Math 75 Linear Class Notes

Prof. Erich Holtmann

For use with Elementary , 7th ed., Larson

Revised 21-Nov-2015

p. i

Contents Chapter 1: Systems of Linear 1 1.1 Introduction to Systems of Equations. 1 1.2 Gaussian Elimination and Gauss-Jordan Elimination. 9 1.3 Applications of Systems of Linear Equations. 15 Chapter 2: Matrices. 19 2.1 Operations with Matrices. 19 2.2 Properties of Matrix Operations. 23 2.3 The Inverse of a Matrix. 27 2.4 Elementary Matrices. 31 2.5 Applications of Matrix Operations. 37 Chapter 3: . 41 3.1 The of a Matrix. 41 3.2 Determinants and Elementary Operations. 45 3.3 Properties of Determinants. 51 3.4 Applications of Determinants. 55 Chapter 4: Vector Spaces. 61 8.1 Complex (Optional). 61 8.2 Conjugates and Division of Complex Numbers (Optional). 67 4.1 Vectors in Rn. 71 4.2 Vector Spaces. 77 4.3 Subspaces of Vector Spaces. 82 4.4 Spanning Sets and Linear Independence. 86 4.5 Basis and Dimension. 97 4.6 Rank of a Matrix and Systems of Linear Equations. 105 4.7 Coordinates and Change of Basis. 117 Chapter 5: Inner Product Spaces. 121 5.1 Length and Dot Product in Rn. 121 5.2 Inner Product Spaces. 129 5.3 Orthogonal Bases: Gram-Schmidt Process. 137 5.4 Mathematical Models and Least Squares Analysis (Optional). 145 5.5 Applications of Inner Product Spaces (Optional). 151 8.3 Polar Form and De Moivre's Theorem. (Optional) 157 8.4 Complex Vector Spaces and Inner Products. 163

p. i Chapter 6: Linear Transformations. 169 6.1 Introduction to Linear Transformations. 169 6.2 The and Range of a Linear Transformation. 175 6.3 Matrices for Linear Transformations. 183 6.4 Transition Matrices and Similarity. 191 6.5 Applications of Linear Transformations. 193 Chapter 7: Eigenvalues and Eigenvectors. 201 7.1 Eigenvalues and Eigenvectors. 201 7.2 Diagonalization. 209 7.3 Symmetric Matrices and Orthogonal Diagonalization. 215 7.4 Applications of Eigenvalues and Eigenvectors. 223 8.5 Unitary and Hermitian Spaces. 223

p. ii Chapter 1: Systems of Linear Equations 1.1 Introduction to Systems of Equations.

Chapter 1: Systems of Linear Equations

1.1 Introduction to Systems of Equations.  Objective: Recognize and solve mn systems of linear equations by hand using Gaussian elimination and back-substitution.

a1x1 + a2x2 + … + anxn = b is a linear in standard form in n variables xi.

The first nonzero coefficient ai is the leading coefficient. The constant term is b.

Compare to the familiar forms of linear equation s in two variables y = mx + b and x = a.

Example: Linear and Nonlinear Equations

2 (sin ) x1 – 4x2 = e sin x1 + 2x2 – 3x3 = 0 Linear Nonlinear

An mn system of linear equations is a set of m linear equations in n unknowns.

Example: Systems of Two Equations in Two Variables

Solve and graph each 22 system.

a. 2x – y = 1 b. 2x – y = 1 c. 2x – y = 1 x + y = 5 –4x + 2y = –2 –4x + 2y = 5

 For a system of linear equations, exactly one of the following is true. 1) The system has exactly one solution (consistent, nonsingular system). 2) The system has infinitely many solutions (consistent, singular system). Use a free parameter or free variable (or several free parameters) to represent the solution set. 3) The system has no solution (inconsistent, singular system).

p. 1 Chapter 1: Systems of Linear Equations 1.1 Introduction to Systems of Equations. To solve mn systems of linear equations (when m and n are large) we use a procedure called Gaussian elimination to find an equivalent system of equations in row-echelon form. Then we use back-substitution to solve for each variable.

Row-echelon form means that the leading coefficients of 1 (called “pivots”) and the zero terms below them form a stair-step pattern. You could walk downstairs from the top left. You might have to move more than one column to the right to reach the next step, but you never have to step down more than one row at a time.

