uv is linear. What is the
kernel of pu ? 37. Let u n . Show that if uv = 0 for every v n , then u = 0 . 38. (Pythagorean Theorem in n ) Let uv n . Show that uv+ 2 = u 2 + v 2 if and only if uv = 0 . 39. (Parallelogram Law in n ) Let uv n . Show that: uv+ 2 +2uv– 2 = u 2 + 2 v 2 40. Let uv Rn . Prove that uv= if and only if uv+ and uv– are orthogonal.
n 41. Prove that if v1v2 vn is such that vi vj = 0 if 1 ijn , then n v1v2 vn is a basis for .
n 42. Let uv1 v2 vm . Prove that if u is orthogonal to each vi , 1 im , then u is
orthogonal to every v v1v2 vm . 7.1 Dot Product 291
43. (Cauchy-Schwarz Inequality in n ) Show that if uv n , then uv uv . Suggestion: (If u = 0 , then equality holds). For u 0 , use the fact that 0 ruv+ ruv+ = uu x2 ++2uv x vv to conclude that the discriminant of the quadratic polynomial uu x2 ++2uv x vv cannot be positive. 44. Use the above Cauchy-Schwuarz Inequality to show that for any nonzero vectors uv n : ------uv 1 . uv 45. Establish the following properties for uvw 3 and rs : (a) ru sv = rs uv (b) –u v ==uv – –uv (c) uvw + = uv + uw (d) uvw = uv w 46. (Metric Space Structure of n ) Define the distance between two vectors uv n to be duv = uv– . Prove that for all uvw n : (a)duv 0 . (b)duv = 0 if and only if uv= . (c) duv = dvu (d)duw duv + dvw Suggestion: Use the Cauchy-Schwuarz Inequality of Exercise 41. 47. (PMI) Use the principle of mathematical induction to show that for any n uv1 v2 vm and any a1a2 am : .
u a1v1 +++a2v2 amvm = a1uv 1 +++a2uv 2 amuv m
PROVE OR GIVE A COUNTEREXAMPLE
48. Let uvw n . If u 0 and if uv = uw , then vw= . 49. Let uv n . If vw = uw for every w n , then uv= .
n 50. Let uv . If vw= 1 + w2 and vz= 1 + z2 with w1 and z1 multiples of v, and if w2 and
z2 are orthogonal to u, then w1 = z1 and w2 = z2 . 51. Let uvz n , with u 0 . If u is orthogonal to both v and z, then v = cz for some c . 52. The function N: n given by Nv = v is linear. 53.uv = vu for all uv 3 . 292 Chapter 7 Inner Product Spaces
7
§2. INNER PRODUCT
As you know, a vector space V comes equipped with but two opera- tions: addition, and scalar multiplication. We now enrich that algebraic structure by adding another binary function on V — one inspired by the dot-product properties of Theorem 7.1, page 279:
While the scalar product DEFINITION 7.6 An inner product on a vector space V is a rv assigns a vector to a INNER PRODUCT function which assigns a real number uv scalar r and a vector v, to any two vectors uv V , such that: the inner product u v assigns a real number to positive-definite axiom: (i)vv 0 , and vv = 0 only if v = 0 a pair of vectors. commutative axiom: (ii) uv = vu
homogeneous axiom: (iii) ruv = r uv
distributive axiom: (iv) uv+ w = uw + vw
INNER PRODUCT SPACE A vector space together with an inner product is said to be an inner product space. The Euclidean vector space n with dot product uv = uv is called the Euclidean inner product space of dimension n. There are other inner products that can be imposed on n , among them:
EXAMPLE 7.8 For any positive real numbers c1c2 cn : Why are we requiring WEIGHTED u1u2 un v1v2 vn the c’s to be positive? EUCLIDEAN INNER PRODUCT SPACE = c1u1v1 +++c2u2v2 cnunvn is an inner product on n .
SOLUTION: We show that the distributive axiom (iv) holds and invite you to establish the remaining axioms in the exercises:
For uu===1u2 un v v1v2 vn w w1w2 wn :
uv+ w = u1 + v1u1 + v1 u1 + v1 w1w2 wn
= c1u1 + v1 w1 +++c1u1 + v1 w1 c1u1 + v1 w1
= c1u1w1 +++c2u2w2 cnunwn + c1v1w1 +++c2v2w2 cnvnwn = uw + vw 7.2 Inner Product 293
CHECK YOUR UNDERSTANDING 7.11
Verify that 2 2 a2x ++a1xa0 b2x ++b1xb0 = a2b2 ++a1b1 a0b0
is an inner product on P2 . Answer: See page B-33. Generalize the above to obtain an inner product on P3
The following theorem extends the Euclidean dot-product results of Exercise 32, page 290 to general inner product spaces: THEOREM 7.4 For every u, v, and w in an inner product space V: In the exercises you are (a) 0 v ==v 0 0 asked to establish the fol- lowing generalization and (b) uv + w = uv + uw combination of (b) and (c). (c) u rv = r uv For uv1 v2 vn V
and c1 c2 cn : (d) uv– w = uw – vw n n (e) uv – w = uv – uw u civi = ci uv i i = 1 i = 1 (f) –v w ==vw – – vw
PROOF: We verify (d), and leave it for you to establish the rest.
Definition 2.7, page 55: uv– w = uv+ – w Axiom (iv): = uw + –vw Theorem 2.11 (x), page 56: = uw + –1vw Axiom (iii): = uv – uw
CHECK YOUR UNDERSTANDING 7.12
Answer: See page B-33. Prove: ru rv = r2 uv
DISTANCE IN AN INNER PRODUCT SPACE In the previous section we defined the norm (or magnitude) of a vec- tor v inn in terms of the dot product: vvv= . The dot product was also used to describe the distance between two vectors u and v in n :uv– = uv– uv– . Replacing “dot-product” with “inner product” enables us to extend the notion of magnitude and dis- tance to any inner product space: 294 Chapter 7 Inner Product Spaces
DEFINITION 7.7 The norm (or magnitude) of a vector v in an inner product space V, denoted by v , is NORM AND given by: DISTANCE vvv= The distance between two vectors u and v in V is given by uv– .
CHECK YOUR UNDERSTANDING 7.13
Show that for any vectors u and v in an inner product space V and any r : Answer: See page B-33. (a) rv = r v (b) uv+ 2 = u 2 ++2 uv v 2
EXAMPLE 7.9 Find the distance between the two vectors 2 p1x = 2x –1x + and p2x = 3x + 4 in the inner product space P2 of CYU 7.11
SOLUTION: Utilizing the inner product: 2 2 a2x ++a1xa0 b2x ++b1xb0 = a2b2 ++a1b1 a0b0 we have: 2 2 p1x –2p2x ==x –1x + –23x + 4 x –34x –
2 2 p1 – p2 ==p1 – p2 p1 – p2 2x –34x –2 x –34x – ==22 ++–4 2 –3 2 29
CHECK YOUR UNDERSTANDING 7.14
With reference to the weighted inner product space: u1u2 u3 v1v2 v3 = 5u1v1 ++2u2v2 4u3v3 on 3 (see Example 7.8), determine: (a) The magnitude of the vector 35– 8 . (b) The distance between the vector 35– 8 and the vector Answer: (a) 710 (b) 645 102 . 7.2 Inner Product 295
THE CAUCHY-SCHWARZ INEQUALITY
The following theorem will enable us to extend the concept of an angle between two vectors in n to vectors in an inner product space: THEOREM 7.5 For any two vectors u and v in an inner product CAUCHY- space: SCHWARZ uv uv
PROOF: If either u = 0 or v = 0 , then uv = 0 and we are done. For u 0 and v 0 , we first show that uv – uv : 1 1 1 1 Definition 7.5(i): ------u + ------v ------u + ------v 0 The proof sketched out in u v u v Exercise 43, page 291, can 1 1 1 1 1 1 also be used to establish 7.5(iv): ------u + ------v ------u + ------u + ------v ------v 0 this result. 7 u v u u v v .5 ( iv ) : 1 1 1 1 1 1 1 1 ------u ------u +++------v ------u ------u ------v ------v ------v 0 u u v u u v v v 1 2 1 CYU 7.10: ------uu ++0------uv ------vv u 2 uv v 2 2 1 ++0------uv 1 uv 2 ------uv –2 uv uv – uv In the following Check Your Understanding box you are asked to show that uv uv . Putting the two inequalities together, we come up with – uvuv uv , which is to say: uv uv
CHECK YOUR UNDERSTANDING 7.15
Verify: uv uv 1 1 1 1 Answer: Seepage B-34. Suggestion: Begin with ------u–------v ------u–------v 0 . u v u v 296 Chapter 7 Inner Product Spaces
Here are norm properties that are reminiscent of absolute value prop- erties in : THEOREM 7.6 Let V be an inner product space. For all uv V and r : (a)v 0 , and v = 0 if and only if v = 0 (b) rv = r v
TRIANGLE INEQUALITY (c) uv+ uv+
PROOF: (a) A consequence of Definition 7.5(i) and vvv= .
(b) rv ===rv rv r2 vv r2 vv =r v
CYU 7.11
CYU 7.12(b) (c) uv+ 2 = u 2 ++2 uv v 2 Cauchy-Schwarz inequality: u 2 ++2 uv v2 = uv+ 2 Taking the square root of both sides of uv+ 2 uv+ 2 yields the desired result. We now extend the angle concept of Definition 7.3, page 281, to inner product spaces:
The Cauchy-Schwarz ine- DEFINITION 7.8 The angle between two nonzero vectors u quality plays a hidden role ANGLE BETWEEN and v in an inner product space is given by: in this definition. (Where?) VECTORS –1 uv = cos ------uv
EXAMPLE 7.10 Find the angle between the two vectors 2 p1x = 2x –1x + and p2x = 3x + 4 in the inner product space P2 of CYU 7.10 SOLUTION: While it is admirable that we were able to extend the geometrical notion of the angle between vectors in the plane to vectors in an arbi- trary inner product space, the real benefit of that generalization surfaces in the next section, where the concept of orthogonality takes center stage. 7.2 Inner Product 297
–1 2x2 –1x +3 x + 4 = cos ------2x2 –1x +3x + 4 –1 20 + –1 314+ = cos ------ 2x2 –1x +2 x2 –1x + 3x +34 x + 4 –1 1 = cos ------22 ++33–1 –1 11 + 44
–1 1 = cos ------ 84 615
CHECK YOUR UNDERSTANDING 7.16
With reference to the weighted inner product space:
u1u2 u3 v1v2 v3 = 5u1v1 ++2u2v2 4u3v3 Answer: 3 –1 –49 on (see Example 7.7), determine the angle between the vectors cos ------ 119.1 351 29 35– 8 and 102 . 298 Chapter 7 Inner Product Spaces
EXERCISES
Exercises 1-8. With reference to the weighted inner product space:
u1u2 u3 v1v2 v3 = 5u1v1 ++2u2v2 4u3v3 of Example 7.7, determine: 1. The magnitude of the vector 12– 3 . 2. The magnitude of the vector 320 . 3. The distance between the vectors 12– 3 and 102 . 4. The distance between the vectors 320 and 12– 3 . 5. The angle between the vectors 12– 3 and 102 . 6. The angle between the vectors 320 and 12– 3 . 7. Verify that the Cauchy-Schwarz inequality holds for the vectors 12– 3 and 102 . 8. Verify that the Cauchy-Schwarz inequality holds for the vectors 320 and 12– 3 . Exercises 9-16. Referring to the inner product space: 2 2 a2x ++a1xa0 b2x ++b1xb0 = a2b2 ++a1b1 a0b0
on P2 of CYU 7.11, determine: 9. The magnitude of the vector 2x2 –3x + . 10. The magnitude of the vector – x2 + x – 5 . 11. The distance between the vectors 2x2 –3x + and – x2 + x – 5 . 12. The distance between the vectors 3x2 + 1 and 2x – 5 . 13. The angle between the vectors 2x2 –3x + and – x2 + x – 5 . 14. The angle between the vectors 3x2 + 1 and 2x – 5 . 15. Verify that the Cauchy-Schwarz inequality holds for the vectors 2x2 –3x + and – x2 + x – 5 . 16. Verify that the Cauchy-Schwarz inequality holds for the vectors 3x2 + 1 and 2x – 5 .
a11 a12 b11 b12 17. For A = B = in the vector space M22 , define: a21 a22 b21 b22
AB = a11b11 +++a12b12 a21b21 a22b22
Show that the above operator is an inner product on M22 . 7.2 Inner Product 299
Exercises 18-22. Referring to the inner product space on Exercise 17, determine:
18. The magnitude of the vector 13 . 02
19. The magnitude of the vector 21– . 10
20. The distance between the vectors 13 and 21– . 02 10
21. The angle between the vectors 13 and 21– . 02 10
22. Verify that the Cauchy-Schwarz inequality holds for the vectors 13 and 21– . 02 10
n n n i i 23. Verify that aix bix = aibi is an inner product on the polynomial space Pn . i = 0 i = 0 i = 0
24. (Calculus Dependent) (a) Show that Cab = f: ab Rf is continuous is a sub- set of the function vector space Fab of Theorem 2.4, page 44. b (b) Show that fg = fxgxdx is an inner product on Cab (called the standard a inner product on Cab ). Exercises 24-35. Calculus Dependent) Referring to the inner product space on Exercise 24, determine: 25. The magnitude of the vector 2x2 –3x + in the inner product space C01 . 26. The distance between the vectors 2x2 –3x + and – x2 + x – 5 in the inner product space C01 . 27. The angle between the vectors 2x2 –3x + and – x2 + x – 5 in the inner product space C01 . 28. The magnitude of the vector ex in the inner product space C01 . 29. The distance between the vectors ex and x in the inner product space C01 . 30. The angle between the vectors ex and x in the inner product space C01 . 31. The magnitude of the vector sinx in the inner product space C– . 32. The distance between the vectors sinx and cosx in the inner product space C– . 33. The angle between the vectors sinx and cosx in the inner product space C– . 300 Chapter 7 Inner Product Spaces
34. Verify that the Cauchy-Schwarz inequality holds for the vectors ex and x in the inner product space C01 . 35. Verify that the Cauchy-Schwarz inequality holds for the vectors sinx and cosx in the inner product space C– . 36. Prove that ordinary multiplication in the set of real numbers R is an inner product on the vec- tor space . 37. Prove Theorem 7.3(a). 38. Prove Theorem 7.3(b). 39. Prove Theorem 7.3(c). 40. Prove Theorem 7.3(e). 41. Prove Theorem 7.3(f).
42. Let uv V , V an inner product space. Show that uv+ 2 –4uv– 2 = uv 43. Let uv V , V an inner product space. Show that uv+ uv– = u 2 – v 2 . 44. Let uv V , V an inner product space. Show that uv+ 2 = u 2 + v 2 if and only if uv = 0 . 45. Let u V , V an inner product space. Show that v V uv = 0 is a subspace of V. 46. (PMI) Let V be an inner product space.Use the principle of mathematical induction to show
that for any uv1 v2 vn V and any a1a2 am :
u a1v1 +++a2v2 amvn = a1 uv 1 +++a2 uv 2 am uv n and:
a1v1 +++a2v2 amvn u = a1 uv 1 +++a2 uv 2 am uv n
PROVE OR GIVE A COUNTEREXAMPLE
47. Let uvw V , V an inner product space. If u 0 and if uv = uw , then vw= .
48. Let uv V . If vw = uw for every w V , then uv= .
49. There exists an inner product on 3 for which 111 = 1 .
50. There exists an inner product on 3 for which 111 = 222 .
51. There exists an inner product on 3 for which 111 222 . 7.3 Orthogonality 301
7
§3. ORTHOGONALITY Having extended the concept of angles between vectors in n to vec- tors in inner product spaces, we can now extend the definition of orthogonality to those spaces: DEFINITION 7.9 Two vectors u and v in an inner product ORTHOGONAL VECTORS space V are orthogonal if uv = 0 . A set S of vectors in an inner product space ORTHOGONAL SET V is an orthogonal set if uv = 0 for every uv S , with uv .
EXAMPLE 7.11 Verify that: S = 2x2 + 2x – 1 –2x2 ++x 22 x2 –2x + is an orthogonal set in the inner product space
P2 of CYU 7.11, page 293, wherein 2 2 a2x ++a1xa0 b2x ++b1xb0
= a2b2 ++a1b1 a0b0
SOLUTION: All pairs of vectors from S are orthogonal: 2x2 + 2x – 1 –2x2 ++x 2 ==21– +22 + – 12 0 (check) 2x2 +22x – 1 x2 –2x + ==22 +21– +–1 2 0 (check) –2x2 ++2x 2 x2 –2x + ==–1 221+– +22 0 (check)
You can check directly that the set S in the above example is a lin-
early independent in P2 . In general:
THEOREM 7.7 If v1v2 vn is an orthogonal set of non- zero vectors in an inner product space V , then v1v2 vn is a linearly independent set in V. PROOF: Let: c1v1 ++c2v2 cnvn = 0 c1v1 ++c2v2 cnvn = 0 For each vi , 1 in , we have: Therem 7.4(a), page 277: vi c1v1 ++civi ++cnvn ==vi 0 0 c1 vi v1 ++ci vi vi ++cn vi vn = 0
vi vj = 0 if ij : ci vi vi = 0 Definition 7.5 (i), page 276: c = 0 each ci = 0 i 302 Chapter 7 Inner Product Spaces
CHECK YOUR UNDERSTANDING 7.17
Let v1v2 vm be an orthogonal set of vectors in an inner prod- uct space V, and let u V be such that uv i = 0 for 1 im . Answer: See page B-34. Prove that u is orthogonal to every vector in Spanv1v2 vm .
DEFINITION 7.10 A unit vector in an inner product space is UNIT VECTOR a vector v of magnitude 1.
NORMALIZATION To normalize a nonzero vector v in an inner product space simply mul- 1 tiply it by ------: v 1 1 ------v ==------v 1 v v CYU 7.13(a), page 294
DEFINITION 7.11 An orthonormal set of vectors in an inner ORTHONORMAL SET product space is a set of orthogonal unit vectors.
The standard basis S = e1e2 en of page 94 can easily be shown to be an orthonormal set in the Euclidean inner product space n . Moreover, for any v = c1 ci cn : vve= 1 e1 ++ve i ei ++ve n en since: ve i ==c1 ci cn 0 1 0 ci ith entry This nicety extends to any orthonormal basis in any inner product space:
THEOREM 7.8 If = v1v2 vn is an orthonormal basis in an inner product space V, then, for any v V :
vvv= 1 v1 +++vv 2 v2 vv n vn n
PROOF: Let v = cjvj . We show ci = vv i for 1 in : j = 1 n n 2 vv i ===== cjvj vi cj vj vi ci vi vi ci vi ci j = 1 j = 1
1 if ji= Exercise 46, page 300. vj vi = vi vi = 1 0 if ji 7.3 Orthogonality 303
CHECK YOUR UNDERSTANDING 7.18
Let = v1v2 vn be an orthonormal basis for an inner prod- uct space V. Show that for any a1v1 +++a2v2 anvn and n
w = b1v1 +++b2v2 bnvn , vw = aibi . Answer: See page B-34. i = 1 The following theorem spells out a procedure that can be used to con- struct an orthogonal basis in any given finite dimensional inner product space. Basically, the construction process is such that each newly added vector in an evolving basis is orthogonal to all of its predecessors.
THEOREM 7.9 Let = v1v2 vn be a basis for an inner product space V, GRAHM-SCHMIDT and let V be the following subspaces of V: PROCESS i
u1 = v1 V1 = Spanv1
u1 v2 u2 = v2 – ------u1 V2 = Spanv1 v2 u1 u1
u1 v3 u2 v3 u3 = v3 – ------u1 – ------u2 V3 = Spanv1v2 v3 . u1 u1 u2 u2 . . . u1 vn u2 vn un – 1 vn un = vn – ------u1 – ------u2 – – ------un – 1 Vn = Spanv1v2 vn u1 u1 u2 u2 un – 1 un – 1
Then, u1 ui is an orthogonal basis for Vi . In particular,
u1u2 un is an orthogonal basis for V. PROOF: By Induction on the dimension of V:
Since u1 = v1 and since u1 is an orthogonal set, the claim is seen to hold for Vi . Assume that u1 uk is an orthogonal basis for Vk , for kn .