1x1  5x2 2x3 1x4  3x5   7 1x  1 Row-echelon form 3 1x4  4x5  8

1x5   9

1x1  5x2 2x3 1x4  3x5   7 1x  1 Row-echelon form 3 0  0 0  0

1x1  5x2 2x3 1x4  3x5   7 1x  3x  2x  1 Not row-echelon form 3 4 5 1x3  2x4  4x5  8

1x4  3x5   9

 The goal of Gaussian elimination is to find an equivalent system that is in row-echelon form. The three operations you can use during Gaussian elimination are 1) Swap the order of two equations. 2) Multiply an equation on both sides by a non-zero constant. 3) Add a multiple of one equation to another equation.  In Gaussian elimination, you start with Equation 1 (the first equation of your mn system). 1) Find the leading coefficient in the current equation. (Sometimes you need to swap equations in this step.) 2) Eliminate the coefficients of the corresponding variable in all of the equations below the current equation. 3) Move down to the next equation and go back to Step 1. Repeat until you run out of equations or you run out of variables.  Solve using back-substitution: solve the last equation for the leading variable, then substitute into the preceding (i.e. second-to-last) equation and solve for its leading variable, then substitute into the preceding equation and solve for its leading variable, etc. Variables that are not leading variables are free parameters, and we often set them equal to t, s, ….

p. 2 Chapter 1: Systems of Linear Equations 1.1 Introduction to Systems of Equations. Examples: Gaussian Elimination and Back-Substitution on 33 Systems of Linear Equations.

p. 3 Chapter 1: Systems of Linear Equations 1.1 Introduction to Systems of Equations.

p. 4 Chapter 1: Systems of Linear Equations 1.1 Introduction to Systems of Equations.

p. 5 Chapter 1: Systems of Linear Equations 1.1 Introduction to Systems of Equations.

p. 6 Chapter 1: Systems of Linear Equations 1.1 Introduction to Systems of Equations. Example: Chemistry Application

Write and solve a system of linear equations for the chemical reaction

(x1)CH4 + (x2)O2  (x3)CO2 + (x4)H2O

Solution: write a separate equation for each element, showing the balance of that element.

C: 1x1 + 0x2 = 1x3 + 0x4 so 1x1 + 0x2 – 1x3 + 0x4 = 0 H: 4x1 + 0x2 = 0x3 + 2x4 so 4x1 + 0x2 + 0x3 – 2x4 = 0 O: 0x1 + 2x2 = 2x3 + 1x4 so 0x1 + 2x2 – 2x3 – 1x4 = 0

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Chapter 1: Systems of Linear Equations 1.2 Gaussian Elimination and Gauss-Jordan Elimination.

1.2 Gaussian Elimination and Gauss-Jordan Elimination.  Objective: Use matrices and Gauss-Jordan elimination to solve mn systems of linear equations by hand and with software.  Objective: Use matrices and Gaussian elimination with back-substitution to solve mn systems of linear equations by hand and with software.  Use row-echelon form or reduced row-echelon form to determine the of solutions of a homogeneous system of linear equations, and (if applicable) the number of free parameters.

A matrix is a rectangular array of numbers, called matrix   aaaa entries, arranged in horizontal rows and vertical columns. 11 12 13 1n   aaaa Matrices are denoted by capital letters; matrix entries are  21 22 23 2n  denoted by lowercase letters with two indices. In a given A   31 32 33  aaaa3n  matrix entry a , the first index i is the row, and the second   ij   index j is the column. The entries a , a , a , … compose the 11 22 33   aaaa main diagonal. If m = n then A is called a square matrix.  mmm321 mn 

11 1 12 2 13 3  1 nn  bxaxaxaxa 1   bxaxaxaxa A linear system 21 1 22 2 23 3 2 nn 2  

m m m 332211  mn  bxaxaxaxa mn can represented either by a coefficient matrix A and a column vector b

 1211 13  aaaa 1n  b1    aaaa  b   2221 23 2n   2 

A   3231 33  aaaa 3n  and b  b3                 mmm 331  aaaa mn  bm  or by an augmented matrix M, which I will sometimes write as [A | b]

 11 12 13  n baaaa 11    baaaa   21 22 23 n 22 

M   31 32 33  n baaaa 33          mmm 331  mn baaaa m  (The book, Mathematica, and the calculator do not display the dotted vertical .) To create M in Mathematica from A and b, type m=Join[a,b,2]

To create M on the TI-89 from A and b, type

Matrixaugment(AB.M

p. 9 Chapter 1: Systems of Linear Equations 1.2 Gaussian Elimination and Gauss-Jordan Elimination. In a similar manner to that used for an mn system of linear equations, we can use a Gaussian elimination on the coefficient side A of the augmented matrix [A | b] to find an equivalent augmented matrix [U | c] in row-echelon form. Then we use back-substitution to solve for each variable. U is called an upper triangular matrix because all non-zero entries are on or above the main diagonal.

row-echelon form row-echelon form not row-echelon form

1  5 2 1 3 1  5 2 1 3 1  5 2 1 3  0 0 1 0 5 0 0 1 0 5 0 0 1 3  2       0 0 0 1 4 0 0 0 0 0 0 0 1 2 4        0 0 0 0 2 0 0 0 0 0 0 0 0 1  3

 Example: Use Gaussian elimination and back-substitution to solve. The three elementary row operations you can use during Gaussian elimination are 1) Swap the two rows. 2) Multiply a row by a non-zero constant. 3) Add a multiple of one row to another row.

2 1 4 6x  4y  3z  8 6 4 3 8 1 3 2 3  3x  2y 1z  3  3 2 1 3  0 1 3 2    2  1x 1y 1z  2 1 1 1 2 0 0 1 2 z  2  3 3  y  2 z  2 so y  2  2 (2)  1  2 1 4 4 2 1 x  3 y  2 z  3 so x  3  3 (1)  2 (2)  1 Steps:

p. 10 Chapter 1: Systems of Linear Equations 1.2 Gaussian Elimination and Gauss-Jordan Elimination. Instead of using back-substitution, we can take the row-echelon form [U | c] and eliminate the coefficients above the pivots by adding multiples of the pivot rows. The result [R | d] is called reduced row-echelon form.

reduced row-echelon form reduced row-echelon form

1  5 0 0 11 1  5 0 1 3  0 0 1 0 5  0 0 1 3  2     0 0 0 1 4  0 0 0 0 4      0 0 0 0 2  0 0 0 0  3

 Example: Use Gauss-Jordan elimination to solve.

2 1 4 6x  4y  3z  8 6 4 3 8 1 3 2 3  1 0 0 1  x  1  3x  2y 1z  3  3 2 1 3  0 1 3 2  0 1 0 1  y 1    2      1x 1y 1z  2 1 1 1 2 0 0 1 2 0 0 1 2  z  2

p. 11 Chapter 1: Systems of Linear Equations 1.2 Gaussian Elimination and Gauss-Jordan Elimination.  Example Using software to find the Reduced Row-Echelon Form (Exercise 1.2 #39)

Mathematica Go to the Palettes Menu and open the Basic Math Assistant. Under Basic Commands, open the Matrix Tab. Type a= and click Use the Add Row and Add Column buttons to expand the matrix, so you can enter the augmented matrix

(If you make the matrix to large, use  to remove rows and columns.) Press  (or  on the number pad) when you are finished entering the augmented matrix. Another way to enter the augmented matrix is to type (or download) a={{1,-1,2,2,6,6},{3,-2,4,4,12,14},{0,1,-1,-1,-3,-3},{2,- 2,4,5,15,10},{2,-2,4,4,13,13}} MatrixForm[a]  You can download this matrix from http://holtmann75.pbworks.com. Click on Electronic Data Sets and open 1133110878_323834.zip/DataSets/Mathematica/0102039.nb (Ch. 01, Section 02, Problem 039). Notice that A and a are different variables! User-defined variables should always begin with a lower-case letter, because Mathematica’s built-in fuctions and commands begin with capital letters. (For example, N and C are already defined by Mathematica.) On the Basic Math Assistant palette, click on MatrixForm RowReduce and type a so you have MatrixForm[RowReduce[a]]

Converting back to a system of linear equations, we have x1 = 2 x4 = 5 x2 = 2 x5 = 1 x3 = 3

p. 12 Chapter 1: Systems of Linear Equations 1.2 Gaussian Elimination and Gauss-Jordan Elimination. TI-89: Type  Data/Matrix Editor New... Type: Matrix Folder: main Variable: A  (A is above . If you type = instead of a, use . If a already exists, then use Open... on the previous screen instead of New....) Use  and the arrow keys to type in the coefficient matrix

(If you need to insert or delete a row or column, use ) When you are finished, press . Another way to enter the matrix is to type (from the home screen) 1,1,2,2,6,6 3,2,4,4,12,14 0,1,1,1,3,3 2,2,4,5,15,10 2,2,4,4,13,13 .A

After the matrix is entered, type A

 Matrix  rref( A

Converting back to a system of linear equations, we have x1 = 2 x2 = 2 x3 = 3 x4 = 5 x5 = 1

p. 13 Chapter 1: Systems of Linear Equations 1.2 Gaussian Elimination and Gauss-Jordan Elimination. Mathematica can help you perform row operations. If your augmented matrix is in a, then a[[2]] is row 2 of the matrix. To view a in the usual format, type MatrixForm[a] To swap rows 2 and 3 of a, type a[[{2,3}]]=a[[{3,2}]] To multiply row 2 of a by 7, type a[[2]]=7*a[[2]] To add 7 times row 2 to row 1, type a[[1]]=a[[1]]+7*a[[2]]

Mathematica performs operations in the order that you type them in, not as they appear on the screen. If you go back and edit your work, you can use Evaluate Notebook under the Evaluation menu to recalculate the notebook in the order on the screen. The TI-89 can also help you perform row operations. If your augmented matrix is in a, then … To swap rows 2 and 3 of a and store the result in a1, type Matrix Row opsrowSwap(A,2,3,).A1 To multiply row 2 of a1 by 7 and store the result in a2, type Matrix Row opsmRow(7,A1,2).A2 To add 7 times row 2 of a2 to row 1 and store the result in a3, type Matrix Row opsmRowAdd(7,A2,1,3).A3 1 2 1 2 1 2 About notation:   and   are matrices, but is a determinant (Ch. 3). 3 4 3 4 3 4  Theorem 1.1. Every homogeneous (constants on right-hand side are all zeroes) system of linear equations is consistent. The number of free parameters in the solution set is the number of variables minus the number of pivots (leading coefficients). If there are zero free parameters, then there is exactly one solution.