We show that u1 uk + 1 is an orthogonal basis for Vk + 1 , where: u v u v 1 k + 1 k k + 1 uk + 1 = vk + 1 – ------u1 – – ------uk (*) u1 u1 uk uk
We are assuming that u1 uk is an orthogonal set. Conse- quently, to establish orthogonality of u1 uk + 1 , we need but show that ui uk + 1 = 0 for 1 ik : 304 Chapter 7 Inner Product Spaces
u1 vk + 1 uk vk + 1 ui uk + 1 = ui vk + 1 – ------u1 – – ------uk u1 u1 uk uk u v u v u v 1 k + 1 i k + 1 i k + 1 = ui vk + 1 – ------ui u1 – – ------ui ui – – ------ui uk u1 u1 ui ui ui ui u v ------i ------k + 1 = ui vk + 1 – ui ui since ui uj = 0 if ij ui ui
ui ui ===ui vk + 1 – ------ui vk + 1 ui vk + 1 –01 ui vk + 1 ui ui
Being an orthogonal set, u1 uk + 1 is linearly independent. To show that Span u1 uk + 1 = Spanv1v2 vk + 1 we need but show that Spanv1v2 vk + 1 Span u1 uk + 1 (why?). Let’s do it: k + 1 k + 1 k a v Spanv v i i 1 k + 1 a v = a v + a v i = 1 i i i i k + 1 k + 1 i = 1 i = 1 k Induction Hypothesis: = biui + ak + 1vk + 1 i = 1 k k uk vk + 1 from (*): = b u + a u + ------u i i k + 1k + 1 k uk uk i = 1 i = 1 k + 1 k + 1 aivi Spanu1 uk + 1 = ciui i = 1 i = 1
u1 v2 u2 v3 where: c1 = b1 + ak + 1------c2 = b2 + ak + 1------ ck + 1 = ak + 1 u1 u1 u2 u2
Note: To obtain an orthonormal basis for an inner product space, simply normalize the orthogonal basis generated by the Gram-Schmidt process.
EXAMPLE 7.12 Extend 120 to an orthogonal basis for the Euclidean inner product space 3 . 7.3 Orthogonality 305
SOLUTION: ONE APPROACH: Extend 120 to a basis for 3 : 120 100 001 :
v1v2 v3 and then apply the Grand-Schmidt process to the above basis:
u1 = 120 u v Multiplying any ui in the 1 2 120 100 u2 ==v2 – ------u1 100 – ------120 Gram-Schmidt process u1 u1 120 120 by a nonzero constant 1 4 2 will not alter that vectors ==100 – ---120 ---–--- 0 42– 0 21– 0 “orthogonality-feature,” 5 5 5 but will simplify subse- see margin quent calculations. u1 v3 u2 v3 u3 = v3 – ------u1 – ------u2 u1 u1 u2 u2 120 001 21– 0 001 = 001 – ------ 120 – ------ 21– 0 120 120 21– 0 21– 0 0 0 ==001 – ---120 – ---21– 10 001 5 5 The above process led us to the following orthogonal basis:
u1u2 u3 = 120 21– 0 001
ANOTHER APPROACH: Since we are dealing with a vector space of dimension 3, we can sim- ply roll up our sleeves and construct an orthogonal bases by “brute This brute force approach is not always practical. force.” First, add a nonzero vector, abc , to 120 with: Software, such as Maple and MATLAB, include the abc 120 = 0 Gram-Schmidt process as a ++2b 0 c = 0 a built-in procedure. Yes, the Gram-Schmidt process This can be done in many ways. One way: abc = 001 , works off of a basis for the brings us to the orthogonal set 120 001 . We still need inner product space, but another nonzero vector abc — one that is orthogonal to both that is not a problem: if you randomly choose n 120 and 001 : vectors in an n dimen- sional space, even if abc 120 ==0 and abc 001 0 n = 100 , there is little a +2b +0 c = 0 and 0 a ++00 b c = chance that those vectors a +2b = 0 and c = 0 end up being linearly dependent! How about abc = 630 ? Sure. Leading us to the orthogonal basis 120 001 630 306 Chapter 7 Inner Product Spaces
CHECK YOUR UNDERSTANDING 7.19 Apply the Gram-Schmidt Process to construct an orthonormal basis for the subspace S of the Euclidean inner product space of 4 spanned by the vectors 2110 , 1010 , 3120 , Answer: See page B-35. 0101 .
ORTHOGONAL COMPLEMENT The set of vectors orthogonal to any subspace W of 3 , denoted by the symbol W , is itself a subspace of 3 . More specifically:
W W 0 3 Line W passing through the origin. Plane passing through the origin with normal W. Plane W passing through the origin. Line passing through the origin orthogonal to W. 3 0 In a more general setting, for W a subspace of an inner product space ORTHOGONAL V we define the orthogonal complement of W to be: COMPLEMENT W = u V uw = 0 for every w W We then have: THEOREM 7.10 If W a subspace on an inner product space V, then: (i)W is a subspace of V. (ii) WW = 0 (iii) Every vector in V can be uniquely expressed as a sum of a vector in W and a vector in W .
(iv) If is a basis for W and is a basis W W for W , then is a basis for V. W W PROOF: (i) For u1 u2 W , r , and w W , we have:
ru1 + u2 w ===r u1 w + u2 w r 00+ 0 The result now follows from Theorem 2.13, page 61. (ii) Let v WW . Being in both W and W , vv = 0 . It fol- lows, from Axiom (i) of Definition 7.6 (page 292) that v = 0 . In this part of the theo- (iii) Let w1w2 wm be an orthonormal basis for W. For v V , rem we assume that W is let: finite dimensional (the v = vw w +++vw w vw w W result does, however, W 1 1 2 2 m m hold in general). We show that the vector vW = vv– W is in W by showing that
vW wi = 0 , for 1 im (see CYU 7.17): 7.3 Orthogonality 307
vW wi ==vv– W wi vw i – vW wi
= vw i – vw 1 w1 +++vw 2 w2 vw m wm wi
===vw i – vw i wi wi vw i –0vw i since w w = 0 for ij i j since wi wi = 1 At this point, we have shown that v can be expressed as a sum of a vector in W and a vector in W : vv= W + vW . Uniqueness of the decomposition follows from part (ii) of this theorem, and Exercise 43, page 67. As for (iv): CHECK YOUR UNDERSTANDING 7.20 Answer: See page B-35. Establish Theorem 7.10(iv) Lets highlight an important observation lurking within the proof of Theorem 7.10(iii):
THEOREM 7.11 Let w1w2 wm be an orthonormal basis for a subspace W of an inner product space V. For any v V there exist unique vectors v W and v W such that v W W v W vv= + vW , where Compare with Theorem W v = vw w +++vw w vw w 7.2, page 283. W 1 1 2 2 m m
W v W and vW = vv– W .
Note: vW is said to be the orthogonal projection v of v onto W, and we write: vW = projW .
EXAMPLE 7.13 Let W = Span = v1v2 vn , in the Euclidean inner product space 4 . (a) Find a basis for W . (b) Express 1234 as the sum of a vector in W and a vector in W . SOLUTION: (a) Since 1002 and 1110 are linearly inde- pendent, they constitute a basis for W. To say that abcd W is We know that W will to say that: turn out to be of dimen- sion 2. How? abcd 1002 = 0 and abcd 1110 = 0 a +02d = abc+ +0= d = –--a- ba= – – c 2 308 Chapter 7 Inner Product Spaces
Choosing a and c to be our free variables, we have: a W = aa– – c c –--- a c 2 First setting a = 0 and c = 1 , and then setting c = 0 and a = 2 leads us to the basis: 0 – 110 22– 01– for W . (b) One approach: Simply express 1234 as a linear combination of the basis 1002 1110 0 – 110 22– 01– :
1234 = a1002 ++b1110 c0 – 110 +d22– 01–
ab++2d = 1 bc–2–2d = 3 3 3 a ====--- b --- c --- d –1 bc+3= 2 2 2 2ad–4= Bringing us to: in W in W 3 3 3 1234 = ---1002 + ---1110 + ---0– 110 – 22– 01– 2 2 2
= 3--3- --3- 3 + –2--1- --3- 1 2 2 2 2
Another Approach: First apply the Gram-Schmidt method on 1002 1110 to obtain an orthonormal basis for W:
u1 = 1002 1110 1002 u = 1110 – ------1002 2 1002 1002 1 4 2 Since u2 is orthogonal ==1110 – ---1002 ---11–--- Or: 455– 2 to u , so is 5u 5 5 5 1 2 see margin Orthonormal basis for W: 1 1 w1 w2 = ------1002 ------455– 2 5 70 Applying Theorem 7.11 we know that
1234 = 1234 W + 1234 W where:
1234 W = 1234 w1 w1 + 1234 w2 w2 1 2 1 2 4 5 5 –2 4 5 5 –2 = 1234 ------00------------00------+ 1234 ------------------------5 5 5 5 70 70 70 70 70 70 70 70 9 1 2 21 4 5 5 –2 3 3 ==------------00------+ ------------------3------3 55 5 7070 70 70 70 2 2 and: 3 3 1 3 1234 ===1234 – 1234 1234 – 3------3 –2------1 W W 2 2 2 2 7.3 Orthogonality 309
Note that both approaches lead to the same decomposition, as must be the case: 3 3 1 3 1234 = 3------3 + –2------1 2 2 2 2 in W in W
CHECK YOUR UNDERSTANDING 7.21 Find the orthogonal projection of the vector v = 201 onto the subspace W = Span 101 120 of the Euclidean inner Answer: See page B-35. product space 3 . The shortest distance between a vector v in an inner product space V and Consider Example 7.3, any vector w in a subspace W of V turns out to be the distance between page 284. v and vW : THEOREM 7.12 Let W be a subspace of the inner product space V, and let v V . Then: v – proj v vw– for every w W W
vW
ROOF v P : Let vW = projW . For any w W : vw– 2 = vw– vw–
= vv– W + vW – w vv– W + vW – w Theorem 7.4(b), page 293: = vv– W vv– W +++vv– W vW – w vW – w vv– W vW – w vW – w
2 2 = vv– W +++vv– W vW – w vW – w vv– W vW – w Since vW – w W and vv– W W , the two middle terms in the above expression are 0, bringing us to: 2 2 2 vw– = vv– W + vW – w
To complete the proof we need but note that vW –0w .
CHECK YOUR UNDERSTANDING 7.22
Find the shortest distance between the vector v =33x2 + x and the 2 3 subspace W = Span x + 1 x + 1 in the inner product space P3 of CYU 7.11, page 293: 3 2 3 2 3 a0 ++a1xa2x +a3x b0 ++b1xb2x +b3x = aibi . Answer: 3 i = 0 310 Chapter 7 Inner Product Spaces
EXERCISES
Exercises 1-7. Determine if the given set of vectors is an orthogonal set in the given inner product space. If so, modify the set to arrive at an orthonormal set. 1.111 –12– 1 –101 in the Euclidean inner product space 3 . 2.111 –12– 1 –101 in the weighted inner product space of Example 7.8, page 292, with u1u2 u3 v1v2 v3 = 5u1v1 ++2u2v2 4u3v3 . 3.111 –12– 1 105 in the weighted inner product space of Example 7.8, page 292, with u1u2 u3 v1v2 v3 = 5u1v1 ++u2v2 u3v3 . 4.x2 ++2x 13 x2 + x – 511 x2 –58x + in the polynomial inner product space of CYU 7.11, page 293. 12 21– 00 5.1 1 in the inner product space of Exercise 17, page 298. 34 00 --- –--- 3 4
4 6. (Calculus Dependent) x2 –1---x + in the inner product space C01 of Exercise 24, 3 page 299. 4 7. (Calculus Dependent) x2 –1---x + in the inner product space C12 of Exercise 24, 3 page 299. 8. Use Theorem 7.8 to express 352 in the Euclidean inner product space 3 as a linear 1 1 combination of the vectors in the orthonormal basis ------210 001 ------120– . 5 5
9. Use Theorem 7.8 to express 2x2 + 3x – 1 in the polynomial inner product space of CYU 7.11, page 293, as a linear combination of the vectors in the orthonormal basis x2 x 1 x2 2x 1 x2 1 ------++------------– ------+ ------– ------. 3 3 3 6 6 6 2 2 10. Find all values of a for which 132 1a 1 is an orthogonal set in the Euclidean inner product space 3 . 11. Find all values of a and b for which 13b 1a 1 is an orthogonal set in the Euclid- ean inner product space 3 . 12. Find all values of a and b for which 11a –12b –1 –101 is an orthogonal set in the weighted inner product space of Example 7.8, page 292, with u1u2 u3 v1v2 v3 = 2u1v1 ++u2v2 u3v3 . 7.3 Orthogonality 311
13. Find all values of a and b for which 111 –12– 1 –101 is an orthogonal set in the weighted inner product space of Example 7.8, page 292, with u1u2 u3 v1v2 v3 = au1v1 ++bu2v2 u3v3. 14. Find all values of a, b, and c for which 111 –12– 1 105 is a n orthogonal set in the weighted inner product space of Example 7.8, page 292, with u1u2 u3 v1v2 v3 = au1v1 ++bu2v2 cu3v3. 15. (Calculus Dependent) Find all values of a and b for which ax2 + 1 – x + b is an orthog- onal set in the inner product space C01 of Exercise 24, page 299. 16. (Calculus Dependent) Find all values of a, and b for which x2 –1x + is an orthogonal set in the inner product space Cab of Exercise 24, page 299. Exercises 17-26. Find an orthonormal basis for the given inner product space. 17.Span201 –120 in the Euclidean inner product space 3 . 18.Span111 –12– 1 in the Euclidean inner product space 3 . 19.Span1111 –12– 12 in the Euclidean inner product space 4 20.Span1010 0202 2000 in the Euclidean inner product space 4 . 21.Span201 –120 in the weighted inner product space of Example 7.8, page 292, with u1u2 u3 v1v2 v3 = 2u1v1 ++3u2v2 u3v3 . 22.x2 + 1 x – 5 in the polynomial inner product space of CYU 7.11, page 293. 2x ++03yz– w = 23. The solution set of in the Euclidean inner product space 4 . 4x +03y –22z – w =
x ++03y – 2z w = 24. The solution set of in the Euclidean inner product space 4 . 4x +03y – 2z = 25. (Calculus Dependent) Spanx2 2x + 1 in the inner product space C01 of Exercise 24, page 299. 26. (Calculus Dependent) Spanx3x + 1 x2 – 1 in the inner product space C–11 of Exercise 24, page 299.
27. Find an orthonormal basis for a 2a 0 aR in the Euclidean inner product space 3 .
28. Find an orthonormal basis for aba– b 2b ab in the Euclidean inner product space 4 . 29. Find an orthonormal basis for a2aca– c ac in the Euclidean inner product space 4 . 30. Find an orthonormal basis for abc abc++= 0 in the weighted inner product space
of Example 7.8, page 292, with u1u2 u3 v1v2 v3 = –5u1v1–2u2v2 + u3v3 . 31. Find an orthonormal basis for ax3 +++ab+ x2 cx 2a abc R in the polynomial inner product space of CYU 7.11, page 293. 312 Chapter 7 Inner Product Spaces
Exercise 32-36. (a) Find a basis for the orthogonal complement of the given Euclidean inner product subspace W . (b) Express the given vector v as a sum of a vector in W and a vector in W . (c) Determine the distance from v to W. 32.W = Span10 11 , v = 13 . 33.W = Span102 , v = 13– 2. 34.W = Span102 13– 2 , v = 111 . 35.W = Span1020 0101 , v = 4 – 133 . 36.W = Span1023 2101 1101 , v = 13– 22 . 37. Find a basis for the orthogonal complement of the subspace W = Span201 –120 in the weighted inner product space of Example 7.8, page 292, with u1u2 u3 v1v2 v3 = u1v1 ++2u2v2 u3v3 , and express the vector v = 12– 1 as a sum of a vector in W and a vector in W .
38. Find a basis for the orthogonal complement of the subspace W = Spanx2 + 1 x – 5 in the polynomial inner product space of CYU 7.11, page 293, and express the vector v = 2x2 – x as a sum of a vector in W and a vector in W .
39. Prove that the standard basis e1e2 en of page 94 is an orthonormal basis in the Euclidean inner product space n . 40. Prove that xnxn – 1 x 1 is an orthonormal basis in the polynomial inner product space of CYU 7.11, page 277. 41. Let V be an inner product space. Prove that V = 0 and that 0 = V .
42. Let W = Spanw1w2 wm in an inner product space V. Prove that v W if and only if vw i = 0 for all wi , 1 im .
43. Let v1v2 vk vk + 1 vm be an orthogonal set in an inner product space V. Show that if w Spanv1v2 vk and zv k + 1 vm , then wz = 0 .
44. Let W be a subspace in an inner product space V. Prove that W = W . 45. Let w be a vector in an inner product space V of dimension n. Prove that w = v V vw = 0 is a subspace of V of dimension n – 1 . 46. Let S be a subset of an inner product space V. Prove that S = v V vw = 0 for all w S is a subspace of V. 47. Let S be a subspace of an inner product space V of dimension n. Prove that dimS + dimS = n
48. Let = v1v2 vn be an orthonormal basis in an inner product space V. Show that for any uv V uv = u v . (See Definition 5.9, page 178.) 7.3 Orthogonality 313
Exercises 51-59 (Orthogonal Matrices) AM nn is an orthogonal matrix if the columns of A is an orthonormal set in the Euclidean inner product space n . (Orthogonal matrices would better have been named “orthonormal matrices,” no?) 49. Sow that the following are equivalent:
(i) AM nn is orthogonal. (ii) ATAI= . (See Exercise 19, page 162). (iii)AXX= for every X n (iv)AX AY = XY for every XY n . 50. Prove that every orthogonal matrix is invertible, and that its inverse is also orthogonal. 51. Prove that a product of orthogonal matrices (or the same dimension) is again orthogonal. 52. Prove that if A is orthogonal, then detA = 1 . 53. Prove that if A is orthogonal then the rows of A also constitute an orthonormal set. 54. Prove that if A is orthogonal, and if B is equivalent to A, then B is also orthogonal.
55. Prove that every 22 orthogonal matrix is of the form ab– or ab where ba ba– a2 +1b2 = .
56. Show that every 22 orthogonal matrix is of the form cos –sin or cos sin . sin cos sin –cos 57. Show that every 22 orthogonal matrix corresponds to either a rotation or a reflection about a line through the origin in 2 .
58. (a) Prove that the null space of AM mn is the orthogonal complement of the row space of A. T (b) Prove that the null space of A Mmn is the orthogonal complement of the column space of A. (See Exercise 19, page 162.)
(c) Verify directly that the null space of A = 1320 is the orthogonal complement of the –0121 row space of A. (d) Verify directly that the null space of AT = 1 320 is the orthogonal complement of –1 012 the column space of A. (See Exercise 19, page 162.) 314 Chapter 7 Inner Product Spaces
59. (Bessel’s Equality) Let v1v2 vn be an orthonormal basis for an inner product space n 2 2 V. Prove that for any w V : wv i = w . i = 1
PROVE OR GIVE A COUNTEREXAMPLE
60. If v1v2 vm is an orthogonal set in an inner product space V, then
a1v1a2v2 amvm is an orthogonal set for all scalars a1a2 am .
61. If v1v2 vm is an orthonormal set in an inner product space V, then
a1v1a2v2 amvm is an orthonormal set for all scalars a1a2 am .
62. Let W be a subspace of an inner product space V. If wv = 0 with w W , then v W .
63. Let v1v2 vk vn be a basis for an inner product space V such that each vj for kj n is orthogonal to every vm for 1 mk . If W = Spanv1v2 vk , then W = Spanvk + 1 vn .
64. Let v1v2 vk vn be an orthogonal basis for an inner product space V. If W = Spanv1v2 vk hen W = Spanvk + 1 vn . 7.4 The Spectral Theorem 315
7
§4. THE SPECTRAL THEOREM
We begin by recalling Definition 6.13 of page 264: In other words, the ith row The transpose of a matrix A = aij Mmn is the th of A is the i column of T matrix A = bij Mnm , where bij = aji AT . For example:
12 103 T If A = A = 04 DEFINITION 7.12 T 245 A Mnn is symmetric if A = A. 35 SYMMETRIC MATRIX As it turns out:
THEOREM 7.13 If A Mnn is a (real) symmetric matrix, then its eigenvalues are real numbers. An outline of a proof for the above theorem, which involves a bit of complex number terminology, is relegated to the exercises.