1 1  5 0 1 1  5 2  4 0  2 5 3  2 1 0       0 5 1 1 1 0   Examples: 1 0  2 0    6 15 4  5  4 0    1 21 9  5 5 0  8  20  7 7 3 0 2 1 1 0      5 20 28 14 12 0  0 0 15 3  21 0 1 0  2 0 Solutions:   0 1  3 0 0 0 0 0

p. 14 Chapter 1: Systems of Linear Equations 1.3 Applications of Systems of Linear Equations.

1.3 Applications of Systems of Linear Equations.  Objective: Set up and solve a system of equations to fit a to a set of data points.  Objective: Set up and solve a system of equations to represent a network.

 Polynomial Curve Fitting.

Given m data points (t1, y1), (t2, y2), …, (tm, ym). We want to find a polynomial of degree m–1 that passes through these points. 2 m 2 m–1 p(t) = c0 + c1t + c2t + … + cmt Notice that yj = c0 + c1(tj)+ c2(tj) + … + cm–1(tj)

m1 1 1  tt1 )(   c0   y1   m1      1  tt )( c y This produces an mm linear system:  2 2   1  =  2            m1      1 m  ttm )(  cm1  ym  that you can solve using Gauss-Jordan elimination.

Example 1.3#7

t  21012 y 28  0860

p. 15 Chapter 1: Systems of Linear Equations 1.3 Applications of Systems of Linear Equations. Example 1.3#9

t 2006 2007 2008

y 5 7 12

 Network Analysis: write a system of linear equations using Kirchoff’s Laws. 1) Flow into each node (also called vertex) equals flow out. 2) In an electrical network, the sum of the products IR (I = current and R = resistance) around any closed path of edges (lines) is equal to the total voltage in the loop from the batteries.

A resistor is represented by the symbol Resistance is measured in ohms (). 1 k = 1000 

A battery is represented by the symbol If the current flows through the battery from the short line (–) to the long line I (+), then the voltage is positive.

Current is measure in amps (A). 1 mA = 0.001 A.

Flow into a node is positive. Flow out of a node is negative. To write a system of equations,

1) Pick a direction (at random) for each current I. 2) For each node, write an equation for the current input and output. 3) For each loop, write the V = IR equation.

p. 16 Chapter 1: Systems of Linear Equations 1.3 Applications of Systems of Linear Equations. Example: 1.3.#32. Solve for the currents.

Example: 1.3.29. The figure shows traffic flow in vehicles/hour through a network of streets.

a) Solve for x1, x2, x3, and x4. b) Find the traffic flow when x4 = 0. c) Find the traffic flow when x4 = 100.

p. 17

Chapter 2: Matrices. 2.1 Operations with Matrices.

Chapter 2: Matrices.

2.1 Operations with Matrices.  Objective: Determine whether or not two matrices are equal.  Objective: Add and subtract matrices and multiply a matrix by a scalar.  Objective: Multiply two matrices.  Objective: Write solutions to a system of linear equations in column vector notation.  Objective: Partition a matrix and perform block multiplication.

 11 12  aaa1n    21 22  aaa2n Three ways to represent the same matrix are A = [aij] =   .      mm31  aaamn 

 Two matrices A = [aij] and B = [bij] are equal if and only if they have the same dimensions or order (mn) and aij = bij for all 1  i  m and 1 j  n.

Comment on logic:  An example of “P if Q” (also written P  Q) is “It is cloudy if it is raining.”  An example of “Q only if P” (i.e. “if Q then P,” also written Q  P) is “It is raining only if it is cloudy.”  “R iff S” is shorthand for “R if and only if S” (also written R  S). R iff S means that R is equivalent to S.

 To add (or subtract) two matrices A = [aij] and B = [bij] that have the same dimensions, add (or subtract) corresponding matrix entries.

A + B = [aij + bij] A – B = [aij – bij] The sum (and difference) of two matrices with different dimensions is undefined. To multiply a matrix by a scalar (number), multiply each entry by that scalar.

cA = [caij]

Examples:

 421    341   50  Let A =   , B =   , and D =   . Find A + B, A – B, A + D, and 4A.   103   231   31 

p. 19 Chapter 2: Matrices. 2.1 Operations with Matrices. In Mathematica, use a+b, a–b, and 4*a or 4a to add matrices, subtract matrices, and multiply  50 scalars c by matrices. Remember, you can enter D =   from the Palettes Menu::Basic  31 Math Assistant::Basic Commands::Matrix Tab.

On the TI-89, use ab, a b, and ca or 4a to add matrices A and B, subtract matrices,

and multiply scalars c by matrices. Remember, you can enter D = from the home screen

as [0,5;-1,3].D or from  Data/Matrix Editor. (Be sure to use Type: Matrix)

Matrix multiplication is defined in a much more complex manner. For a 11 system, we can write ax = b. We want a definition of matrix multiplication that allows us to write an mn system

11 1 12 2 13 3  1 nn  bxaxaxaxa 1   bxaxaxaxa 21 1 22 2 23 3 2 nn 2 as Ax = b,  

m m m 332211  mn  bxaxaxaxa mn

a11 a12  a1n  a a  a  where A is the coefficient matrix A   21 22 2n  ,         am1 am2  amn 

 x1  b1  x  b  and x and b are column matrices (or column vectors): x   2  and b   2  .           xn  bm 

Observe that each row of A was multiplied by the column x to give the corresponding row of b.

 If A = [aij] is an mn matrix and B = [bij] is an np matrix, then the product AB is an mp matrix A = [cij] where

n [cij] = aik bkj = ai1b1j + ai2b2j + ai3b3j + … ainbnj k 1 If the column dimension of A does not match the row dimension of B, then the product is undefined

p. 20 Chapter 2: Matrices. 2.1 Operations with Matrices. Examples  2 3   3 2   4 1 0       3  3  2 1  5    2  5

In Mathematica, use a.b to multiply matrices.(Do not use a*b)

On the TI-89, use ab multiply matrices.

 Example: Write solutions to a system of linear equations in column vector notation.

 x1   2 1 2 1 3 0 x  1 Solve    2  =    = t   1 1 0 1 0 x3   1  0 1 1 2   0   x4   1 

 Block multiplication on partitioned matrices works whenever the dimensions are OK.

Examples 0 1 0 0 0 1 0 0 A A    11 12  with A11    A12    and A  1 0 0 0    1 0  0 0 A21 A22  0 0 2 1 A21  0 0 A22  2 1

 1 2 0   1 2 0   B11  B12  1 1 0 B B  1 1 0 B     11 12 with       4 3 1  4 3 1  B21 B22       B21    B22     5 6  2 5 6  2

 1 1 0  A B  A B A B  A B   A B  0B A 0  0B  Then AB = 11 11 12 21 11 12 12 22 = 11 11 21 11 22 =        1 2 0 A21B11  A22B21 A21B12  A22B22 0B11  A22B21 0 0  A22B22 13 12 0

p. 21 Chapter 2: Matrices. 2.1 Operations with Matrices.

a11 a12  a1n   x1   a11x1  a12x2  a1n xn  a11   a12  a1n  a a  a  x   a x  a x  a x  a  a  a  Ax=  21 22 2n   2  =  21 1 22 2 2n n  =  21 x   22 x  2n x                1    2    n             am1 am2  amn  xn  am1 x1  am2 x2  amn xn  am1  am2  amn 

In block matrix notation, we write [ a1 | a2 | … | an ] = where the ai are

m1 column matrices. The Ax = [ a1 | a2 | … | an ] = a1x1 + a2x2 + … + anxn

Similarly, if B is an lm matrix (so BA is defined), then BA = B[ a1 | a2 | … | an ] = [ Ba1 | Ba2 | … | Ban ], i.e. the columns of BA are Bai. Notice that B is lm, ai is m1, Bai is l1, and BA is ln.

On the other hand, we could partition A into 1m row matrices ri

a11 a12  a1n  r1  a11 a12  a1n   x1  r1  r1x  a a  a  r  a a  a  x  r  r x A =  21 22 2n  =  2  . Then Ax =  21 22 2n   2  =  2  x =  2  .                                     am1 am2  amn  rm  am1 am2  amn  xn  rm  rm x

Notice that ri is 1n, x is n1, rix is 11 (i.e. a number), and Ax is m1.

e11 e12  e1m  e1  e1  e1 A e e  e  e  e  e A If E =  21 22 2m  =  2  then EA =  2  A =  2  .                        el1 el 2  elm  e l  e l  el A

Notice that ei is 1m, A is mn, eiA is 1n, and EA is ln

If e = [ e1 | e2 | … | em], then eA = [ e1 | e2 | … | em] = e1r1 + e2r2 + … + emrm

Notice that e is 1m, A is mn, ei is a number, ri is 1n, and eA is 1n.

p. 22 Chapter 2: Matrices. 2.2 Properties of Matrix Operations.