2 1 n n We remind you that we are using to denote Mn 1 . For XY 3 4 = 235 140 5 0 we now define XY to be the dot product of the corresponding vertical = 21 + 34 + 50 n-tuples (see margin). It is easy to show that n ,with XY defined to = 14 be XY , is an inner product space (see Definition 7.6, page 292). Note, that the above dot product can also be effected by means of matrix mul- 2 1 235 1 tiplication: = 3 4 4 XY = XTY (see margin) 5 0 0 = 21 + 34 + 50 THEOREM 7.14 Any two eigenvectors in the inner product ==14 14 space n corresponding to distinct eigen- values of a symmetric matrix A Mnn are orthogonal.
PROOF: Let 1 2 be distinct eigenvalues of a symmetric matrix A Mnn , with corresponding eigenvectors XY , so that: AX = 1X and AY = 2Y We show that XY = 0 : T 1XY ==1X Y AX Y Exercise 19(f), page 162 = XTAT Y Definition 7.12: ==XTA YXTAY T T ==X 2Y 2X Y = XY 2 Then: 1XY = 2XY 1 – 2 XY = 0
Since 1 2 , XY = 0 ; which is to say: X and Y are orthogonal. 316 Chapter 7 Inner Product Spaces
EXAMPLE 7.14 Verify the result of Theorem 7.14 for the sym- metric matrix: 11–0 A = –211 – 01–1
SOLUTION: To determine the eigenvalues of A, we turn to Theorem 6.8, page 219, and calculate the determinant of A – I :
1 –1 –0 detA – I ==det –21 –1 – –– 1 – 3 01–1– details omitted
We see that A has three distinct eignevalues: 1 = 0 , 2 = 1 and 3 = 3 , with corresponding eigenspaces:
11–0 Note: E0 ==nullA – 0I null –211 – =aaa aR 11–0 10– 1 01–1 rref –211 – = 01– 1 margin 01–1 00 0 01–0 01–0 101 E1 ==nullA – 1I null –111 – =–0bb bR rref –111 – = 010 01–0 01–0 000 –12 –0 –12 –0 10– 1 E3 ==nullA – 3I null –11 –1– =c–2c c cR rref = –11 –1– 01 2 01–2– 01–2– 00 0 You can easily verify that for 1 ij 3 , every vector in Ei is orthogonal to every vector Ej . For example: aaa c–2c c ==ac – 2ac +0ac
CHECK YOUR UNDERSTANDING 7.23
211 Verify the result of Theorem 7.13 for the matrix A = 121 . Answer: See page B-37. 112
THEOREM 7.15 A Mnn is symmetric if and only if AX Y = XAY
for all vectors XY Rn . 7.4 The Spectral Theorem 317
PROOF: If A is symmetric, then: AX Y ==AX TYXTAT YX =TATY e e 2 3 symmetry: T Exercise 19(f), page 161 ==X AY XAY Conversely, assume that Aa= is such that 135 0 0 3 0 ij 267 1 0 = 6 0 (*) AX Y = XAY 4 5 2 0 1 5 1 th Turning to the n-tuple ek , with 1 as its k entry and 0 elsewhere, we = 5 show that A is symmetric, by showing that aij = aji : a32 aij = Aej ei (see margin) By (*): = ej Aei Theorem 7.1(ii), page 279: ==Aei ej aji
CHECK YOUR UNDERSTANDING 7.24
111 (a) Let A = 121 . Show directly that AX Y = XAY for 113 every XY 3 .
(b) Write down an arbitrary non-symmetric matrix AM 33 and Answer: See page B-37. exhibit XY 3 for which AX Y XAY .
SYMMETRIC OPERATORS As you know, there is an intimate relation between matrices and lin- ear maps; bringing us to: DEFINITION 7.13 Let V be an inner product space. A linear SYMMETRIC OPERATOR operator T: VV is symmetric if Compare with Theorem 7.15. Tv w = v Tw for all vectors vw V .
Here is the linear operator version of Theorem 7.14: THEOREM 7.16 Let T: VV be a symmetric linear opera- tor on an inner product space V. If v1 and v2 are eigenvectors associated with distinct eigenvalues 1 and 2 , then v1 and v2 are orthogonal. 318 Chapter 7 Inner Product Spaces
PROOF: Tv1 v2 –0Tv1 v2 =
Definition 7.13: Tv1 v2 –0v1 Tv2 =
Definition 6.5, page 224: 1v1 v2 –0v1 2v2 =
Theorem 7.4, page 293: 1 v1 v2 –02 v1 v2 =
1 – 2 v1 v2 = 0
Since 1 2 , v1 v2 = 0 .
The matrix representation T of a symmetric linear operator need not be symmetric for every basis (see Exercise 18). However: THEOREM 7.17 If T: VV is a symmetric linear operator
on an inner product space V, then T is a symmetric matrix for any orthonormal basis of V.
PROOF: Employing Theorem 7.15, we show that for any two (column) n n-tuples vv= 1v2 vn , ww= 1w2 wn in
T v w = v T w :
T v w = T v w
Theorem 5.22, page 180: = Tv w CYU 7.18, page 303: = Tv w By symmetry: ==v Tw v Tw =v T w
CHECK YOUR UNDERSTANDING 7.25
For 3 the Euclidean inner product space, let T: 3 3 be the linear transformation given by: Tabc = ab– – a + 2bc– – b + c (a) Show that T is symmetric.
(b) Verify that T is symmetric for the orthonormal basis Answer: See page B-38. = 010 001 100 of 3 . Theorem 7.9, page 303, assures us that every finite dimensional inner product space contains an orthonormal basis . It follows that every symmetric linear operator T: VV has a symmetric matrix represen- tation T . Indeed, can be chosen so that T is a diagonal matrix: 7.4 The Spectral Theorem 319
Note that V contains an THEOREM 7.18 A linear operator T: VV on an inner orthonormal basis if and THE SPECTRAL product space V is symmetric if and only if only if it contains a nor- THEOREM mal basis. V contains an orthonormal basis of eigen- vectors of T.
PROOF: Assume that = v1v2 vn is an orthonormal basis of eigenvectors of T with corresponding eigenvalues 12 n . We show that T is symmetric: n n Let vw V , with v = aivi and w = bivi . Then: i = 1 i = 1 n n
Tv w = aiTvi bivi i = 1 i = 1 n n
= aivi bivi i = 1 i = 1 n n n
== aibi aivi bivi =v Tw i = 1 i = 1 i = 1 1 if ij= Since vi vj = [along with Theorem 7.4, page 293] 0 if ij To establish the converse, we apply the Principle of Mathematical Induction on the dimension n of V: v I. Let dimV = 1 . For any v 0 , ----- is an orthonormal v v v basis for V. Since T ----- V , and since ----- is a basis for V, v v v v T ----- = ----- for some . v v II. Assume that the claim holds for dimV = k . III. We establish validity for dimV = k + 1 : Applying the Grahm-Schmidt process, we can obtain an ortho- normal basis for V. By Theorem 7.17, T is symmetric. Let be a (real) eigenvalue of T (Theorem 7.13). Let n v be an eigenvector associated with , and let vn be such
that vn = v . Then: 320 Chapter 7 Inner Product Spaces
T v = v T vn = vn Theorem 5.21, page 181: Tvn = vn
Tvn = vn
Let W = Spanvn . Theorem 7.10, page 306 tells us that: W = v V vv n = 0 is a subspace of V of dimension k – 1 . Nothing that for any v W :
Tv vk ====v Tvk v vk vv k 0 we conclude that Tv W for every v W . Let TW:W W denote the restriction of T to W . The lin-
earity of T assures us that TW is linear. Moreover, since Tv w = v Tw holds for every vw V , it must cer- tainly hold for every vw W . Invoking the induction hypothesis (II) to the symmetric linear operator TW:W W , we let v1v2 vk – 1 denote an orthonor- mal basis of eigenvectors of TW . Since each vector in that basis is orthogonal to the eigenvector vk , the set
vk v1v2 vk – 1 ------is seen to be an orthonormal basis for V vk consisting of eigenvectors.
CHECK YOUR UNDERSTANDING 7.26 Find an orthonormal basis of eigenvectors for the symmetric linear Answer: See page B-38. operator Tabc = ab– –2a + bc– – b + c of CYU 7.25.
MATRIX VERSION OF THE SPECTRAL THEOREM Here is the link between symmetric matrices and symmetric linear operators:
THEOREM 7.19 AM nn is symmetric if and only if n n TA: given by TAX = AX is a symmetric linear operator. 7.4 The Spectral Theorem 321
n PROOF: Assume that AM nn is symmetric. For any XY : Recall that for XY n : TAX Y ==TAX Y AXY XY = XY Theorem 7.15: (see page 307) ==X AY X TAY =X TAY
Conversely, if TA is symmetric, then:
AXY ==TAX Y TAX Y
Definition 7.13: ===X TAY X TAY X AY
DEFINITION 7.14 AM nn is an orthogonal matrix if the ORTHOGONAL AND columns of A constitute an orthogonal set in ORTHONORMAL the Euclidean inner product space n . MATRICES A is an orthonormal matrix if its columns constitute an orthonormal set in n .
THEOREM 7.20 Every orthogonal matrix is invertible.
PROOF: If AM nn is orthogonal, then the columns of A constitute a linearly independent set of vectors in n (Theorem 7.7, page 301), and are therefore a basis for n (Theorem 3.11, page 99). The result now follows from Exercise 37, page 175. Yes, every orthogonal matrix is invertible; but more can be said for orthonormal matrices:
THEOREM 7.21 AM nn is orthonormal if and only if A–1 = AT
T T PROOF: Let Aa= ij , A = aij , and AA = cij . Then: n n
cij == aiaj aiaj = 1 = 1
the dot product of the ith column of A with the jth column of A It follows that A–1 = AT , if and only if AAT = I , if and only if n 1 if ij= aiaj = 0 if ij = 1 if and only if the columns of A constitute an orthonormal set in n .
CHECK YOUR UNDERSTANDING 7.27
Prove that the product of any two orthonormal matrices in Mnn is Answer: See page B-39. again orthonormal. 322 Chapter 7 Inner Product Spaces
Note: In the literature the DEFINITION 7.15 AM is if there exists an orthonor- term orthogonally diago- nn nalizable is typically ORTHONORMALLY mal matrix P and a diagonal matrix D such used to refer to what we DIAGONALIZABLE that: are calling . P–1AP= D As it turns out:
THEOREM 7.22 AM nn is orthonormally diagonalizable THE SPECTRAL if and only if it is symmetric. THEOREM n n PROOF: If AM nn is symmetric, then TA: given by TAX = AX is a symmetric operator (Theorem 7.19). Employing Theorem 7.18, we chose an orthonormal basis = X1X2 Xn of eigenvectors of TA with corresponding eigenvalues 12 n . n For S = e1e2 en the standard basis of we have: See Theorem 5.26, page 193 TA = I STA SSI S (*) We now show that: (1), (2), (3) and (*) tell us (1) T is a diagonal matrix with the s along its diagonal. that: A i P–1AP is a diagonal matrix, (2) TA SS = A with PI= an orthonor- S (3)The columns of I are the X s — an orthonormal set. mal matrix. S i In particular: (1):A consequence of Definition 5.10, page 179, and the fact that A is ! TAXi ==AXi iXi . Moreover: P–1AP= PTAP (2):A consequence of Definition 5.10 and the fact that TAei S is th th with Xi the i column of P. the i column of A. (3):A consequence of Definition 5.10 and the fact that
IXi S ==Xi S Xi . Conversely, assume that A is orthogonally diagonalizable. Let P be an orthogonal matrix and D a diagonal matrix with: P–1AP= D APDP= –1 Theorem 7.21: APDP= T By Then: AT = PDPT T Exercise 24(f), page 1643: = PT TDTPT Exercise 24(a), page 164: ===PDTP–1 PDP–1 A Since every diagonal matrix is symmetric Since AT = A , A is symmetric. 7.4 The Spectral Theorem 323
CHECK YOUR UNDERSTANDING 7.28 Find an orthonormal diagonalization for the symmetric matrix: 2 1 1 121 Answer: See page B-40. 112 324 Chapter 7 Inner Product Spaces
EXERCISES
Exercises 1-4. Verify that the given matrix is orthonormal.
1 1 1 001 3 2 6 –------3------–------2 3. 100 7 7 7 1. 2 2. 2 2 1 1 010 4. 2 6 3 --1------3- –------–------2 2 2 2 7 7 7 6 3 --- –--2- --- 7 7 7
Exercises 5-8. Verify that the given matrix AM nn is symmetric. Show directly that Av w = v Aw for every vw n .
5.51 6. 73– 010 21–0 15 –43 7.101 8. –311 012 012 Exercises 1-4. Find an orthonormal diagonalization for the symmetric matrix of: 9. Exercise 5. 10. Exercise 6. 11. Exercise 67 12. Exercise 8.
Exercises 9-12. Verify that the given linear operator T: n n on the Euclidean (dot product) inner product space Rn is symmetric. Determine T where S denotes the standard basis in Sn Sn n n . 13.Tab = 2ab+ a + 2b 14.Tab = – a +33b a + 5b
15.Tabc = a ++2b 3c 2ab+ 3a + 2c
16.Tabc = a ++22bc a + 2ba + 3c
17. Verify that Tab = 3a 2ab+ is a symmetric operator on the weighted inner product 1 space R2 with ab cd = 4ac+ bd . Verify that = ---0 01 is an orthonor- 2
mal basis in this inner product space, and determine T . 18. (a) Verify that Tax2 ++bx c = a ++2bcx2 ++2ab x 3a + 2c is a symmetric oper- 2 2 ator on the standard inner product space P2 : ax ++bx c ax ++bx c = aa+ bb + cc . (b) Use the Grahm-Schmidt process of page 303 on the basis = x2 ++x 1 x + 1 x2 + 1 1 1 1 1 1 1 1 to arrive at the orthonormal basis = ------x2 ++------x ------– x2 ++------– ---x + --- . Verify that 3 3 3 x 2 4 4 T is not symmetric, and that T is symmetric. 7.3 The Spectral Theorem 325
19. Let 2 denote the standard Euclidean dot product inner product space. Find a symmetric lin-
2 2 32 ear operator T: and a basis for which T = . 21 20. Let 2 denote the weighted inner product space R2 with ab cd = 3ac + 2bd . Find
2 2 32 a symmetric linear operator T: and a basis for which T = . 21
21. Let P1 denote the standard inner product space P1 : ax+ b ax+ b = aa+ bb + cc . 32 Find a symmetric linear operator T: P1 P1 and a basis for which T = 21
T T 22. Show that for any A Mnn both AA+ and AA are symmetric.
23. Show that if AB Mnn are orthonormally diagonalizable, then so is: (a) cA for every c . (b) AB+ (c) A2
n 24. (PMI) Show that if A Mmm is orthonormally diagonalizable, then so is A for any posi- tive integer n.
25. (PMI) Show that if Ai Mmm is orthonormally diagonalizable for 1 in , then so is A1 +++A2 An .
26. Show that if A Mmm is an invertible orthonormally diagonalizable matrix, then so is A–1 . 27. Prove that if A is a real symmetric matrix, then the eigenvalues of A are real. Suggestion: For Av = v , show that Av = v , where here denotes the (complex) conju- gate or and v is the n-tuple obtained by taking the conjugate of each entry in the n-tuple v. Proceed to show that vTv = vTv .
PROVE OR GIVE A COUNTEREXAMPLE
T 28. If A Mnn is a symmetric matrices, then so is A .
–1 29. If A Mnn is a symmetric matrices, then so is A .
30. If AB Mnn are symmetric matrices, then so is AB+ .
31. If AB Mnn are symmetric matrices, then so is AB . 326 Chapter 7 Inner Product Spaces
32. If AB Mnn are orthonormally diagonalizable, then so is AB .
T 33. If A Mnn is orthonormally diagonalizable, then so is A .
–1 34. If A Mnn is orthonormally diagonalizable, then so is A . 35. Let V be an inner product space. If T: VV is a symmetric operator, then so is cT for every c . 36. Let V be an inner product space. If T: VV and L: VV are symmetric operators, then so is TL+ . 37. Let V be an inner product space. If T: VV and L: VV are symmetric operators, then so is LT . Chapter Summary 327
CHAPTER SUMMARY
OT RODUCT D P The dot product of u = u1u2 un and v = v1v2 vn , denoted by uv , is the real number:
uv = u1v1 +++u2v2 unvn
PROPERTIES Let uvw n , and r . Then: positive-definite property: (i)vv 0 , and vv = 0 only if v = 0
commutative property: (ii) uv = vu
homogeneous property: (iii) ruv = ruv
distributive property: (iv) uv+ w = uw + vw
n NORM IN The norm of a vector v = v1v2 vn , denoted by v , is given by: vvv= Denotes length of vector.
ANGLE BETWEEN The angle between two nonzero vectors uv n is given by: VECTORS –1 uv = cos ------uv
ORTHOGONAL VECTORS Two vectors u and v in n are orthogonal if uv = 0 .
VECTOR Let v n and let u be any nonzero vector in n . Then: DECOMPOSITION vv= – projuv + projuv where: vu proj v = ------u and v – proj v proj v = 0 u uu u u
INNER PRODUCT SPACE An inner product on a vector space V is an operator which assigns to any two vectors, u and v in V, a real number uv , satisfying the following four axioms: positive-definite axiom: (i)vv 0 , and vv = 0 only if v = 0 commutative axiom: (ii) uv = vu
homogeneous axiom: (iii) ruv = r uv
distributive axiom: (iv) uv+ w = uw + vw 328 Chapter 7 Inner Product Spaces
PROPERTIES For every u, v, and w in an inner product space V: (a) 0 v ==v 0 0 (b) uv + w = uv + uw (c) u rv = r uv (d) uv– w = uw – vw (e) uv – w = uv – uw (f) –v w ==vw – – vw
NORM AND The norm (or magnitude) of a vector v in an inner product space V, DISTANCE denoted by v , is given by: vvv= The distance between two vectors u and v in V is given by uv– .
CAUCHY-SCHWARZ For any two vectors u and v in an inner product space: INEQUALITY uv uv
PROPERTIES Let V be an inner product space. For all uv V and r : (a)v 0 , and v = 0 if and only if v = 0 (b) rv = r v (c) uv+ uv+
ANGLE BETWEEN The angle between two nonzero vectors u and v in an inner prod- VECTORS uct space is given by: –1 uv = cos ------uv
ORTHOGONAL VECTORS Two vectors u and v in an inner product space V are orthogonal if uv = 0 . ORTHOGONAL SET A set S of vectors in an inner product space V is an orthogonal set if uv = 0 for every uv S , with uv .
THEOREM If v1v2 vn is an orthogonal set of non-zero vectors in an inner
product space V , then v1v2 vn is a linearly independent set in V.
UNIT VECTOR A unit vector in an inner product space is a vector v of magnitude 1.
ORTHONORMAL SET An orthonormal set of vectors in an inner product space is an orthog- onal set of unit vectors. Chapter Summary 329
THEOREM If = v1v2 vn is an orthonormal basis in an inner product space V, then, for any v V :
vvv= 1 v1 +++vv 2 v2 vv n vn
GRAHM-SCHMIDT An algorithm (page 303) for generating an orthogonal base in any finite PROCESS dimensional inner product space.
ORTHOGONAL The orthogonal complement of a subspace W of an inner product COMPLEMENT space: W = u V uw = 0 for every w W
PROPERTIES If W a subspace on an inner product space V, then: (i)W is a subspace of V. (ii) WW = 0 (iii) Every vector in V can be uniquely expressed as a sum of a vec- tor in W and a vector in W . (iv) If is a basis for W and is a basis for W , then W W W W is a basis for V.
VECTOR Let w1w2 wm be an orthonormal basis for a subspace W of an DECOMPOSITION inner product space V, and let v V . Then, there exists a unique vec- tor w W and u W such that: v vw= + u u where:
w = vw 1 w1 +++vw 2 w2 vw m wm w W and: u = vw– .
SYMMETRIC MATRIX T AM nn is symmetric if A = A .
THEOREMS If AM nn is a (real) symmetric matrix, then its eigenvalues are real.
Any two eigenvectors in the inner product space n corresponding to
distinct eigenvalues of a symmetric matrix AM nn are orthogo- nal.
AM nn is symmetric if and only if Av w = v Aw for all vectors vw Rn (in column form). 330 Chapter 7 Inner Product Spaces
SYMMETRIC Let V be an inner product space. A linear operator T: VV is sym- OPERATOR metric if Tv w = v Tw for all vectors vw V .
THEOREMS Let T: VV be a symmetric linear operator on an inner product space V. If v1 and v2 are eigenvectors associated with distinct eigen- values 1 and 2 , then v1 and v2 are orthogonal. If T: VV is a symmetric linear operator on an inner product space
V, then T is a symmetric matrix for any orthonormal basis
= o1o2 on of V.
SPECTRAL THEOREM (Linear Operator) A linear operator T: VV on an inner product space V is symmetric if and only if V contains an orthonormal basis of eigenvectors of T.
(Matrix) AM nn is symmetric if and only if there exists a diago- nal matrix D and a matrix P with columns an orthonormal set in n such that P–1AP= D . PRINCIPLE OF MATHEMATICAL INDUCTION A-1
APPENDIX A PRINCIPLE OF MATHEMATICAL INDUCTION We introduce a most powerful mathematical tool, the Principle of Mathematical Induction (PMI). Here is how it works: (PMI) Let Pn denote a proposition that is either true or false, depend- ing on the value of the integer n. If: I.P1 is True. And if, from the assumption that: II. Pk is True one can show that: III.Pk+ 1 is also True. then the proposition Pn is valid for all integers n 1 Step II of the induction procedure may strike you as being a bit strange. After all, if one can assume that the proposition is valid at nk= , why not just assume that it is valid at nk= + 1 and be done with it? Well, you can assume whatever you want in Step II, but if the proposition is not valid for all n you simply are not going to be able to demonstrate, in Step III, that the proposition holds at the next value of n. Its sort of like the domino theory. Just imagine that the propositions P1 P2 P3 Pk Pk+ 1 are lined up, as if they were an infinite set of dominoes:
P(1 P(2 P P P(6) P(7) P(8) P(9) P(10) ) ) (3) (4) P(5) ......
If you knock over the first domino (Step I), and if when a domino falls (Step II) it knocks down the next one (Step III), then all of the domi- noes will surely fall. But if the falling kth domino fails to knock over the next one, then all the dominoes will not fall.
The Principle of Mathemati- To illustrate how the process works, we ask you to consider the sum cal Induction might have been of the first n odd integers, for n = 1 through n = 5 : better labeled a Principle of Mathematical Deduction; for: n Sum of the first n odd integers Sum n Sum Inductive reasoning is a pro- 11 1 1 cess used to formulate a 2 1 + 3 4 2 4 hypotheses or conjecture, 1 + 3 + 5 9 3 9 while deductive reasoning is 4 16 a process used to rigorously 1 + 3 + 5 + 7 4 16 establish whether or not the 1 + 3 + 5 + 7 + 9 25 5 25 conjecture is valid. 6 ? Figure 1.1 A-2 Principle of Mathematical Induction
Looking at the pattern of the table on the right in Figure 1.1, you can probably anticipate that the sum of the first 6 odd integers will turn out to be 62 = 36 , which is indeed the case. In general, the pattern cer- tainly suggests that the sum of the first n odd integers is n2 ; a fact that we now establish using the Principle of Mathematical Induction.
Let Pn be the proposition that the sum of the first n odd integers equals n2 . I. Since the sum of the first 1 odd integers is 12 , P1 is true. The sum of the first 3 odd 2 integers is: II. Assume Pk is true; that is: 135+++ +2k – 1 = k see margin 135++ 2 3 – 1 III. We show that Pk+ 1 is true, thereby completing the proof: The sum of the first 4 odd integers is: the sum of the first k + 1 odd integers
1357+++ 2 4 – 1 2 2 Suggesting that the sum of 135+++ +2k – 1 + 2k + 1 ==k + 2k + 1 k + 1 the first k odd integers is: induction hypothesis: Step II 13++ +2k – 1 (see Exercise 1).
EXAMPLE 1.1 Use the Principle of Mathematical Induction to establish the following formula for the sum of the first n integers: nn+ 1 123+++ +n = ------(*) 2
SOLUTION: Let Pn be the proposition: nn+ 1 123+++ +n = ------ 2 11+ 1 I.P1 is true: 1 =------ Check! 2 kk+ 1 II. Assume Pk is true: 123+++ +k = ------2 III. We are to show that Pk+ 1 is true; which is to say, that (*) holds when nk= + 1 : k + 1 k + 1 + 1 k + 1 k + 2 123+++ ++kk+ 1 ==------------------- 2 2 Let’s do it: 123+++ ++kk+ 1 = 123+++ +k + k + 1 kk+ 1 induction hypothesis: = ------+ k + 1 2 ==k------k + 1------+ 2k------+ 1------k + 1 ------k + 2 2 2
PRINCIPLE OF MATHEMATICAL INDUCTION A-3
The “domino effect” of the Principle of Mathematical Induction need not start by knocking down the first domino P1 . Consider the fol- lowing example where domino P0 is the first to fall. EXAMPLE 1.2 Use the Principle of Mathematical Induction to establish the inequality n 2n for all n 0 .
SOLUTION: Let Pn be the proposition n 2n . I.P0 is true: 02 0 , since 20 = 1 . II. Assume Pk is true: k 2k III: We need to show that III. We show Pk+ 1 is true: n n 2 holds for k +221 2k + 12 k + 2k ==k 2k + 1 nk= + 1 ; which is to II k say, that: k + 12 k + 1 : 12
Recall that:. EXAMPLE 1.3 Use the Principle of Mathematical Induction to 2 n!12= n show that n! n for all integers n 4 .
SOLUTION: Let Pn be the proposition n! n2 :
I.P4 is true: 4!1234== 24 42 . II. Assume Pk is true: k! k2 (for k 4 ) III. We show Pk+ 1 is true; namely, that k + 1 ! k + 1 2 : k + 1 ! = k!k + 1 k2k + 1 II Now what? Well, if we can show that k2k + 1 k + 1 2 , then we will be done. Let’s do it: Since k 4 (all we need here is that k 2 ): k2 k + 1 Multiplying both sides by the positive number k + 1 : k2k + 1 k + 1 2 . A-4 Principle of Mathematical Induction
Our next application of the Principle of Mathematical Induction involves the following Tower of Hanoi puzzle: Start with a number of washers of differing sizes on spindle A, as is depicted below:
A B C The objective of the game is to transfer the arrangement cur- rently on spindle A to one of the other two spindles. The rules are that you may only move one washer at a time, without ever placing a larger disk on top of a smaller one. EXAMPLE 1.4 Show that the tower of Hanoi game is winna- ble for any number n of washers.
SOLUTION: If spindle A contains one washer, then simply move that washer to spindle B to win the game (Step I). Assume that the game can be won if spindle A contains k washers (Step II —the induction hypothesis). We now show that the game can be won if spindle A contains k + 1 washers (Step III): Just imagine that the largest bottom washer is part of k + 1 wahsers the base of spindle A. With this sleight of hand, we are looking at a situation consisting of k washers on a modified spindle A (see margin). By the induction . . k washers hypothesis, we can move those k washers onto spindle } B. We now take the only washer remaining on spindle { new
base A (the largest of the original k + 1 washers), and move it to spindle C, and then think of it as being part of the base of that spindle. Applying the induction hypotheses one more time, we move the k washers from spindle B onto the modified spindle C, thereby winning the game.
CYU SOLUTIONS B-1
APPENDIX B CHECK YOUR UNDERSTANDING SOLUTIONS CHAPTER 1 MATRICES AND SYSTEMS OF LINEAR EQUATIONS
9 10713 100--- 1 10 7 13 5 10 7 13 ------R R 3R + R R 4 15 1 3 01–4 3– 3 2 2 010 --- –7R3 + R1 R1 4 CYU 1.1 01–4 3– 5 010--- 8 5 0015 24 00 1 --- 8 5 001 --- 8 5 001--- 5
CYU 1.2 (a) Yes (b) No [fails (ii)] (c) Yes (d) Yes
x y z x y z xyz++= 6 11 1 6 1001 x = 1 CYU 1.3 rref 3x +42yz– = 32–4 1 0102 y = 2 3xy++2z = 11 31211 0013 z = 3
CYU 1.4 (a) Inconsistent: The last row 0002 corresponds to the equation 0x ++0y 0z = 2 which clearly has no solution. free variable x y z (b) 10– 2 1 x +10y – 2z = x = 12+ z 01 5 4 : 12+ r 45– rr r 0xy++5z = 4 y = 45– z 00 0 0
free variables
1 2 01 1 x +++2y 0zw= 1 x = 12– y – w 0 014 –2 : 0x +++0yz4w = – 2 z = –42 – w (c) 0000 0 12– r – sr –42 – ssrs R B-2 CYU SOLUTIONS
CYU 1.5
7 4 7 4 100a + ------b – ---c xa= + ------b – ---c (a) 4x – 2y + z = a 18 9 18 9 42–1a –42x ++y 2z = b rref 11 5 –422 b 010 a + ------– ------: ya= + -----11- – -----5- 18 9c 18 9c 5xy–4+ z = c 51–4c 1 2 1 2 001– a – ---b + ---c za= – – ---b + ---c 3 3 3 3 Solution for all a, b, and c
10– 5a + ------b – 2a- 4 x –44y – z = a 14–4– a after two cycles b – 2a (b) 2x + 8y – 12z = b 28– 12b 01–--1------ 4 16 –12x ++y 2z = c –1221 c 00 0 c ++3ab The system is consistent if and only if c ++3ab= 0
3x + 7yz– = a 37– 1 100 rref coefS CYU 1.6 (a) S: 13x –24y + z = b coef(S) 13–2 4 010 2x –24y + z = c 24–2 001
does not contain a row of zeros: system has a solution for all values of a, b, and c
x – 3y + w = a 13–01 100 10 3xy–2+ z – 3w = b coef(S) 31– 23– rref coefS 010 3 (b) S: xz+ – 5w = c 101 –5 001–15 2xy–3+ z – 2w = d 21– 1 2 000 0 contain a row of zeros: system does not have a solution for all values of a, b, c, and d CYU SOLUTIONS B-3
CYU 1.7 x y z w x y z w 1 0 0 –2 2x +++3y 4z 5w = 0 234 5 coef S rref 0 1 0 1 S : 3xy++4zw += 0 314 1 3 0 0 1 --- x +++7y 4z 11w = 0 17411 2 Setting the free varialble w to r we arrive at the system: x –02r = yr+0= 3 Solution set: 2rr – –---r r rR 3 2 z +0---r = 2 = 4r –2r–23r r rR
CHAPTER 2 VECTOR SPACES
CYU 2.1 rv +2sw ===32– 2 + –3 –310 64– 4 + 93– 0 15 1– 4
CYU 2.2 (iv): vv+ – = 0 (in 2 ): PofR
If vv= 1 v2 , then: vv+ – v1 v2 + –v1 –v2 v1 – v1 v2 – v2 = 00 0 Definition 2.5 Definition 2.3 Definition 2.4 n (in ): If vv= 1v2 vn , then:
vv+ – v1v2 vn + –v1–v2 –vn v1 – v1v2 – v2 vn – vn = 000 0
CYU 2.3 For A = aij Mmn and rs :
rsA rsaij rsaij rsaij = rs aij rs aij rs A
n n n n n n i i i i i i CYU 2.4 aix + bix ai + bi x = bi + ai x bix + aix i = 0 i = 0 i = 0 i = 0 i = 0 i = 0
CYU 2.5 rs+ f x rs+ fx = rfx+ sfx rf x + sf x B-4 CYU SOLUTIONS
CYU 2.6 The Zero Axiom: We need to find 0 = abc such that for any xyz V :
xyz + abc = xyz , which is to say: xa+ – 1 = x a = 1 xa+ – 1 yb++2 zc+ – 3 = xyz yb++2 = y b = –2 zc+ – 3 = z c = 3 Let’s verify directly that 12– 3 + xyz does indeed equal xyz for every xyz V : 12– 3 + xyz ==1 +2x – 1– ++y 2 3+ z – 3 xyz . The Inverse Axiom: For given xyz V we are to find abc such that xyz + abc = 0 , which is to say: xa+1– 1 = ax= –2+ xa+ – 1 yb++2 zc+ – 3 = 12– 3 yb++2 = – 2 by= –4+ zc+3– 3 = cz= –6+ It is easy to verify, directly that: xyz++ + – x + 2– y + 4 – z + 6 = 12– 3 .
CYU 2.7 Since V = 0 must be closed under addition and scalar multiplication we have no choice but to define: 00+ = 0 and r00= for every r . It is easy to see that all eight axioms of Defi- nition 2.6 hold. We establish (v): If uv V , then they must both be 0 . Consequently: ruv+ r00+ r0 0 and ru + rv r0 + r0 000+ . Hence: ruv+ = ru + rv . As for (iii) and (iv), simply note that 0 is certainly the zero vector in V and that it is also its own inverse.
CYU 2.8 vz+ = wz+ CYU 2.9 Start with: r0 = r00+ vz+ + –z = wz+ + –z r0 = r0 + r0 vz+ + – z= wz+ – z r0 + –r0 = r0 + r0 + –r0 v + 0 = w + 0 0 = r0 + r0 + –r0 vw= 0 = r00+ Conclusion:0 = r0 CYU 2.10 (a) rv = rw (b) rv = sv 1 1 rv + –sv = sv + –sv ---rv = ---rw r r rv + –sv = 0 1 1 rs– v = 0 ---r v = ---r w r r Theorem 2.8: rs–0= 1v = 1w rs= Axiom (viii): vw= CYU SOLUTIONS B-5
CYU 2.11 (a) –1–v ====– –1 v –1 –1 v 1vv (b) –r v ===–1 r v –1 rv –rv (c) r–v ====r–1 v r–1 v –1 rv –rv
CYU 2.12 (a) –1vw+ ==– vw+ –1 v + –1 w ==– v + –w – v – w
(b) I. Valid at n = 2 : –v1 + v2 = – v1 – v2 [by part (a)]. II. Assume –v1 ++v2 +vk = – v1 – v2 – – vk . (The induction hypothesis) III. We show that –v1 ++v2 ++vk vk + 1 = – v1 – v2 – – vk – vk + 1 : –v1 ++v2 ++vk vk + 1 = –v1 ++v2 +vk + vk + 1 By I: = – v1 ++v2 +vk – vk + 1 By II: ==– v1 – v2 – – vn – vk + 1 – v1 – v2 – – vk – vk + 1
CYU 2.13 Since 00 S , S . S is closed under addition: 00
For any a 2a b 2b S: a 2a + b 2b = ab+ 2ab+ S –0a –0b –0a –0b –0ab+ S is closed under scalar multiplication:
For any a 2a S and r : r a 2a = ra 2ra S –0a –0a –0ra
CYU 2.14 Since 000 S , S . For any xyz abc S and r : rxyz + abc = rx+ a ry+ b rz+ c is back in S, since:
rx+ a ++ry+ b rz+ c ===rxyz++ + abc r00+ 0
CYU 2.15 Since the zero of that vector space is the zero function Z: which maps every number to zero, and since S = f F f9 = 0 , S does not contain the zero vec- tor and is therefor not a subspace of F .
CYU 2.16 Since 000 S , S . For any 16a – 2b 4a – 17b11a 11b 16x – 2y 4x – 17y11x 11y S and r : r16a – 2b 4a – 17b11a 11b + 16x – 2y 4x – 17y11x 11y = 16ra+ x – 2rb+ y 4ra+ x – 17rb+ y 11ra+ x 11ra+ x ===16A – 2B 4A – 17B 11A 11B , where Arax+ and B ra+ x B-6 CYU SOLUTIONS
CYU 2.17 False. The subsets S = 01 and T = 02 are not subspaces of since neither is close under addition (nor under scalar multiplication), yet ST = 0 is a subspace of .
CYU 2.18 (a) For v ==2135– – 12 – and u = 23 : u + rv r ==23 + r12 – 2 +32r – r
(b) Since 1 +52r – r = 2 +32r – 1 – r – 1 : 1 +52r – r r = 2 +32r – r r
CYU 2.19 Choosing r = –1 and r = 1 in L = 135 + r21– 1 r we obtain the two points 135 –121– 1 = –26 , 135 +21– 1 = 344 on L. Preceding as in Example 2.14 we arrive at a direction vector v ==344 – –126 42– 2 . Select- ing u = 344 as our translation vector, we have: L ==344 + r42– 2 r 34+42r+42r – r r , which we now show to be equal to the set L ==135 + r21– 1 r 12+3r+5r – r r : r – 1 34+42r+42r – r ==12+3r+5r – r , were r ------2
CYU 2.20 Choosing 41– 3 to play the role of w, instead of 32–2 we obtain:
P = w ++ru sv rs = 41– 5 ++r–175– s13– 5 rs = 4 – r +5s 17++r 3s – – 5r – 5s rs
which we show to be equal to the set P = w ++ru sv rs : = 32– 2 ++r–175– s13– 5 rs = 3 – r + s – 2 ++7r 3s 25– r – 5s rs
1 : 4 – r +3s = – r + s Equating the first two components of P and P we obtain: . 2 : 1++ 7r 3s = –7 2++r 3s Multiplying equation (1) by –3 and adding it to equation (2) we find that rr= . Substituting in (1) we then find that ss= – 1 ; bringing us to: 4 – r +5s 17++r 3s – – 5r – 5s = 3 – r + s – 2 ++7r 3s 25– r – 5s CYU SOLUTIONS B-7
CHAPTER 3 BASES AND DIMENSIONS
a b a b a +22b = – CYU 3.1 (a) No: 12 – 2 augS rref 10 0 S: 3a +35b = – 35– 3 01 0 8a +84b = No solution! 848 00 1
a b a b (b) Yes: a +22b = – 12 – 2 augS rref 10 2 S: 3a +45b = – 35– 4 01– 2 8a +84b = 848 000
–2 –4 – 8 = 2138 – 2254
CYU 3.2 (a) We are to show that for any given matrix ab there exist scalars xyzw for which cd
(*) x 12 +++y 10 z 01 w 04 = ab: 34 10 01 20 cd
xy+2xz++4w = ab 3xy++2w 4xz+ cd
xy++0z +0w = a 1100 1000 2x +++0yz4w = b coef(S)2014 [rref(coef(S)] 0100 S: 3xy++0z +2w = c 3102 0010 4x +++0yz0w = d 4010 0001 Since rref coefS does not contain a row consisting entirely of zeros, the system S [which stems from (*)] has a solution for any given abcd .
xy++0z +0w = – 1 (b) 1100– 1 1000 2 2x +++0yz4w = 5 aug(S) 2014 5 rref 0100–3 S: 3xy++0z +2w = 1 3102 1 0010 5 4x +++0yz0w = 13 401013 0001–1
from the above we see that: 2 12–3 10 + 5 01–1 04 = –51 34 10 01 20 113 B-8 CYU SOLUTIONS
CYU 3.3 We are to find the set of vectors abc for which there exist scalars x, y, z, such that: abc = x215 ++y12– 2 z051 :
12–5 b 2xy++0z = a 210a 12–5b 12–5 b a – 2b 01– 2 ------x –52y + z = b 12–5b 210a 05– 10a – 2b 5 5x ++2yz= c 521c 521c 012– 24c – 5b 00 0c – 5b – 12------a – 2b 5 Note that in the above we didn’t bother to reduce the coefficient matrix to its row-reduced-echelon forms with 0’s above and below leading ones. Rather, we obtained its row-echelon form with 0’s only below the leading ones [see Exercises18-22, page 12]. This still enables us to determine Span215 12– 2 051 , for it consists of all vectors abc for which 12 a – 2b c – 5b – ------= 0 , which is equivalent to: 12ab+0– 5c = or b = 5c – 12a . 5 Conclusion: Span215 12– 2 051 = abc b = 5c – 12a . Any vector abc for which b 5c – 12a , say (1,2,3), will not be in the spanning set.