2.2 Properties of Matrix Operations.  Objective: Know and use the properties of matrix operations (matrix addition and subtraction, scalar multiplication, and matrix multiplication), and of the zero and identity matrices.  Objective: Know which properties of fields do not hold for matrices (commutativity of matrix multiplication and existence of multiplicative inverses).  Objective: Find the transpose of a matrix and know properties of the transpose.

The real numbers R, together with the operations of addition and multiplication, is an example of a mathematical . (The complex numbers C with addition and multiplication is another example.) Fields and their operations have all of the usual properties.

1) Closure under addition: if a and b  R, then a + b  R. 2) Addition is associative: (a + b) + c = a + (b + c) 3) Addition is commutative: a + b = b + a 4) Additive identity (zero): R contains 0, which has the property that a + 0 = a for all a. 5) Additive inverses (opposites): every a  R. has an opposite –a, such that a + (–a) = 0. We define subtraction a – b as a + (–b). 6) Closure under multiplication: if a and b  R, then ab  R. 7) Multiplication is associative: (ab)c = a(bc) 8) Multiplication is commutative: ab = ba 9) Multiplication distributes over addition: a(b + c) = ab +ac 10) Multiplicative identity (one): R contains 1, which has the property that 1a = a for all a. 11) Multiplicative inverses (reciprocals): every a  R. has an inverse a–1, such that aa–1 = 1. We define division a  b as ab–1.

 Matrix addition (and subtraction) has all of the usual properties: closure, associativity, commutativity, zero matrices, and opposites.  A zero matrix has zero in all entries, but because the matrix can have any dimensions mn, we have many zero matrices 0mn.

 The opposite of a matrix [aij] is –[aij] = [–aij].  Multiplication of a scalar by a matrix has all of the usual properties: closure, associativity, commutativity, multiplicative identity (the scalar 1), and distribution. For distribution, we have both c(A + B) = cA + cB and (c + d)A = cA + dA.

Examples:   98    98   98    295   295   000  (–1)   = –   =   ,   +   =   = 023   97    97   97    356    356   000 

p. 23 Chapter 2: Matrices. 2.2 Properties of Matrix Operations.  Multiplication of matrices is closed, is associative, and distributes over matrix addition. The multiplicative identity matrices are square matrices with ones on the main diagonal and zeros 1 0  0 0 1  0 everyplace else: Inn =   . If A is mn, then Imm Amn = Amn and Amn Inn = Amn       0 0  1 k  We define exponents for square matrices A and positive integers k: A = AAA . Also, we k times define A0 = I.  Multiplication of matrices is not commutative in general.  Many matrices do not have multiplicative inverses. Division of matrices is undefined.

Examples:

1 0 0 1 0  295       = 0 1 0 = 0 1   356   0 0 1

 98   = =  97

1 0  1 0    =   =  2 1   2 1   3 1  3 1

Theorem 2.5: For a system of linear equations, exactly one of the following is true. 1) The system has no solution (inconsistent, singular system). 2) The system has exactly one solution (consistent, nonsingular system). 3) The system has infinitely many solutions (consistent, singular system). Use a free parameter or free variable (or several free parameters) to represent the solution set.

Proof using matrix operations:

Given a system of linear equations Ax = b. Exactly one of the following is true: the system has no solution, the system has exactly one solution, or the system has at least two solutions (call them x1 and x2).

p. 24 Chapter 2: Matrices. 2.2 Properties of Matrix Operations.

If the system has two solutions, then Ax1 = b and Ax2 = b so A(x1 – x2) = Ax1 – Ax2 = b – b = 0. Let xh = x1 – x2, so xh is a nonzero solution to the homogenous equation Ax = 0. Then for any scalar t, x1 + txh is a solution of Ax = b because

A(x1 + txh) = Ax1 + tAxh = b+ t 0 = b. Thus, in the last case, the system has infinitely many solutions with a parameter t.

 The transpose AT of a matrix A is formed by writing its columns as rows. For example,

a11 a12 a13  a1n  a11 a21 a31  am1  a a a  a  a a a  a   21 22 23 2n   12 22 32 m2  T if A  a31 a32 a33  a3n  then A  a13 a23 a33  am3  .                       am1 am2 am3  amn  a1n a2n a3n  amn  T Equivalently, the transpose of A is formed by writing rows as columns. The ij entry of A is aji.