CYU 3.4 For given vV we are to find scalars A,B,C such that vAv= 1 ++Bv1 + v2 Cv1 ++v2 v3 . Since v1v2 v3 span V, there exist scalars a,b,c such that vav= 1 ++bv2 cv3 . Equating these two expressions for v we have: Av1 ++Bv1 + v2 Cv1 ++v2 v3 = av1 ++bv2 cv3
ABC++ v1 ++BC+ v3 Cv3 = av1 ++bv2 cv3
ABC++ = a Cc= Bringing us to: BC+ = b with solution: BbC==– bc– Cc= AaB==– – C abc– – – c =ab–
CYU 3.5 ax2 ++b2x2 + x cx– 3 = 0 a ++2b 0c = 0 12 0 100 2 rref a + 2b x + bc+ x – 3c = 0 0abc++= 0 01 1 010 0a +00b – 3c = 00– 3 001
No free variable Linearly independent
CYU 3.6 av1 ++bv1 + v2 cv1 ++v2 v3 +dv1 +++v2 v3 v4 = 0 (we are to show abcd====0 abcd+++ v1 +++bcd++ v2 cd+ v3 dv4 = 0
Since v 1 v 2 v 3 v 4 is abcd+++= 0 linearly independent: bcd++= 0 a ====bcd0 cd+0= d = 0 CYU SOLUTIONS B-9
CYU 3.7 Aa1a2 a3 a4 ++++Bb1b2 b3 b4 Cc1c2 c3 c4 Dd1d2 d3 d4 Ee1e2 e3 e4 = 0
a1 b1 c1 d1 e1
a2 b2 c2 d2 e2 rref must have a free variable (5 variables and 4 equations ) a3 b3 c3 d3 e3
a4 b4 c4 d4 e4
2511 rref 103 CYU 3.8 Linear dependent since: 10 3 011 . 3211 000
2 5 11 8 1034 8412 Span213 502 11311 since: 10 3 4 rref 0110 . 3 2 11 12 0000
From the above rref-matrix we see that a213 ++b502 c11311 = 8412 for any a +43c = a,b, and c for which . Letting c = 0 we get a = 4 and b = 0 , giving us the linear bc+0= combination 8412 = 4213 ++0502 011311 . Letting c = 1 we get a = 1 and b = –1 , giving us another linear combination 8412 = 213 –502 + 11311 .
CYU 3.9 x3 + x and –7 are clearly linearly independent. Since x2 cannot be “built” from those vectors, x3 +7x – x2 is linearly independent. Can x3 be “built” by those three vectors? No; so x3 +7x– x2 x3 is linearly independent. Just to make sure: a b c d 1001 1000 ax3 + x +++b–7 cx2 dx3 = 0 0010 rref 0100 3 2 ad+ x +++cx ax b–7 = 0 1000 0010 07–00 0001
3 (Incidentally, if you throw in any two randomly chosen vectors from P3 into x +7x – chances are really good that you will end up with a linearly independent set. Try it.) B-10 CYU SOLUTIONS
CYU 3.10 For spanning: x 21 ++y 11 z –03 +w 04 = ab 30 22 –55 – 15 cd
21 11 –03 04 00 For linear independence: x ++y z +w = 30 22 –55 – 15 00
21–0 3 1000 Spans: Does not contain a row consisting entirely of zeros. coefficient matrix 1104 rref 0100 32–1 5 0010 Linearly Independent: each row 02–5 5 0001 has a leading one.
CYU 3.11 If S = v1v2 vn is a basis, then it spans V insuring us that every vector in V can be expressed as a linear combination of the vectors in S. Being a basis, S is also linearly independent, and Theorem 3.6, page 89 insures us that the representation is unique. Conversely, if every vector in V can uniquely be expressed as a linear combination of the vectors in S then S certainly spans V. To show that it is also linearly independent consider c1v1 +++c2v2 cnvn = 0. Since 0v1 +++0v2 0vn = 0 , and since we have unique repre- sentation: ci = 0 for 1 in.
CYU 3.12 We show that S = 10 01 is a basis for the space V of Example 2.5. S spans V: For ab V can we find rs such that ab = r10 + s01 ? Yes: ab = r10 + s01 Since rxy = rx–1 r – ry+ r – 1 : = rr–1+ r – 1 + –1s + ss+ – 1 = 1 r – 1 + –1s +2 s – 1 Since xy + x y = xx+ – 1 yy++ 1 : = –1s + r –21 ++s –11 = –1s +2 sr+ – 1 equating coefficients as= –1+ s ==–a + 1 and r b + 2a – 1 b = 2sr+ – 1 Check: r10 + s01 = b + 2a – 1 10 + –1a + 01 = b –12a – –1b + 2a – 1 + b + 2a – 1 – 1 + –1–1a + + –1a + – 1 ==1 b + 2a – 2 +1a –12a + + a – 1 b + 2a – 2 + a –12a + – 1 =ab S is linearly independent: Recalling that 11 – is the zero vector in S we start with the equation a10 +11b01 = – and go on to show that ab==0 : a10 +11b01 = – 1 a – 1 +11–1b + bb+ – 1 = – 1 a – 1 +11–1b +2 b – 1 = – 1 –1b = solution: ab==0 1 –11b + – a –21 ++b –11 = 11 – a –21 +1b = – 1 – b a –21 + b = 11 – CYU SOLUTIONS B-11
CYU 3.13 (a) Knowing that M22 has dimension 4, we simply have to add two vectors to 21 22 L = without rupturing linear independence. By now, you may be convinced that 12 11 if you add any two randomly chosen vectors, say 15 and 01 , chances are good that you –11 –62 will end up with a set of four independent vectors, and therefore a basis for M22 . Let’s make sure that we do: a b c d 22 1 0 1000 21 22 15 01 00 rref a ++b c +d = 12 5 1 0100 12 11 –11 –62 00 11–2 1– 0010 21 1 6 0001
(b) Here is a brute force approach to obtain a basis for Span(S). We start with the first vector in S = 312 – –93– 6 12– 2 –54– 6 624 – : 31– 2 . Since the second vector is easily seen to be a multiple of the first, we discard it and turn our attention to the third vector 12– 2 . Since that vector is not a multiple of 31– 2 , we throw it into that set to obtain the two independent vectors 31– 2 12– 2 . Can the vector –54– 6 be built from those two independent vectors? Yes:
31– 5 10– 2 –241 rref 01 1 –12312 – + 12– 2 = –54– 6 22–6– 00 0 Since we are looking for a maximal independent set, we discard –54– 6 and turn our attention to the last remaining vector 624 – . Is it independent of the vectors 31– 2 12– 2 ? Clearly not, since 624 – = 2312 – . Conclusion: 31–2 12– 2 is a basis for Span(S). Since 3 is of dimension 3, S does not span 3 .
CYU 3.14: first vector
39–15–6 13–02–2 –32421 – rref 00110 Basis for SpanS = 31–2 12– 2 26–2–6–4 00000 third vector B-12 CYU SOLUTIONS
CHAPTER 4 LINEARITY CYU 4.1 The function f: 2 3 given by fab = ab+ 2ba – b is linear. f preserves sums: fab + a b = faa++ b b = aa+ +2bb+ bb+ aa+ – bb+ ==ab+2 ba – b + a+2b b a – b fab + fa b
f preserves scalar products: frab ==fra rb ra+ rb 2rb ra– rb ==rab+2 ba – b rfab
CYU 4.2 Since f00 ==0x2 ++0x 1 1 0 , f is not linear.
CYU 4.3 frab + a b = fra++ a rb b = ra+ a +2rb+ b rb+ b ra+ a – rb+ b = ra+ rb +2a + b rb + 2b ra– rb + a – b = ra+2 rbrb ra– rb + a +2bb a – b ==rab+2 ba – b + a +2bb a – b rfab + fa b
CUU 4.4 A counterexample: The trivial function T: given by Tx= 0 for every x is linear. The set S = + is not a subspace of , but fS= 0 is a subspace of .
CUU 4.5 True: The proof of Theorem 4.5 makes no mention of linear independence.
CUU 4.6 (a) We first express 342 as a linear combination of the basis 100 020 111 :
ac+3= 342 = a100 ++b020 c111 = ac+2bc+ c 2bc+4= c ===2 b 1 a 1 c = 2
Then: T342 = T100 ++020 2111 By linearity: = T100 ++T020 2T111 ==2x2 + x ++2x2 + x 2x – 5 4x2 + 4x – 10 (b) Expressing abc as a linear combination of the given basis we have: AC+ = a abc = A100 ++B020 C111 bc– C ==cB ------A =ac– = AC+2BC+ C 2BC+ = b 2 Cc= CYU SOLUTIONS B-13
bc– Then: Tabc = Tac– 100 ++------020 c111 2 bc– = ac– T100 ++------T020 cT111 2 bc– = ac– 2x2 + x ++------2x2 + x cx– 5 2 b c = 2ab+ – 3c x2 + a + --- – --- x – 5c 2 2
CYU 4.7 (a) 10 11 is easily seen to be linearly independent, and therefore a basis of 2 . 010 110 011 is easily seen to be linearly independent and therefore a basis of 3 . (b) From the given information, we have: LT 10 = LT10 ==L020 L2010 =2L010 ==201 02 To determine LT 11 ==LT11 L101 , we need to express 101 as a linear combination of the basis 010 110 011 . Let’s do it:
b = 1 101 = a010 ++b110 c011 abc++= 0 c ===1 b 1 a –2 = babc++ c c = 1
Then: LT 11 ==LT11 L101 = L– 2010 ++110 011 = – 2L010 ++L110 L011 = –201 ++10 10 = 22 – To determine LT ab , we first express ab as a linear combination of 10 11 : ab ====A10 + B11 ABB+ B b and Aab– ; so: ab = ab– 10 + b11 Putting this together we find that: LT ab = LT ab– 10 + b11 = ab– LT 10 + bLT 10 ==ab– 02 + b22 – 2b 2a – 4b
CYU 4.8 (a):
Trax2 ++bx c + a'x2 ++b'x c' = Tra+ a' x2 ++rb + b' xrc+ c'
==ra + a' rb + b' r ab + a' b' rc + c' ra + a' ca c' a' = rT ax2 ++bx c + Ta'x2 ++b'x c' B-14 CYU SOLUTIONS
(b) The vectors Tx2 = 10 , Tx= 01 , T1 = 00 span ImT (Theorem 4.9), and 01 00 10 they are easily seen to be linearly independent. Consequently: rankT = 3 .
KerT = ax2 ++bx c Tax2 ++bx c = 0 By definition, 2 ab 00 ====ax ++bx c = a bc0 ca 00 Consequently, KerT = 0 and therefore nullityT = 0
CYU 4.9 We first show that KerT = 0 :
2a = 0 Tabc ==02 ab + ccb 0000 bc+0= a ===bc0 c = 0 b = 0
At this point we know that nullityT = 0 and that the kernel has no basis. Since rankT +30 = , Im(T) has dimension 3. It is easy to see that the three vectors T100 ===2000 T010 0101 T001 0110 in the image of T are linearly independent. It follows that 2000 0101 0110 is a basis for Im(T).
CYU 4.10 Let v V be such that Tv ==Tv v v (*) . We establish that T is one -to-one by showing that KerT = 0 [see Theorem 4.11(a)]: Assume that Tv = 0 (we want to show that v = 0 ). Consider the vector v + v . By linearity we have: Tvv+ ===Tv+ Tv 0 + Tv Tv . From (*) and the fact that Tvv+ = Tv we have: vv+ = v v = 0 CYU 4.11 f –1: YX is one-to-one: –1 –1 –1 –1 –1 –1 f a ==f b ffa ffb ff a ==ff ba b –1 –1 f –1: YX is onto: For any xX , f fx ==f f xx .
CYU 4.12 T is one-to-one: Tab ==Ta' b' ab+ xa– a' + b' xa– ' aa= ' Equating coefficients: aa= ' and b = b' ab+ = a' + b' CYU SOLUTIONS B-15
2 T is onto: For given ax+ b P1 , we need to find AB such that: TAB ==AB+ xA– ax+ b
Equating coefficients: –A = b A ===–b and B aA– ab+ AB+ = a
Check: T–b ab+ ==– b ++abxb– – ax+ b.
Determining T –1 : Let T –1ax+ b = AB . Then: Applying T to both sides: ax+ b ==TAB ax+ b AB+ xA– –A = b Equating coefficients: A ===–b and B aA– ab+ AB+ = a
–1 2 From the above: T ax+ b = –b ab+ . Let’s show that the function L: P1 given by Lax+ b = –b ab+ is linear: Lrax+ b + a'xb+ ' = Lraa+ ' xrbb+ + ' = –rb+ b' ra+ a' + rb+ b' ==r–ba + b + –b' a' + b' rL ax+ b + La'xb+ '
4 ab CYU 4.13 We show that T: M22 given by Tabcd = is an isomorphism: cd T is one-to-one:
Tabcd ==Ta'b' c' d' ab a' b' a ====a' b b' c c' d d' cd c' d'
ab ab T is onto: For given M22 , Tabcd = . cd cd
ab ab ra+ a rb+ b ab ab T is linear: Tr+ ==T rT + T . cd cd rc+ c rd+ d cd cd
CYU 4.14 Let dimV ==dimW n . By Theorem 4.15, V n and W n . By Theorem 4.14, VW .
Conversely, suppose that VW . Let T: VW be an isomorphism, and let V = v1v2 vn be a basis for V. We show STv= 1 Tv2 Tvn is a basis for W, thereby establishing that V and W are of the same dimension, n. B-16 CYU SOLUTIONS
S is linearly independent: since is a linearly independent set n n n V a Tv = 0 Tav = 0 a v ==0 a 0 for 1 in i i i i i i i i = 1 i = 1 i = 1 since T is linear Theorem 4.11(a), page 129 n S spans W: Let wW . Since T is onto, there exist va= ivi such that Tv= w . Then: i = 1
n n wTv== Tav =a Tv i i i i i = 1 i = 1 since T is linear
CYU 4.15 By CYU 4.15, we know that the dimension of W equals that of V. We can therefore verify that Lv1 Lv2 Lvn is a basis for W by showing that the n vectors Lv1 Lv2 Lvn span W (see Theorem 3.11, page 99): n
Let w W . Since L is onto, there exist va= ivi such that Lv= w . Since L is n i = 1 linear: wLv== a . i Lvi i = 1
CYU 4.16 Lets move the elements of: S = 2x3 –53x2 + x – 1x3 –8x2 + x – 3 x2 + 11x – 5 –2x3 ++x2 3x – 2
over to 4 via the isomorphism Tax3 +++bx2 cx d = abcd to arrive at the 4-tuples: TS= 23– 51– 1183 – – 0 1 11 – 5 –123– 2 Applying Theorem 3.13, page 103, we conclude that the first
two vectors of TS , 23– 51– and 1183 – – , consti- 210– 1 10–1 1– tute a basis for Span TS . It follows that –13 –12 rref 01 2 1 2x3 –53x2 + x – 1 and x3 –8x2 + x – 3 is a basis for Span(S). 58113 00 0 0 –31 –5–2– 00 0 0
CYU 4.17 (a) fxyz = 2x + 1 xyz+ – is one-to-one:
fx1y1 z1 = fx2y2 z2
2x1 +21 = x2 + 1 x1 = x2 2x1 + 1 x1 + y1 –z1 = 2x2 + 1 x2 + y2 –z2 x1 + y1 = x2 + y2 y1 = y2 –z1 = –z2 z1 = z2 CYU SOLUTIONS B-17 f is onto: For given xyz X we need to find abc 3 such that: fabc = xyz
a = ------x – 1 2a + 1 = x 2 2a + 1ab+ –c = xyz ab+ = y bya==– y – x------– 1 2 –c = z cz= – x – 1 x – 1 From the above formula: f------y – ------–z = xyz , we conclude that: 2 2
–1 x – 1 x – 1 f xyz = ------y – ------–z 2 2
–1 x – 1 x – 1 (b) From Theorem 4.16 and the above formula f xyz = ------y – ------–z we have: 2 2 x – 1 x – 1 x – 1 x – 1 x y z x y z = f ------1 -y – ------1 - –z + ------2 -y – ------2 - –z 1 1 1 2 2 2 2 1 2 1 2 2 2 2 x x x x = f ----1- + ----2-1– y + y – ----1- – ----2-1+ – z – z 2 2 1 2 2 2 1 2 x x x x x x Since fxyz = 2xx + y – z: = 2----1- + ----2-1– ----1- +++----2-1– y y – ----1- –1----2- +–– z – z 2 2 2 2 1 2 2 2 1 2
= x1 + x2 – 2y1 + y2 z1 + z2
x – 1 x – 1 rx– r rx– r and: r xyz ==fr------y – ------–z f ------ry – ------–rz 2 2 2 2 rx– r rx– r rx– r = 2 ------------+ ry – ------––rz 2 Since fxyz = 2xx + y – z: 2 2 = rx– rry rz (c) The zero in the space X: f000 = 100 . The inverse of xyz in X:
–1 x – 1 x – 1 –1x + x – 1 ff– xyz ===f –------y – ------–z f------– y + ------z –2x + –yz 2 2 2 2
–1 x – 1 x – 1 f xyz = ------y – ------–z fxyz = 2x + 1 xyz+ – 2 2 B-18 CYU SOLUTIONS
CHAPTER 5 MATRICES AND LINEAR MAPS
35 36 +3453 + 55 33 37 CYU 5.1 (a) 64 42 ==46 +4423 + 25 30 26 35 90 96 +9403 + 03 54 36 (b) Number of columns in A does not equal the number of rows in B.
2 CYU 5.2 False: 11 12 11 12 11 12 23 23 77 + ===+ + 11 00 11 00 11 00 11 11 34
11 11 11 12 12 12 22 24 12 58 While: ++2 = + + = 11 11 11 00 00 00 22 24 00 46
The columns associated with the leading 13 25–4 3 10–12------ones in rref(A); namely: 2 CYU 5.3 –104 –68– rref 240 – and 5–1 10 01 2– 4 constitute a basis for the column space 012– 4 00 0 0 of A. 13 Since A and rref(A) share a row space, 10–12------ and 0 1 2– 4 constitutes a basis for the 2 row space of A. Note that the first two rows of A are not linearly independent. We are assured, however, that two of A’s rows will constitute a basis for its row space (either rows 1 and 3, or rows 2 and 3, will do the trick).
CYU 5.4 If X1 and X2 are solutions of the homogeneous system of m equations in n unknowns of equations, AX = 0 , and if r , then:
ArX1 + X2 ===rAX1 + AX2 r00+ 0 It follows, from Theorem2.13, page 61, that the solutions set of AX = 0 is a subspace of n .
21 3 0 100– 1 21 3 0 CYU 5.5 From 14–7 2– rref 010– 1 , we see that: null14–7 2– = cc– cc c 30 1– 2 001 1 30 1– 2 A basis: 11– 11
CYU 5.6 Let AM mn . By definition:
nullityA ==dimXAX= 0 dimXTAX = 0 =nullityTA Since nullityA equals the number of free variables in rrefA and since rankA equals the number of leading ones in rrefA : rankA = n – nullityA . By Theorem 4.10, page 126:
rankTA = n – nullityTA . Since nullityA = nullityTA : rankA = rankTA . CYU SOLUTIONS B-19
CYU 5.7 32– ab 10 3a –12c = 3b –02d = 1 2 4 3 ===== and a –--- b --- c –--- d --- 41– cd 01 4ac–0= 4bd–1= 5 5 5 5
From the above, we see that the matrix 32– is invertible, with inverse –2515 . 41– –3545
–1 –1 –1 CYU 5.8 I. Theorem 5.8 tells us that the claim is valid for n = 2 : A1A2 = A2 A1 .
–1 –1 –1 –1 II. Assume validity at nk= : A1A2 Ak = Ak Ak – 1 A1 III. We establish validity at nk=1+ :
–1 –1 A1A2 AkAk + 1 = A1A2Ak Ak + 1 –1 –1 By I: = Ak + 1 A1A2Ak –1 –1 –1 –1 –1 –1 By II: ==Ak + 1Ak Ak – 1 A1 Ak + 1Ak Ak – 1 A1
–1 100 0 –1 100 0 CYU 5.9 01 0 01 0 1 –1 050 0 0 --- 0 0 16 16– 10 0 ===10 0 5 001 0 01 01 00 1 00 1 001 0 000 1 000 1
E 2103 3341 100 001 001 2103 3341 CYU 5.10 (a) R R R R 1326 1 3 1326 010 1 3 010 and 010 1326 = 1326 3341 2103 001 100 100 3341 2103
E (b) 2103 334 1 100 001 001 2103 334 1 2R R 2R R 1326 2 2 26412 010 2 2 0 2 0 and 0 2 0 1326= 26412 3341 210 3 001 100 100 3341 210 3
A A–1 1 0 1 21 0 0 0 1000–586 7 – CYU 5.11 AI = 0 2 1 40 1 0 0 0100–232 3 – 1 1 1 00 0 1 0 0010107–10–8 0 3 1 10 0 0 1 0001–111 1 – B-20 CYU SOLUTIONS
CYU 5.12 Assume AB= I . Multiplying both sides of the equation BX = 0 by A we have: AB XA= 0 IX ==0 X 0 B is invertible (Theorem 5.17) Then: AB== I AB B–1 B–1 ABB–1 ===B–1 A B–1 A–1 B .
120– 1 100– 1 4 –1 –14 CYU 5.13 301 1 rref 010– 3 8 1 = –38
042 2 001 7 4 2 74
CYU 5.14
33 72 11– 26 T ====334 T 272 T–116 T6214 22 11 51 59
12032–6 1 rref 10012–14 3 4 1234–14 30137 1 2 01010–238 1 8 T = 1018–238 04242 6 14 00101 134 54 0 1 13 4 54
T
rref 1203 100 3 4 34 CYU 5.15 13 13 T==312 and 3011 010 9 8 T 98 20 20 0422 001–4 5 –45
37121 1000 6928 69 28 rref 32–6 1 3 0100– 1114 13 = –11 14 21552 0010 1 28 20 128 21190 0001– 1328 –13 28
69 28 1234–14 34 13 –11 14 13 T ===1018–238 98 T 20 128 20 01134 54 –13 28 –45 From CYU 5.15 CYU SOLUTIONS B-21
T 2 CYU 5.16 T : T111 = 1x ++1x 1 100110 100110 110 T110 = 1x2 ++1x 0 010111 010111 T = 111 T100 = 0x2 ++1x 0 002100 0011 200 12 00
2 L L : Lx= 11 01100 10012 012 Lx= 01 L = 11112 01100 100 L2 = 02
2 LT LT : LT 111 ===LT111 Lx++x 1 13 01110 10211 211 LT 110 ===LT110 Lx2 ++x 0 12 LT = 11321 01110 110 LT 100 ==LT100 L0x2 ++x 0 =01
CYU 5.17 Since T is invertible, it must be a square matrix of dimension n. It follows that V
and W are both of dimension n. Let = v1v2 vn , = w1w2 wn , and –1 T = aij . Consider the linear transformation L: WV which maps wi to the vector –1 a1iv1 +++a2iv2 anivn . From its very definition, we see that L ==aij T (note,
for example that Lw1 is the first column of aij ). Applying Theorem 5.22, we see that: –1 LT ===L T T T I . It follows that LT: VV is the identity map, and that therefore T is invertible with inverse L. 23 I
V CYU 5.18 IV and 23 : 0 2 122 rref 105 623 43 31– 213 011 2 11
122rref 104 3 56 23 43 43 23 :. IV 23 ===23 . 213 011 3 12 1 13 1
CYU 5.19 Rotating the standard basis = 10 01 clockwise by 60 leads us to the basis 1 3 1 3 = --- –------ --- ------. Finding I and 13 : 2 2 2 2
I 13 1 1 ------101 101–13 3 2 – 2 2 rref –3232 013 011 3 2 31+
1 3 1 3 Check: 13– --- –------+1331+ --- ------= 2 2 2 2 B-22 CYU SOLUTIONS
2 2 2 CYU 5.20 For T and I . Noting that Tx+ 1 ==–2x + Tx– x – x – x , and T1 = 2 ; and that = x2x2 + x x2 ++x 1 = x2 + 1x2 –1x ; leads us to:
T I T I 1100–0111 1 100–2 1–0122 rref 01–01–1–0011 0101100–1 1– 101202001 001322–2 1–1–
Noting that Tx2 ==–xTx 2 + x x2 – x and Tx2 ++x 1 =x2 –2x + leads us to: T T I I 111011110 100122120 011–1 1–1–01–0 rref 010–1 1–3–1–1–1– 001002101 001002101
122 122 120 –21 –0 Then: I T I ===01–1– –11 –3– –11 –1– –11 –1– T –21 –1– 002 101 101
CYU 5.21 (One possible solution). Choosing to let c ==0 d 8 in system (*) at the bottom of page 197 we obtain the solution a ====–18 b 11 c 0 d 8 . At this point, we know that: –1 –812 = –1118 –1 4–1118 . Preceding as in part (c) of Example 5.13, we arrive –1518 08 8 4 08
at the basis = v1 v2 , with:
v1 = –1812 + 021 = –18 –36 and v2 ==1112 + 821 27 30 . –812 Let’s verify that T = for = –18 –36 27 30 : –1518
T–18 – 36 = –270 –108 T27 30 = 261 162 –18 27– 270 261 rref 10–8 12 –36 30– 108 162 01–15 18 T CYU SOLUTIONS B-23
CHAPTER 6 DETERMINANTS AND EIGENVECTORS
29– 3 93– 23– 29 CYU 6.1 (a-i) det 32–4=75det –det –det6 ==536– 6 –6478+ 9 ––27– 217 –42 34 32– 57– 6
29– 3 34 23– 23– (a-ii) det 32–4=–det9 – 2det –97det ==–18–20– –789212–15+ –+ 217 56– 56– 56– 57– 6
CYU 6.2 By induction on the dimension, n, of Mnn .
I. Claim holds for n = 2 : det a 0 ==ad – 0 c ad . cd II.Assume claim holds for nk= .
III.We establish validity at nk= + 1 : Let Aa= ij Mk + 1 k + 1 . Since all entries in the
first row after its first entry a11 is zero, expanding about the first row of A we have
detA = a11detB , where B is the k by k lower triangular matrix obtained by removing the first row and first column from the matrix A. As such, by II: detB = a22a33 ak + 1 k + 1 . Consequently: detA ==a11detB a11a22a33 ak + 1 k + 1
CYU 6.3 Let the ith and jth row of A be identical, with ij . Multiplying row i by –1 and adding it to row j we obtain a matrix B whose jth row consists entirely of zeros. As such detB = 0 (just expand about the jth row of B). Applying Theorem 6.3(c), we conclude that detA = 0 .
CYU 6.4 Theorem 6.4(a) Theorem 6.4(c) (two times)
2 101 1014 1014 012 2 012 2 012 2 det ==–det –det 101 4 210 1 01– 2 –7 411 3 411 3 0 1 –3 –13
1014 1014 012 2 012 2 15 ==–det –4det ==– –------15 00– 4–9 00– 4 –9 4 15 00– 5 –15 000 –------4 CYU 6.2(b) B-24 CYU SOLUTIONS
CYU 6.5 (a-i) Let BM nn , and let E be the elementary matrix obtained by multiplying row i of
In by c 0 . By Theorem 5.12, EB is the matrix obtained by multiplying row i of B by c. Conse- quently: detEB ==cdetB detE detB Theorem 6.4(b) detE = c [Theorem 6.4(b)] th (a-ii) Let BM nn , and let E be the elementary matrix obtained by adding a multiple of the i th row of In to its j row. By Theorem 5.12, EB is the matrix obtained by adding a multiple of row i of B to its jth row. Consequently: detEB ==detB detE detB Theorem 6.4(b) detE = 1
(b) We use the Principle of Mathematical Induction to show that for any BM nn and elementary
matrices E1E2 Es Mnn : detEs E2E1B ==detEs E2E1 detB detEs detE2 detE1 detB I. Validity for s = 1 follows from Theorem 6.5. II. Assume detEk E2E1B ==detEx E2E1 detB detEk detE2 detE1 detB III. detEk + 1 Ek E2E1B = detEk + 1 Ek E2E1B By I: = detEk + 1 detEk E2E1B By II: = detEk + 1 detEk detE2 detE1 detB
1 CYU 6.6 IAA=det–1 I =1detAA–1 = detA detA–1 detA–1 = ------detA
32 10 62 62 11 3 CYU 6.7 E–3 ==null– –3 null . From rref we see 32– 01 31 31 00
r that E –3 = –--- r rR with basis 13– . 3 CYU 6.8 The eigenvalues are the solutions of the equation: 16 –3 2 detA – I3 ==det –4 3– –8 0 –2 –116 – Which reduces to –0x – 15 2 = . It follows that 0 and 15 are the eigenvalues of A.
16 3 2 100 16 3 2 16 3 2 101 2 rref Then: E0 ==null–384 – – 0 010 null–384 – . From –384 – 01– 2 we –62 –11 001 –62 –11 –62 –11 00 0 r see that E0 = –--- 2rr r with basis –142 . 2 CYU SOLUTIONS B-25
16 3 2 100 132 132 132 rref E15 ==null–384 – – 15 010 null–124 –8– . From –124 –8– 000 we –62 –11 001 –62 –4– –62 –4– 000 see that E15 = – 3r – 2srs rs with basis –310 –201 .
CYU 6.9 The kernel of the linear operator:
T – –3 I 2 ab ==Tab + 3ab 3a ++2b 3a 3a – 2b + 3b =6a + 2b 3a + b is, by definition, the set: ab 6a +32b ab+ = 00 . Equating coefficients, we have:
6a +02b = 62rref 11 3 . It follows that E–3 = –r 3r r with basis 3ab+0= 56– 00 –31 .
CYU 6.10 With respect to the basis = x2 ++x 1 x + 11 : Tx2 ++x 1 = 2x2 ++5x 2 Tx+ 1 = x2 ++3x 1 T1 = x2 ++x 1 T 100211 100 2 1 1 rref 110531 010 3 2 0 111211 001–2 3–0
T 2 – 1 1 2 From detT – I3 ===det 3 2– 0 –02 – we see that the eigenval- –3 –2 – ues are 0 and 2: same as those found in the solution of Example 6.10. Since the determination of the corresponding eigenspaces only depends on T and the eigenvalues, the spaces E0 and E2 are identical to those determined in the solution of Example 6.10.
CYU 6.11 The three vectors 120 122 211 are easily seen to constitute a basis for 3 . Since T120 = 120 , T122 = –122 , and T211 = 2211 , = 120 122 211 consists of eigenvector with corresponding eigenvalues 1 – 1 and 2 , respectively.
T120 = 1120 ++0122 0211 100 From:T 122 = 0120 ++–1 122 0211 we have: T = 01–0 T211 = 0120 ++0122 2211 002 B-26 CYU SOLUTIONS
CYU 6.12 By Theorem 6.12, the n eigenvalues of T are linearly independent. Since V is of dimen- sion n, those n eigenvectors constitute a basis for V. It follows, from Theorem 6.11, that T is diag- onalizable.
CYU 6.13 (a) Characteristic polynomial: – 1 –0 1 detA – I ===det–31 –0 – 3 +++2 2 – – 2 2 ++ 1 –1314 – – From the above, we see that = 2 is the only (real) eigenvalue of A (note that the discriminant of 2 ++ 1 is negative). Determining the dimension of E2 –21 –0 1 E2 ==nullA – 2I null–321 –0 –1314 –2–
–013 10– 1 3 Turning to the homogeneous system of equations: –101 rref 01– 1 3 we conclude –1334 – 00 0 that E2 = rr3r r with basis 113 . It follows that there does not exist a basis for 3 consisting of eigenvectors of A, and that therefore A is not diagonalizable.
3 –2 – 1 (b) det2 6 –2 – ==– 3 +++2 2 – – 2 2 – 8 . –21 – 3 –
1 21– –215 – 12– 1 12– 1 E2 = null2 4 –2 and E8 = null2 –22 – . From 24– 2 rref 00 0 –21 – 1 –21 – –5 –21 –1 00 0
–215 – 101 rref and 22–2– 012 , we see that E2 = – 2r + srs rs and –21 –5– 000 E8 = –r–2r r r , with bases –210 101 and –1 –21 , respectively. It follows that –210101 –1 –21 is a basis for 3 consisting of eigenvectors of A, and that therefore A is diagonalizable. Theorem 6.15 tells us that the matrix P–1AP will turn
–112 – out to be a diagonal matrix with eigenvalues along its diagonal, where P = 10– 2 . 011 CYU SOLUTIONS B-27
–1 –112 – 32– 1–112 – 200 We leave it for you to verify that: 10– 2 26– 2 10– 2 = 020 . 011 –21 –3 011 008
CYU 6.14 From Theorem 6.19 and Example 6.11 we have: 10 10 02–0 2 000 0 0 00 0 0 00 0 0 00 0 110– 1 ==000 0 =0 00 0 –121 –1 002 0 0 0 210 0 0 01024 0 –121 –1 000– 2 0 00– 210 0 0 0 1024
32– 1 CYU 6.15 In the solution of CYU 6.13(b), we found the characteristic polynomial of 26– 2 –21 –3 to be – – 2 2 – 8 with E2 and E8 of dimensions 2 and 1, respectively.
th 12 CYU 6.16 Let sk denote the k element of the sequence (for k 3 ), F = , and 10
sk 3 3 2 3 k – 2 3 Sk = . From: S3 ==FS2 F S4 ==FS3 FF =F Sk =F , we see sk – 1 2 2 2 2 k – 2 3 that sk is the sum of the entries in the first row of F . We now set our sights on finding the 2 matrix Fk , and begin by finding a diagonalization for F:
det 1 –2 == – 2 + 1 ; E2 2rr r ; E–1 =–r r r 1 – It follows that 21 –11 is a basis of eigenvectors for F. Employing Theorem 6.18 we have: –1 FPDP==–1 21– 20 21– =12 11 01– 11 10 steps omitted From Theorem 6.19: –1 1 1 k + 1 1 2 k + 1 21– 2k 0 21– --- 2k + 1 –---–1 --- 2k + 1 + ---–1 Fk ==PDkP–1 =3 3 3 3 11 k 11 01– ************** *************** steps omitted B-28 CYU SOLUTIONS
1 1 k – 1 1 2 k – 1 k – 2 3 --- 2k – 1 –---–1 --- 2k – 1 + ---–1 3 Thus: F = 3 3 3 3 2 2 ************** ***************
2 k – 1 4 5 1 2k – 1 – –1 k – 1 ++--- 2 ---–1 k – 1 --- 2k – 1 + ---–1 k – 1 ==3 3 3 3 ****************************** *************** 5 1 Conclusion: s = --- 2k – 1 + ---–1 k – 1 . k 3 3
CYU 6.17 –11101001 1100 1011 is a basis of eigenvalues for 02–0 2 110– 1, with 0 the eigenvalue associated with –1110 and 1001 , 2 the eigenvalue –121 –1 –121 –1 associated with 1100 , and –2 the eigenvalue associated with 1011 (Example 6.11, page 237). Applying Theorem6.25 we conclude that: – c ++c c e2x +c e–2x –1 1 1 1 1 2 3 4 2x 0x 1 0x 0 2x 1 –2x 0 c1 + c3e c1e +++c2e c3e c4e = –2x 1 0 0 1 c1 + c4e 0 1 0 1 –2x c2 + c4e is the general solution of the given system of equations.
2x –2x y10 = 0 y1 = – c1 ++c2 c3e +c4e 2x y20 = 1 CYU 6.18 From CYU 6.16: y2 = c1 + c3e Since : –2x y30 = 2 y3 = c1 + c4e –2x y40 = 3 y4 = c2 + c4e
– c1 +++c2 c3 c4 = 0 –1 1110 1000 2 c1 +1c3 = aug(S) 1 0101 0100 3 S: rref c1 +2c4 = 1 0012 0010– 1 0 1013 0001 0 c2 +3c4 =
2x 2x Solution: y1 = 1 – e y2 = 2 – e y3 = 2 y4 = 3 . CYU SOLUTIONS B-29
5 1 5 1 --- –------– –--- 5 2 1 5 1 3 T t = 2 4 Tt det 2 4 ===0 --- – – --- 0 --- – --- = CYU 6.19 . Ft 5 Ft 5 2 4 2 2 2 –1 --- –1 --- – 2 2
5 1 5 1 ---3– –------2– –--- E3 ==null 2 4 –r 2r r and E2 ==null 2 4 r 2r r –1 --5-3– –1 --5-2– 2 2 Choosing –12 and 12 as eigenvectors for the eigenvalues 3 and 2, respectively, we have:
–c e3t + c e2t Tt ==c e3t –1 + c e2t 1 1 2 . Turning to the initial conditions Tt = 120 : Ft 1 2 2 2 3t 2t Ft 200 2c1e + 2c2e
30 20 –c + c = 120 –c1e + c2e 120 1 2 === c1 –10 and c2 110 30 20 200 2c1 +2002c2 = 2c1e + 2c2e
10e3t + 110e2t Bringing us to:Tt = . Setting Tt= Ft we have: Ft 3t 2t –20e + 2202e 3t 2t 3t 2t 10e +20110e = – e + 2202e 30e3t = 110e2t ln30e3t = ln 110e2t ln30 + lne3t = ln 110 + lne2t ln30+ 3t = ln 110+ 2t t =ln110 – ln 30 1.3years
CYU 6.20 Let D denote the state that a student is living in the dorm, and C denote the state that current state
D C next state .8 .1 D 858 the student is a commuter. Then:T = with S0 = . Then: .2 .9 C 702
------3783- .8 .1 858 5 757 D S1 == 803 .2 .9 702 ------4017- C 5
.8 .1 757 686 D .8 .1 686 636 D S2 = and S2 = .2 .9 803 874 C .2 .9 874 924 C Conclusion: 757, 686, and 636 of the current freshmen will live in the dorm in their sophomore, junior, and senior year, respectively. B-30 CYU SOLUTIONS
.73x ++.32y .09z = x CYU 6.21 .73 .32 .09 x x .21 .61 .04 y = y .21x ++.61y .04z = y .06 .07 .87 z z .06x ++.07y .87z = z
479 100------–.27x ++.32y .09z = 0 1157 297 .21x –.39 +0 .04z = rref 010------ 1157 .06x +0.07y – .13 = 381 001------xyz++= 1 1157 0000 see solution of Example 6.15.
479 297 381 Conclusion: Eventually ------------, or approximately 41%, 26%, 33% of the population, 1157 1157 1157 will vote democratic, republican, green, respectively.
CHAPTER 7 INNER PRODUCT SPACES
n n
CYU 7.1 ruv ====ru1u2 un v1v2 vn rui vi uirvi u rv i = 1 i = 1
CYU 7.2
(a) cv ==cv1v2 vn cv1v2 vn cv1cv2 cvn cv1cv2 cvn
n n 2 2 2 2 == c vi c vi =c v i = 1 i = 1 (b) uv– 2 ==uv– uv– uuv – – vuv – ==uu – uv – vu + vv u 2 – 2uv + v2
–1 uv –1 120 –131 –1 5 CYU 7.3 ==cos ------cos ------=cos ------ 83 uv 14+ 191++ 55 CYU SOLUTIONS B-31
CYU 7.4 For uu v and r : ruu+ v = ruv + uv ==r0 +00 . We see that ruu+ v . It follows, from Theorem 2.13 (page 61), that v is a subspace of n .
uv 0241 301– 1 3 6 12 3 CYU 7.5 proj v ==------u ------0241 =------0241 =0------u uu 0241 0241 21 21 21 21
6 12 3 6 9 24 and: v – proj v ==301– 1 – 0------3–------– ------. u 21 21 21 21 21 21 6 9 24 6 12 3 301– 1 = 3–------– ------+ 0------21 21 21 21 21 21
CYU 7.6 (a) A direction vector for the line L passing through 12 – and 24 : u ==24 – 12 – 16 . The vector from the point 12 – on L to P = 25 : v ==25 – 12 – 17 . Applying Theorem 7.2, we have: uv 16 17 43 proj v ==------u ------16 =------16 u uu 16 16 37
43 258 6 1 62 + 12 1 Hence: v – projuv ====17 – ------ ------–------ ------. 37 37 37 37 37 37
(b) u ==1221 – 1201 0020 . v ==1013 – 1201 02– 12 . uv 0020 02– 12 1 proj v ==------u ------0020 =---0020 =0010 u uu 0020 0020 2
Hence: v – projuv ====02–12–020010 –02 16 4 .
CYU 7.7 A normal to the desired plane will have the same direction is that of the line passing through the two given points; namely: n ==021 – 110 –111 . Normal form for the plane: –111 x – 1y – 3 z + 2 = 0 .