9 3  7 0  Example: Find the transpose of A =   3 0  8  4 1 5 2  8

Mathematica: to take the transpose of a matrix, either type Transpose[a] (also available on the Basic Math Assistant palette) or type a followed by tr (four separate keystrokes). After the first three keystrokes, you will see atr. After the fourth keystroke, atr will change to aT. Of course, at the end, type 

TI-89: AMatrixT

 Properties of the transpose: 1) (AT)T = A 2) (A + B)T = AT +BT * 3) (cA) T =cAT 4) (AB) T = BTAT

p. 25 Chapter 2: Matrices. 2.2 Properties of Matrix Operations. Example of Property #4:

Consider A = and B =

The 32 entry of (AB)T is the 23 entry of AB =

which is a21b13 + a22b23 + a23b33

If we look at AT and BT, we much reverse the order of multiplication to obtain row  column.

BTAT =

T T T T T The 32 entry B A is b13a21 + b23a22 + b33a23. You can see that (AB) = B A

A matrix M is symmetric iff MT = M. 10  6  5 Example: M =   is symmetric. Prove that AAT is symmetric for any matrix A.  6 4 2   5 2 5  Proof:

p. 26 Chapter 2: Matrices. 2.3 The Inverse of a Matrix.

2.3 The Inverse of a Matrix.  Objective: Find the inverse of a matrix (if it exists) by Gauss-Jordan elimination.  Objective: Use properties of inverse matrices.  Objective: Use an inverse matrix to solve a system of linear equations.

A square nn matrix is invertible (or nonsingular) when there exists an nn matrix A–1 such that

–1 –1 AA = Inn and A A = Inn

–1 where Inn is the identity matrix. A is called the (multiplicative) inverse of A. A matrix that does not have an inverse is called singular (or noninvertible). Nonsquare matrices do not have inverses.

Theorem 2.7: If A is an invertible matrix, then the inverse is unique.

Proof: let B and C be inverses of A. Then BA = I an AC = I. So B = BI = B(AC) = (BA)C = IC =C. Therefore, B = C and the inverse of A is unique.

 53  Example: Find the inverse of A =   .  21 

 xx 1211   01  Solution: We need to solve the system AX = I, or   =   ,  xx 2221   10  xxxx  053153 which gives four equations 11 21 12 22 11 21 12 xxxx 22  121021

To solve these four equations, we take the reduced row echelon forms

 153   201   053    501  –1   52       and      , so A = X =    021   110   121   310   31 

Since the row operations performed to find the reduced row echelon form depend only on the  53  coefficient part of the augmented matrix   , we could solve all four equations  21  simultaneously by using a doubly augmented matrix

 0153    5201  –1 [ A | I ] =      = [ I | A ]  1021    3110 

To create the doubly augmented matrix in Mathematica from A, type m=Join[a, IdentityMatrix[2],2] To create the doubly augmented matrix on the TI-89 from A, type

p. 27 Chapter 2: Matrices. 2.3 The Inverse of a Matrix. Matrixaugment(A,Matrixidentity(2)).M  We used IdentityMatrix[2] and identity(2) because we wanted a 22 matrix.

 To find the inverse of an nn matrix A by Gauss-Jordan elimination, find the reduced row echelon form of th n2n augmented matrix [ A | I ]. If the nn block on the left can be reduced to –1 I, then the nn block on the right is A * [ A | I ]  [ I | A–1]

If the nn block on the left cannnot be reduced to I, then A is not invertible.

1 1 1 Example: Invert A =   using Gauss-Jordan elimination. 1 2 2 1 2 3

1 1 1 Example: Invert A =   using Gauss-Jordan elimination. 1 2 2 2 3 3

a b Example: Invert M =   using Gauss-Jordan elimination. c d

You can also use software. First clear the variables using Clear[a,b,c,d] or Clear a-z, then type Inverse[m] or M^1

(also on Basic Math Assistant palette)

p. 28 Chapter 2: Matrices. 2.3 The Inverse of a Matrix. Theorem 2.8 Properties of Inverse Matrices If A is an invertible matrix, k is a positive integer, and c is a nonzero scalar, then A–1, Ak, cA, and AT are invertible, and 1) (A–1) –1 = A * 2) (Ak)–1 = (A–1)k –1 1 –1 3) (cA) = c A 4) (AT)–1 = (A–1)T

Proof:

1)

2) Proof by Induction (see the Appendix)  When k = 1,

 If (Ak)–1 = (A–1)k,

 By mathematical induction, we conclude that (Ak)–1 = (A–1) for all positive integers k.

3)

4)

p. 29 Chapter 2: Matrices. 2.3 The Inverse of a Matrix.

 Theorem 2.9 Inverse of a Product * If A and B are invertible nn matrices, then (AB)–1 = B–1A–1.

Proof:

 Theorem 2.10 Cancellation Properties Let C be an invertible matrix. * 1) If AC = BC, then A = B. (Right cancellation property)

2) If CA = CB, then A = B. (Left cancellation property)

 Theorem 2.11 Systems of Equations with Unique Solutions.* If A is an invertible matrix, then the system Ax = b has a unique solution x = A–1b.