CYU 7.8 Let A= xyz 3xy+6– 2z = , and B ==200 ++r021 s203 rs R 22+2s rr + 3s rs R. BA : xyz = 22+2s rr + 3s . 3xy+322– 2z ===+ s +662r – 2r + 3s ++s 2r –62r – s 6 AB : If xyz is such that 3xy+6– 2z = (*) , can we find rs R such that x = 22+ s , y = 2r , and zr= + 3s ? Yes: x – 2 y In order for x = 22+ s , s = ------. In order for y = 2r , r = --- . We show that for 2 2 those particular values of r and s, zr= + 3s : z ==------3xy+ ------– 6------32+ 2------s + 2------r – 6- =------66++s ------2r – 6-2 =s + 2 2 2 2 (*) B-32 CYU SOLUTIONS
CYU 7.9 Let A= xyz be on the plane ax++ by cz = d with normal n = abc .
Determine the vector v from A to x0y0 z0 : v ==xyz – x0y0 z0 xx– 0yy– 0 zz– 0 . Applying Theorem 7.2, we have: vn ax– ax + by– by + cz– cz proj v ==------n ------0------0 ------0- abc n nn a2 ++b2 c2 ax – ax + by – by + cz – cz = ------0------0 ------0- abc a2 ++b2 c2
ax0 ++by0 cz0 – d ax ++by cz – d Since ax++ by cz = d: ==------a2 ++b2 c2 ------0 ------0 ------0 2 2 2 a ++b c a2 ++b2 c2
CYU 7.10 (a)AB ==25– 3 – 32–2 –175– and AC ==41 – 3 – 32–2 13– 5 .
ijk Here is a normal to the plane: n ==det –751 – –1020i –10j – k =–20 – 10 –10 . 13– 5 1 Here is a “nicer” normal: n ==–------–20 – 10 –10 211 . 10 Choosing the point A = 32–2 on the plane, we arrive at the general form equation of the plane: 211 x – 3y + 2 z – 2 ==0 or: 2xyz++– 6 0. (b) In Example 2.15, page 72, we found the vector form representation for the above plane: P = 3 – r + s –72 ++r 3s 25–5r – s rs . We are to show that: P ===3 – r + s –72 ++r 3s 25–5r – s rs xyz 2xyz++= 6 Q . If xyz = 3 – r + s –72 ++r 3s 25–5r – s P , then: 2zyz++= 23– r + s ++–72 ++r 3s 25–5r – s Thus: PQ . ==62–2r +s –72 +++r 3s 25–5r – s 6 If xyz is such that 2xyz++= 6 can we find real numbers r and s such that: 3 – r + s ==x –72 ++r 3s y and 2–5 5r – s ==z 62– x – y since 2xyz++= 6 which is to say: – r + s ==x –73 r + 3s y +2 and –5 5r – s =– 2x –4y + Turning to the system of equation: r s 1 0 –------3x ++y 19 – r + s = x – 3 –11 x – 3 10 augmented 7r + 3s = y + 2 rref 7xy+ – 19 matrix 73 y + 2 0 1 ------ 10 –5r –25s = – x –4y + –55 –2– x –4y + 00 0 steps ommited 7xy+ – 19 We see that rx= –3+ and s = ------does the trick. Thus: QP . 10 CYU SOLUTIONS B-33
CYU 7.11 (i) For v = ax2 ++bx c , vv ==ax2 ++bx c ax2 ++bx c a2 ++b2 c2 0 and vv = 0 only if abc===0 . 2 2 (ii) For u ==a2x ++a1xa0 v b2x ++b1x b0 : 2 2 uv ==a2x ++a1xa0 b2x ++b1x b0 a2b2 ++a1b1 a0b0
= ba2 ++b1a1 b0a0 ==b x2 ++b x b a x2 ++a x a vu 2 1 0 2 1 0 2 2 (iii) For u ==a2x ++a1xa0 v b2x ++b1xb0 , and r : 2 2 ruv = ra2x ++a1xa0 b2x ++b1xb0 2 2 = ra2x ++ra1xra0 b2x ++b1xb0
===ra2b2 ++ra1b1 ra0b0 ra2b2 ++a1b1 a0b0 r uv 2 2 2 (iv) For u ==a2x ++a1xa0 v b2x ++b1xb0 and z = c2x ++c1x c0 : 2 2 2 uv+ z = a2x ++a1xa0 + b2x ++b1xb0 c2x ++c1xc0 = a + b x ++a + b xa+ b c x2 ++c xc 2 2 2 1 1 0 0 2 1 0 = a2 + b2 c2 + a1 + b1 c1 + a0 + b0 c0
==a2c2 ++a1c1 a0c0 + b2c2 ++b1c1 b0c0 uz + vz
CYU 7.12 ru rv ==r u rv rruv =r2 uv
Definition 7.5(iii) Theorem 7.4 (c)
CYU 7.13 (a) rv ===rv rv r2 vv r2 vv =r vv CYU 7.12 Definition 7.5(iv) (b) uv+ 2 ==uv+ uv+ uu + v + vu + v =uu +++uv vu vv Definition 7.6 and 7.5(i): = u 2 +++uv uv v 2 = u 2 ++2 uv v 2
CYU 7.14 (a) 35– 8 ==35– 8 35– 8 532 ++552 58– 2 = 710.
(b) 35– 8 – 102 = 25– 10 25– 10 ==25– 10 25– 10 522 ++552 510– 2 = 645 B-34 CYU SOLUTIONS
CYU 7.15 1 1 1 1 1 1 1 1 1 1 ------u–------v ------u–------v 0 ------u – ------v ------u –0------u – ------v ------v u v u v u v u u v v 1 1 1 1 1 1 1 1 ------u ------u – ------v ------u – ------u ------v +0------v ------v u u v u u v v v 1 2 1 ------uu – ------uv +0------vv u 2 uv v 2 2 2 1 – ------uv +01 2 ------uv uv uv uv uv
–1 35– 8 102 CYU 7.16 = cos ------35– 8 102
–1 531 ++250 48– 2 = cos ------533 ++255 48– –8 511 ++200 423 –1 –49 = cos ------ 119.1 351 29
m
CYU 7.17 For any v = civi Spanv1v2 vm : i = 1 m m
uv ===u civi ci uv i 0 , since uv i = 0 for 1 im . i = 1 i = 1
CYU 7.18 Exercise 46, page 300
vw = a1v1 ++ anvn b1v1 ++ bnvn
= a1v1 ++ anvn b1v1 ++a1v1 +++a2v2 anvn bnvn
n n n n == aivi b1v1 ++ aivi bnvn aib1 vi v1 ++ aibn vi vn i = 1 i = 1 i = 1 i = 1 n 1ifij= since vi vj = : ==a b ++ a b a b 0ifij 1 1 n n i i i = 1 CYU SOLUTIONS B-35
CYU 7.19 The first order of business is to determine a basis for the space S spanned by the vec- tors 2110 , 1010 , 3120 , 0101 . Applying Theorem 3.13, page 103 we see that first, second, and fourth of the above four vectors constitute a basis: v1 = 2110 , 2130 1010 1011 0110 v2 = 1010 , v3 = 0101 of S: . We now apply the Grahm- 1120 0001 0001 0000
Schmidt Process to that basis v1v2 v3 to generate an orthonormal basis u1u2 u3 for S:
u1 ==v1 2110 u v 1 2 2110 1010 u2 ==v2 – ------u1 1010 – ------2110 u1 u1 2110 2110 3 1 1 ==101 – ---2110 0–------0 6 2 2
1 1 0–------0 0101 u1 v3 u2 v3 2110 0101 2 2 1 1 u3 ==v3 – ------u1 – ------u2 0101 – ------2110 – ------0 –------0 u u u u 2110 2110 1 1 1 1 2 2 1 1 2 2 0–------0 0–------0 2 2 2 2 1 –12 1 1 1 1 1 ==0101 – ---2110 – ------0–------0 –---------1 6 12 2 2 3 3 3
1 1 1 1 3 1 1 1 Conclusion: ---2110 ---0–------0 ---–---------1 is an orthonormal basis for 6 2 2 2 4 3 3 3 Span2110 1010 3120 0101 .
CYU 7.20 A consequence of Theorem 7.10(iii) and CYU 3.11, page 98.
CYU 7.21 We first use the Grahm-Schmidt Process to determine the orthonormal basis
w1 w2 of W stemming from the given basis v1 v2 = 101 120 : 1 u1 ==v1 101 : w1 = ------101 2 u v 1 2 101 120 1 1 2 1 1 ------120 – ------101 ---2–---------2–--- u2 ==v2 – u1 =: w2 = u1 u1 101 101 2 2 3 2 2
Turning to Theorem 7.11, we determine the orthogonal projection, vW , of the vector v = 201 onto W: B-36 CYU SOLUTIONS
vW = vw 1 w1 + vw 2 w2 1 1 2 1 1 2 1 1 = 201 ------101 ------101 + 201 ---------2 –---------2 –--- 2 2 3 2 2 3 2 2 3 1 1 1 7 5 ==---101 + ------2 –--- ---1 --- 2 22 2 4 4
CYU 7.22 NOTE: It is easy to see that the function 2 n fa0a1 an = a0 ++a1xa2x ++ anx from the standard (dot-product) inner product space n of Theorem 7.1 (page 279) to the polyno- mial inner product space Pn of Exercise 23 (page 299) is an isomorphism which ALSO preserves the inner product structure of the two spaces: n a0a1 an b0b1 bn ==fa0a1 an fb0b1 bn aibi i = 0
As such, we could translate the given P3 -problem: Find the shortest distance between the vector v =33x2 + x and the 2 3 subspace W = Span x + 1 x + 1 in the inner product space P3 into the following 4 -form: Find the shortest distance between the vector v = 03 30 and the sub- space W = Span 0101 1001 in the inner product space 4 . You are invited to take the above 4 -approach. For our part, we will deal directly within the inner product space P3 : 2 Employing the Grahm-Schmidt Process we go from the basis v1 v2 , with v1 = x + 1 and 3 2 3 v2 = x + 1 , to an orthonormal bases w1 w2 for W = Span x + 1 x + 1 : u 2 1 1 2 u1 ==v1 x + 1 : w1 ==------x + 1 . u1 2
u1 v2 x2 + 1 x3 + 1 u ==v – ------u x3 + 1 – ------x2 + 1 2 2 1 2 2 u1 u1 x + 1 x + 1 u 1 1 1 2 2 1 1 ==x3 + 1 – ---x2 + 1 x3 – ---x2 + --- : w ==------x3 – ---x2 + --- 2 2 2 2 2 u2 3 2 2 Turning to Theorems 7.11 we determine the projection vW of v =33x + x onto W: vW = vw 1 w1 + vw 2 w2 1 1 2 1 1 2 1 1 = 3x2 + 3x ------x2 + 1 ------x2 + 1 + 3x2 + 3x ---x3 – ---x2 + --- ---x3 – ---x2 + --- 2 2 32 2 32 2 3 1 3 2 1 1 ==------------x2 + 1 – ------x3 – ---x2 + --- x3 ++x2 1 22 232 2 CYU SOLUTIONS B-37
Appealing to Theorem 7.12 we calculate the shortest distance between the vector v =33x2 + x and the subspace W = Span x2 + 1 x3 + 1 :
2 3 2 vv–3W = x + 3x – x ++x 1 ==–2x3 –31x2 ++ –2x3 –31x2 ++ –2x3 –31x2 ++
===–1 2 ++–2 2 3 2 +1 2 93
CYU 7.23 Determining the eigenvalues of A:
2 – 11 2 detA – I ==det 1 2 – 1 –4 – 4 – 1 Eignevalues: == 1 112– details omitted Determining the corresponding eigenspaces: –112 –112 10– 1 E4 ==nullA – 4I null 12–1 =aaa a since rref 12–1 =01– 1 11– 2 11– 2 00 0
111 111 111 E1 ===nullAI– null 111 – c – dcd cd since rref 111 =000 111 111 000 As is indicated in Theorem 7.14: aaa – c – dcd ==ac– – d ++ac ad 0
CYU 7.24 (a) v w 111 1 1 121 v w = v1 ++v2 v3v1 ++2v2 v3 v1 ++v2 3v3 w1w2 w3 2 2 113 v3 w3
= w1v1 ++v2 v3 ++w2v1 ++2v2 v3 w3v1 ++v2 3v3 v w 1 111 1 v 121 w = v1v2 v3 w1 ++w2 w3w1 ++2w2 w3 w1 ++w2 3w3 2 2 v3 113 w3
= v1w1 ++w2 w3 ++v2w1 ++2w2 w3 v3w1 ++w2 3w3 = w v ++v v ++w v ++2v v w v ++v 3v 1 1 2 3 2 1 2 3 3 1 2 3 B-38 CYU SOLUTIONS
111 0 1 (b) One possible answer: 000 1 0 ==100 100 1 000 0 0
0 111 1 while: 1 000 0 ==010 100 0 0 000 0
CYU 7.25 (a) Tabc ABC = ab– – a + 2bc– – b + c ABC = Aa– b ++Ba–2+ bc– Cb– + c abc TABC = abc AB– – A + 2BC– – B + C = aA– B ++bA–2+ BC– cB– + C = Aa– b ++Ba–2+ bc– Cb– + c (b) For Tabc = ab– – a + 2bc– – b + c and = 010 001 100 :
11–0 T = –211 – 01–1
11–0 CYU 7.26 Consider the symmetric matrix AT== –211 – of CYU 7.25(b). 01–1 Employing Theorem 6.10 of page 226, we find a basis of eigenvectors for the linear operator Tabc = ab– –2a + bc– – b + c :
1 – –01 detA – I ==det –1 2 – –1 –0– 1 – 3 Eignevalues: === 1 3 01–1– Here are the associated eigenspaces: CYU SOLUTIONS B-39
11–0 11–0 10– 1 E0 ==nullA null –211 – =aaa a 3 since rref –211 – =01– 1 01–1 01–1 00 0
01–0 01–0 101 E1 ==nullAI– null –111 – =–a0 a a 3 since rref –111 – =010 01–0 01–0 000
–12 –0 –12 –0 10– 1 E3 ==nullA – 3I null –11 –1– =a–2a a a 3 since rref –11 –1– =01 2 01–2– 01–2– 00 0
Letting a = 1 in each of the above eigenspaces we arrive at a normal basis for 3 consisting of eigenvectors of T: 111 –101 12– 1 (Theorem 6.15); which is easily turned 1 1 1 –1 1 1 –2 1 into an orthonormal basis: ------------------0------------. 3 3 3 2 2 6 6 6
CYU 7.27 If A–1 = A T and B–1 = B T , then: AB –1 ===B–1A–1 BTAT AB T Theorem 5.12(iii), page 167 Exercise 19(f), page 162
2 1 1 CYU 7.28 In CYU 7.23 we showed that the matrix 121 has eigenvalues ==4 1 112 with E4 ==aaa a E1 – a – bab ab , with 111 a basis for E4 , and –101 –101 a basis for E1 (set a = 0 and b = 1 , and then set b = 0 and a = 1 ). Applying the Grahm-Schmidt process (Theorem 7.9, page 303) to 1 1 –101 –101 , we arrive at the orthogonal basis –101 –1---–--- of E1 , and 2 2
1 1 1 1 1 1 1 1 to the orthonormal basis of eigenvectors ------–0------–------–------------–------. 3 3 3 2 2 6 6 6 Turning to the marginal comment on page 314 we conclude that:
2 1 1 400 13 – 12 – 16 T P 121P = 010, where P = 13 016 112 001 13 12 –16 B-40 CYU SOLUTIONS ANSWERS C-1
Appendix C Answers to Selected Exercises 1.1 Systems of Linear Equations, page 11. 5xy++4z = 6 1. 33–12 3. 5. 100 7. 100 –32x – y +4z = 55–1 9– 010 010 –43 –10 1 001 001 ---xy–0= 2 000
9. x = 1 y ==0 z 2 11. x1 = 1 x2 = 2 x3 = 2 x4 = 2 x5 = –1 1 1 1 3 13. x ==0 y –24 z = 15. x = --- y = --- z = –--- w = --- 2 4 2 4 19. No: first non-zero entry in last row is not 1. 23. x ==5 y –22 z = 25. x = 12 y = –5 z = 1 w = 0
1.2 Consistent and Inconsistent Systems of Equations, page 23. 1. r 2 r 3. –22–2– 5. –s – 3t r 12– t 12– sst rst
12+ r 36+ r 7. 11–6 5r–3+ r r r 9. 79– 6 11. 2 – r ------------rr 5 5 13. Yes. 15. No. Solutions if and only if a, b, c, satisfy the equation 4ab–2+0c = .
3 17. No. 19. Yes. 23. –---rr 0 rR 25. –11r – s r + 6srs rs R 2 27. No. 29. No. 31. a 1 33. None. 35. ab 1 37. ad– bc 0 2.1 Vectors in the Plane and Beyond, page 38.
1. 3. 5. 23 3_ B = –32 . 3_ 3 _ _ 2 –32 2 2 . A = 11 . B = 01 . 1_ . 1 1 2 3 – 2 _ –3 –2 –1 1 2 -3 -2 -1 1 2 3 –1 -1 AB. . _ . –2 -2 A = –12 ANSWERS C-2
7. 20– 2 9. –9111 – 11. 13 – 13 15. –6 6 17. 25 – 7 1 5 1 19. (a) r = –14 s = 10 (b) r = –--- s = ------(c) r = --- s = ------5 10 7 14
19 17 11 5 5 21. r ==–------ s ------ t = ------23. ------ –------6 6 6 2 2 2.2 Abstract Vectors Spaces. page 49. 1. No. 3. No. 5. Yes. 7. No. 9. Yes. 11. No. 13. No. 15. Yes.
2.3 Properties of Vectors Spaces, page 57. (All exercises call for either a proof or a counterexample)
2.4 Subspaces, page 65. 1. Yes. 3. Yes. 5. No. 7. Yes. 9. No. 11. No. 13. Yes. 15. Yes. 17. No. 19. No. 21. Yes. 23. Yes. 25. Yes. 27. Yes. 29. No. 31. Yes. 33. No. 35. Yes. 37. Yes.
2.5 Lines and Planes, page 73.
1. r15 r 3. r51 – r 5. 13 + r17 – r
7. 35 + r02 r 13. 37 + r15 r 15. 37 + r51 – r
17. 37 + r17 – r 19. 37 + r02 r 21. 37 + r51 – r
23. 37 + r15 r 25. 37 + r71 r 27. 37 + r01 r
29. r245 r 31. r–402 r 33. 245 + r134 – –
35. 210 + r13 – 1 41. 12– 1 + r245 r
43. 12– 1 + r–402 r 45. 12– 1 + r13–4– r
47. 12– 1 + r13– 1 r 49.r132 + s211 rs
51. r200 + s020 rs 53. 341 ++r13– 4 s232 rs
55. 24– 3 ++r33–8 s23–2rs ANSWERS C-3
3.1 Spanning Sets, page 84.
1. No. 3. Yes. 5. Yes. 7. No. 9. No. 11. No. 13. cos2x = cos2x – sin2x
15. sin--- – x = sin--- cosx – cos--- sinx 17. Span. 7 7 7 19. Do not span. Just about any randomly chosen four-tuple will not be a linear combination of the given vectors (check it out). 21. Do not span. Just about any randomly chosen four-tuple will not be a linear combination of the given vectors (check it out). 23. Do not span. Just about any randomly chosen four-tuple will not be a linear combination of the given vectors (check it out). 25. Span. 27. All c 0 . 3.2 Linear Independence, page 91.
1. Yes. 3. No. 5. No. 7. Yes. 9. No. 11. No. 13. Yes. 15. No. 17. Yes. 19. Yes. 21. No. 23. No. 25. Yes. 27. No. 29. a = 3 3.3 Bases, page 104.
5 5 1. (a) –3 --- = – 3e + ---e (b) 320 = 3e ++2e 0e 2 1 2 2 1 2 3
19 25 14 3. (a) x2 + 3x – 1 = ------2x2 + 3 – ------x2 – x + ------x – 5 13 13 13 5. No. 7. Yes. 9. Yes. 11. Yes. 13. Yes. 15. No. 17. No. 19. Yes.
13 20 01 21 29. 31. Do not span. –21 11– 12 01
33. A basis for SpanS : 214 –132 . A basis for 3 : 214 –132 111 .
35. A basis for SpanS = 3 : S = 113 –132 32– 1
37. A basis for SpanS = 5 : S = 13132 24142 11202 22111 12345
39. A basisfor SpanS : 5 – x3 – xx 4 ++++x3 x2 x 12 x4 – 2x2 . A basis for 3 : 5 – x3 – xx 4 ++++x3 x2 x 12 x4 – 2x2 1 + x
41. sinxx cos sin2x cos2x sin2x 43. –1 1001010 –1001
45. x3 ++x2 2x x + 1 47. c 0 49. a 0 b 0 and ab ANSWERS C-4
4.1 Linear Transformations, page 120. 1. Yes 3. No 5. No 7. No 9. Yes 11. No 13. Yes 15. No ab+ 17. No 19. (a) 4102– (b) ------ 2aa – + b 2
83 ab+ b 21. (a) (b) 27. b = 0 39. LT ab = 2a + 2b – a – b 13 8 2ab+ ab+
41. LT ab = 3ab+ x + 6a + 2b 43. KLT a = –022aa 4.2 Kernel and Image, page 131. 1. Linear. Nullity: 0, Rank: 1. 3. Linear. Nullity: 0, Rank: 1. A basis for the image space: 11 – 5. Not linear. 7. Linear. Nullity: 0, Rank: 2.
00 11 9. Linear. Nullity: 2, Rank: 2. A basis for the kernel: . 01 10
A basis for the image space: x – 1 –1x2 + . 11. Not linear. 13. Linear. Nullity: 1, Rank: 2. A basis for the kernel: x2 – x . A basis for the image space: 01 11 . 15. Linear. Nullity: 0, Rank: 4.
10 –11 01– 17. Linear. Nullity: 0, Rank: 3. A basis for the image space: . 10 11 01
19. Nullity: 1, Rank: 3. A basis for the kernel: 1 . A basis for the image space: x2x 1 .
21. Nullity: 0, Rank: 4. A basis for the image space: x3 x2x 1
23. Nullity: 0, Rank: 3. A basis for the image space: x2x 1 .
x2 1 x 1 25. Nullity: 2, Rank: 1. A basis for the kernel: ----- – --- --- – --- . A basis for the image space: 1 . 3 3 2 2
27. Nullity: 0, Rank: 3. A basis for the image space: x3x2 x . 29. (a) 033 –
21 22 01 (b) x3 + x and 5. 31. Nullity: 0, Rank: 3. A basis for the image space: . 01 13 10
33. (a) Tab = 9aa3a (b) Tab = 9ab0 (c) None exists. ANSWERS C-5
35. nullityT = n , rankT = 0 ; nullityT = 0 , rankT = n . 37. nullityT = 1 , rankT = 2 ; nullityT = 2 , rankT = 1 ; nullityT = 3 , rankT = 0 . 39. nullityT = 1 , rankT = 4
4.3 Isomorphisms, page 145.
1 1 1. f –1x = –---x 3. f –1ab = b –---a 5. f –1ax2 ++bx c = –cab 5 2 7. Isomorphism. 9. Not an isomorphism. 11. Isomorphism. 13. Not an isomorphism. 15. Isomorphism. 17. Isomorphism. 19. Isomorphism. 21. Not an isomorphism. 23. f –1xy = y + 4 xy–7– . Zero: 34 – and –xy = –6x + –8y – .
5.1 Matrix Multiplications, page 161.
011 002 010 Column: 3 10 4 262 1. 14– 3. 200 and 006 5. Row: 3215 106110 –151 090 600
Column: 1–55 1 –1 032 01–2–1– n 7. 25. 11 = 1 n Row: 1101 –0141– 53– 27 01 01
n n 10 10 27. 10 = 29. 10 = 02 02n 02 02n
5.2 Invertible Matrices, page 175.
2 –1 –--8- --- 1 3 1 5 5 --- 0 --- –--- –124 2 4 2 1 2 1 1. 3. 5. 20– 1 7. Singular. 9. Singular. 11. ------–------1 1 1 2 5 10 0 --- –------–011 5 4 2 5 1 1 --- –--- 3 3
500 001 13. ab 12 15. b 0 and a 1 17. E1 = 010, E2 = 010 001 100
1 19. X = ---ABA–1 21. XA= –1BAB 23. XAB= 2A 2 ANSWERS C-6
5.3 Matrix Representation of Linear Maps, page 186.
--8------17- 2 5 5 5 1. 23 = , T23 = 3. 23 = , T23 = 3 6 –--1- –--4- 5 5
4 7 2 0 5. 231 = 6 , T231 = 7 7. 12 = , T12 = 12 1 –3 –6 –3
0 0 2 1 2 –2 9. x ++x 1 = , Tx++x 1 = 0 2 1 0
11– 2 –1 –33 – –3 11. T = 1 1 , T121 ==T 121 3 13. 21 , T12 ==T 12 2 ------1 --- 2 2 2 –11 –1
01 –1 15.T = 01– ,T2x + 1 ==T 2x + 1 1 13 1
010 –1 17. I = 001 , I121 ==I 121 2 100 0
203 2 5 101 1 12 12 2 19. T = , T==T 516 4 21– 21– 8 –132 –2– –8
000 0100 20 6 100 21. T = , T13 ==T 13 23. (a) T = (b) D = 0020 11 4 010 0003 001
0000 000 100 0100 0100 100 (c) D T = 020 (d) T D = (e) TDT = (f) DTD = 0040 0020 020 003 0009 0003 003
0000 0 0130 3 2 3 2 3 25. D = , D5x + 3x ==D 5x + 3x 27. Tab = –43a + b 2a 0004 5 0000 0
00 –13 – 29. T is the identity map. 31. LT ==L T --2- --1- 3 3 ANSWERS C-7
--1- --1- –1 b –22 – –1 2 2 35. T ax+ b = a --- , T = , T = 2 42 –1 –--1- 2
1001 –011 –1– –1ab 1100 –1 111 1 37. T = ab – ad + cc , T = , T = cd –12 –11– 111 0 01–0 1 201 1
00 0 0 1 1 00 1 1 0 0 00–1 1–1–0 39. L = 00 0 1 1 0 01 1 0 0 0 10 0 0 0 0
5.4 Change of Basis and Similar Matrices, page 199.
12 ------2 5 1. 25 ==I 25 3. 23– 1 ==I 23– 1 3 1 --- –1 5
0 –1 2 2 20 20 1 5. 2x ++x 1 ==I 2x ++x 1 1 7. ==I 11 11 –2 1 2
1 1 –1 --- –--- 01– 2 2 9. T ==I T I 11. T ==IV T IV 10 000 201
0100 0040 13. T ==IV T IV 9 15. 011 110 021 19. I 000--- n 2 0000
0 –--1- 3 –011 – 1 25. T ==I T I 0 --- 27. T ==I T I 110 W V 3 W V 101 1 1 --- 3
–14 –2– 29. T ==IW T IV 211 ANSWERS C-8
6.1 Determinants, page 215. 1. 6 3. 6 5. 0 7. 86 9. 9 11. 9 13. 81 15. k 0 , k 1 17. k 01
6.2 Eigenspaces, page 228. 1. = –14 ; E–1 = –r 2r r , E4 = r 3r r
3. = 45 ; E4 = r 3r r , E5 = r 2r r
5. = 28 ; E2 = – 2r + srs rs , E8 = r –2rr r
7. = 21 ; E2 = 03 rr0 r , E1 = 2r03r 4r r E–1 = 00r 0 r
9. = 123 ; E1 = 32 – rr r , E2 = r00 r , E3 = 0r 0 r
11. = 3 ; E3 = rsrs– rs , E–3 = rr3r r
13. = 01 – ; E0 = r0 rr r , E–1 = r00s rs
15. = 0 , E0 =
17. = 516 – ; E5 = 2rr r , E–16 = r 4r r
19. = 42 – ; E4 = 3rr2r r , E–2 = 3r – 3srs rs
21. = 21 – 3 ; E2 = 3rr34r 9r r , E–1 = 0 –rr3r r E3 = 00 2+ 3rr r , E3 = 00 2– 3rr r
23. = 1 ; E1 = rx+ r r , E–1 = –rxr+ r
25. = 1 , E1 = –rx2 ++sx r rs , E–1 = rx2 + r r
27. = 012 , E0 = rx2 r , E1 = –2rx3 + rx2 r , E2 = –2rx2 + rx r
57r 38r 29. = 4 ; E4 = r 31. = 0 ; E0 = V 84r –72r
39. (a) a2 ++d2 –42ad bc 0 (b) a2 ++0d2 –42ad bc = (c) a2 ++d2 –42ad bc 0 ANSWERS C-9
6.3 Diagonalization, page 243.
1. Not diagonalizable. 3. Diagonalizable. 5. Diagonalizable. 7. Diagonalizable. 9. Not diagonalizable. 11. Diagonalizable. 13. Diagonalizable. 15. Diagonalizable. 17. Not diagonalizable. 19. Diagonalizable. 21. Diagonalizable. 23. Diagonalizable. 25. Not diagonalizable. 27. Diagonalizable. 29. Not diagonalizable. 31. Diagonalizable. 33. Not diagonalizable.
6.4 Applications, page 256.
k k 2 15+ 15+ 1. sk = ------– ------5 2 2
k – 1 k – 1 k – 2 k – 2 1 15+ 15– 15+ 15– 3. sk = ------–2------+ ------– 2 ------5 2 2 2 2
k + 1 –2x –3x k – 1 2 1 1 k y1 –c1e – c2e 5. sk = 32 – 2 7. sk = ------++------–1 13. = 6 y –2x –3x 3 2 2 c1e + 2c2e
6x –7x x –x fx – 2c1 + c2 – c3e x c1e + c2e – 7c3e f1x e3x + ex 15. gx = c – 2c e6x 17. y = c e–7x – 2c ex – 3c e–x 19. = 1 3 1 2 3 3x x f2x – e + e hx 6x z –7x x –x c2 + c3e c1e ++c2e c3e
1 7 –--3-e–7x ++---ex ---e–x 5 8 2 8 –---t fx A-concentration of alcohol: 4.8– 2.8e 2 3 3 21. gx = –---e–7x – ex + ---e–x 23. 8 8 5 hx –---t 3 1 1 2 –---e–7x + ---ex – ---e–x B-concentration of alcohol: 3.2+ 2.8e 8 2 8
6.5 Markov Chains, page 270.
A B A .3 .7 1. Regular 3. No 5. Regular 7. .4 .6 9. A B .6 .9 .1 B .4 ANSWERS C-10
A .5 .2 (a) Probability 1 of ending up at A. 1 .5 (0 probability of ending up in B or C) BC(b) Probability 1 of ending up at A. 11. .5 .6 13. (0 probability of ending up in B or C) .4 .3 (c) Probability 1 of ending up at A D (0 probability of ending up in B or C)
-----3- 2 3 10 ------.40 5 8 1 .62 .64 .66 15. 17. 19. --- 29. (a) (b) (c) 31. .22 3 5 2 .39 .36 .34 ------.38 5 8 --1- 5
.17141 .27767 33. (a) 0.247 (b) 0.265 (c) 0.273 (d) .24065 (e-a) 0.281 (e-b) 0.274 (e-c) 0.274 .12204 .18823 (e-d) The initial state of the system has no bearing on the fixed or final state of the system. Inde- pendent of the initial state, eventually (to five decimal places): 17.141%, 17.767%, 24.065%, 12.204%, and 18.823% of the employees will be enrolled in plans A, B, C, D, and E, respectively.
.33333 35. (a) 0.63 (b) 0.62 (c) 0.59 (d-a) 0.50 (d-b) 0.49 (d-c) 0.48 (e) .22222 .44444
14 37. (a) 0.33 (b) 0.25 (c) 14 14 14 ANSWERS C-11
7.1 Dot Product, page 289.
15 9 1. 33 3. 13 5. 15 7. a = ------9. ac= b = ------for any c 0 2 2c
–1 31 4 8 6 3 11. cos ------ 42 13. ---0 –--- ---3 --- 83 21 5 5 5 5
15. A normal form: 213 x – 1y – 3 z + 1 = 0 ; a general form: 2xy++3z = 1 ; a vector form: 010 ++r120 – s031 – rs .
17. A normal form: 401 x – 1yz+ 3 = 0 ; a vector form: 1 001 ++r10– 4 s010 rs . 19. 2 21. ------5529 19 9 –3a – b 23. ------25 (a) ab------ab (b) 7c–45c c c (c) 000 21 2
27. 3x ++2y 6z = 6
7.2 Inner Product, page 298.
–1 –19 –1 –18 1. 7 3. 29 5. cos ------ 126 9. 14 11. 77 13. cos ------ 158 721 378
1 –1 –1 19. 6 21. cos ------ 96 25 2x2 –3x + 2dx 10.1 221 0
1 2x2 –3x + – x2 + x – 5 dx 1 27. cos–1------0 ------ 174 29. ex – x 2dx 1.2 1 1 0 2x2 –3x + 2dx – x2 + x – 5 2dx 0 0
sinxxxcos d 31. sin2xxd 0.08 33. cos–1------–------= 0 – sin2xxd cos2xxd – – ANSWERS C-12
7.3 Orthogonality, page 310
00 1 1 1 –1 2 –1 –1 0 1 1 1 134 ------------------------------------12------21– ------1. 3. No 5. 1 1 3 3 3 6 6 6 2 2 2 30 34 5 00 5 --- –--- 3 4
4 x2 x 1 5 x2 2x 1 3 x2 1 7. No 9. ------------++------– ------------– ------+ ------+ ------------– ------33 3 3 66 6 6 22 2
–612r + 11. arb== –31 – r for r . 13. a = 1 b = 2 15. br= a = ------for r . –43 + r
1 10 2 2 1 1 1 1 –7 8 –7 8 –------------ ------01 5 ------------ ------------17. 19. 105 105 105 5 5 5 5 5 226 226 226 226
2 1 1 18 4 1 0 2 0 36 –40 –18 6 ------0 ------ –------------------------ ------------21. 23. 5 5 341 341 341 5 5 5 5 3256 3256 3256 3256
25 12 6 1 2 1 1 2 22 2 2 5x2 ------x2 – ------x – ------------0 00------–------ ------------------25. 27. 29. 31 31 31 5 5 2 2 11 11 211211
x3 x2 2 x3 5x2 2 31. x------++------– ------+ ------33. (a) 010 20– 1 6 6 6 30 30 30
3 4 305 (b) 13– 2 = – ---102 + ---20– 1 + 3010 (c) ------5 5 25
9 35. (a) –2 010 01–01 (b) 4133 – = 2141 + 2212 – – (c) ------5
1 1 37. 418 , 12– 1 = ------55 54 41 + ------–832–64– 23 23 7.4 The Spectral Theorem, page 324
123 21 21 13. 15. 210 17. 13 13 302 Index I-1
A E Abstract Vector Space, 40 Eigenvalues, 218, 224 Algebraic Multiplicity, 238 Eigenvectors, 218, 224 Angle Between Vectors, 281, 296 Eigenspaces, 218, 224 Augmented Matrices, 3 Elementary Operations, 2 Elementary Matrix, 168 B Equivalent Matrices, 3 Basis, 94 Equivalent Systems of Equations, 2 Ordered, 177 Consistent, 14 Bijection, 135 Inconsistent, 14 Euclidean Vector Space, 35 C Expansion Theorem, 90, 100 Cauchy-Schwarz Inequality, 295 Change of Basis Theorem, 193 F Characteristic Equation, 219, 225 Fibonacci Numbers, 246 Characteristic Polynomial, 219, 225 Formula for nth term, 248 Closure Axioms, 40 Function Spaces, 44 Coefficient Matrix,18 Fundamental Theorem of Homogeneous Cofactor, 206 Systems of Equations, 20 Column Space, 156 Composition, 117 G Composition Theorem, 117, 182 Gauss-Jordan Elimination Method, 10 Consistent Systems of Equations, 14 Geometric Multiplicity, 238 Coordinate Vector, 177 Golden Ratio, 248 Counterexample, 13 Grahm-Schmidt Process, 303 Cross Product, 287 H D Homogeneous Systems of Equations, 20 Determinant, 205 Cofactor, 206 I Minor, 206 Idempotent Matrix, 162 Diagonal Matrix, 161, 233 Image, 124 Diagonalizable Matrix, 233 Inconsistent Systems of Equations, 14 Diagonalizable Operator, 233 Inner Product, 292 Dimension, 98 Inner Product Space, 292 Dimension Theorem, 126 Distance between vectors, 294 Distance Between Vectors, 280, 294 Norm of a vector, 294 Dot Product, 279 Invertible Matrix, 164 Inverse Function, 136 Isomorphic Spaces, 139 Isomorphism, 138
I-2 INDEX
K Space, 42 Kernel, 124 Symmetric, 161 Trace, 162 L Transpose, 161 Upper Triangle, 207 Laplace Expansion Theorem, 206 Proof, 212 Leading One, 7 N Linear Combination, 77 n-Tuples, 33 Linear Extension Theorem, 115 Nilpotent Matrix, 162 Linear Independent, 86 Norm, 280, 294 Linear Independence Theorem, 22 Normal Vector, 269 Linear Transformation (map),111 Null Space, 155 Image, 124 Nullity, 124 Kernel, 124 M O Markov Chain, 259 One-To-One Function, 129 Regular, 263 Onto Function, 129 Fundamental Theorem, 264 Ordered Basis, 177 Transitional Diagram, 259 Orthogonal Complement, 290 Transitional Matrix, 259, 261 Orthogonal Matrix, 321 Initial State. 260 Orthogonal Vectors, 282, 302 Fixed State, 261 Orthogonal Projection, 307 Matrix Orthogonal Set of Vectors, 301 Augmented, 3 Orthonormal Matrix, 321 Coefficient, 18 Orthonormal Set of Vectors, 302 Cofactor, 207 Orthonormally Diagonalizable, 322 Column Space, 156 P Determinant, 205 Pivoting, 4 Diagonal, 233 Pivot Point, 5 Diagonalizable, 236 Plane Elementary, 168 General Form, 285 Equivalent, 3 Normal Form, 285 Idempotent, 162 Scalar Form, 285 Invertible, 164 Vector Form 71 Minor, 206 Polynomial Space, 43 Multiplication, 151 Properties, 153 R Nilpotent, 162 Rank, 124, 159 Null Space, 155 Recurrence relation, 249 Orthogonal, 321 Reduction Theorem, 100 Orthonormal, 321 Row-Echelon Form, 12 Powers, 159, 167 Row Operations, 2 Rank, 158 Row-Reduced-Echelon Form, 7 Representation of a Linear Map, 179 Row Space, 157 Row Space, 157 Similar, 195 Skew-Symmetric, 162 Index I-3
S U Scalar Product, 34 Unit Vector, 302 Similar Matrices, 195 Upper Triangle Matrix, 207 Skew-Symmetric Matrix, 163 Spanning, 79 V Spanning Theorem, 18 Vector, 31 Spectral Theorem, 319, 322 Abstract, 40 Stochastic Process, 259 Additive Inverse, 31 Subtraction, 55 Addition, 34 Subspaces, 59 Coordinate, 179 Of 2 , 68 Cross Product, 287 Decomposition, 283 3 Of , 70 Normal, 285 Symmetric Matrix, 161, 315 Scalar product, 34 Symmetric Operator, 317 Sum, 25 Systems of Differential Equations, 249 Subtraction, 55 Systems of Linear Equations, 1 Standard Position, 32 Equivalent, 2 Unit, 302 Elementary Operations, 2 Zero, 35 Homogeneous, 20 Vector Space Abstract, 40 T Euclidean, 35 Theorems Function, 44 Cauchy-Schwarz Inequality, 279 Matrix, 42 Change of Basis, 193 Polynomial, 43 Composition Theorem, 182 Properties, 51, 56 Dimension Theorem, 126 Subspaces, 59 Expansion Theorem, 90, 100 Trivial, 48 Fundamental Theorem of Homogeneous systems of Equations, 20 W Grahm-Schmidt Process, 303 Laplace Expansion Theorem, 206 Weighted Inner Product Spaces, 292 Linear Extension Theorem, 115 Linear Independence Theorem, 22 Z For n , 88 Zero Vector, 35Weigh Reduction Theorem, 100 Spanning Theorem, 1 Spectral Theorem, 319, 322 Vector Decomposition Theorem, 267,306 Trace, 162 Translation Vector, 72 Transpose of a Matrix, 161, 264 Triangle Inequality, 296 Trivial Space, 48