Review: What is wrong with the following “proof” that if AB = I and CA = I then B = C?

AB  CA

A A

B = C

What is wrong with the following “proof” that if AB = I and CA = I and A is invertible then B = C?

AB = CA

A–1AB = CAA–1

IB = CI

B = C

p. 30 Chapter 2: Matrices. 2.4 Elementary Matrices.

2.4 Elementary Matrices.  Objective: Factor a matrix into a product of elementary matrices.  Objective: Find the PA = LDU of a matrix.

An elementary matrix is a square matrix of dimensions nn (or of order n) is a matrix that can be obtained from the nn identity matrix by a single elementary row operation.

The three elementary row operations, with examples of corresponding elementary matrices, are

 0001  1000 1) Swapping two rows, e.g. R2  R4 E1 =    0100    0010

 001  2) Multiplying a single row by a nonzero constant, e.g. 3R  R E =   3 3 2  010   300 

 0001   0010  3) Adding a multiple of one row to another row, e.g. –2R1 + R3  R3 E3 =    0102     1000 

Theorem 2.12 Representing Elementary Row Operations

If we premultiply (multiply on the left) a matrix A by an elementary matrix, we obtain the same result as if we had applied the corresponding elementary row operation to A.

Examples:

 0001    0189  r1        1000  1913 r2     =   = R2  R4  0100    9497  r3         0010    2514  r4 

 001    189      = 3R  R  010    913  3 3  300   497 

p. 31 Chapter 2: Matrices. 2.4 Elementary Matrices.

 1 0 0 0  9  8 1 0   0 1 0 0  3 1  9 1      = –2R1 + R3  R3  2 0 1 0  7 9 4  9      0 0 0 1  4 1 5 2 

Gaussian elimination can be represented by a product of elementary matrices. For example,

 0 1  3 2  A =    1  2 1  5  3 6 9 3  0 1 0 R  R E =   1 2 1 1 0 0 0 0 1  1  2 1  5    0 1  3 2   3 6 9 3  1 0 0 3R + R  R E =   1 3 3 2 0 1 0 3 0 1 1  2 1  5    0 1  3 2  0 0 6 12 1 0 0  1 R  R E = 0 1 0  6 3 3 3   0 0 1/ 6 1  2 1  5   0 1  3 2  0 0 1  2

1 0 0  1 0 0 so = E (E (E A)) =     3 2 1 0 1 0  0 1 0 0 0 1/ 6 3 0 1

Two nn matrices A and B are row-equivalent when there exist a finite number of elementary matrices such that B = EnEn–1…E2E1A.

p. 32 Chapter 2: Matrices. 2.4 Elementary Matrices.

A square matrix L is lower triangular if all entries above the main diagonal are zero, i.e. lij = 0 whenever i < j. A square matrix U is upper triangular if all entries below the main diagonal are zero, i.e. uij = 0 whenever i > j. A square matrix D is diagonal if all entries not on the main diagonal are zero, i.e. dij = 0 whenever i  j.

 0 0     0 0 L =   U =   D =     0 0   0  0    0 0  0 0 

Theorem 2.13+ Elementary Matrices are Invertible

If E is an elementary matrix, then E–1 exists and is an elementary matrix. Moreover, if E is lower triangular, then E–1 is also lower triangular. And if E is diagonal, then E–1 is also diagonal.

Examples:

0 1 0 E =   R  R E –1 = R  R 1 1 0 0 1 2 1 1 2 0 0 1 1 0 0  1 0 0 E =   3R + R  R E –1 =   –3R + R  R 2 0 1 0 1 3 3 2  0 1 0 1 3 3 3 0 1  3 0 1 1 0 0  1 0 0 E = 0 1 0  1 R  R E –1 = 0 1 0 6R  R 3   6 3 3 3   3 3 0 0 1/ 6 0 0 6

You can check these using matrix multiplication. 1 0 0 1 0 0 E.g. =   and =   . 0 1 0 0 1 0 0 0 1 0 0 1

Theorem 2.14 Invertible Matrices are Row-Equivalent to the Identity Matrix

A square matrix A is invertible if and only if it is row equivalent to the identity matrix:

A = En…E2E1I = En…E2E1 if and only if A is a product of elementary matrices.

Proof of “If A is invertible, then A is a product of elementary matrices”:

Since A is invertible, Ax = b has a unique solution (namely, x = A–1b). But this means that we can use row operations to reduce [ A | b ] to [ I | c ] (where c = A–1b, of course). If the

p. 33 Chapter 2: Matrices. 2.4 Elementary Matrices.

corresponding elementary matrices are E1, E2, …, En, then I = EnEn–1…E2E1A so –1 –1 –1 –1 A = E1 E2 … En–1 En , which is a product of elementary matrices.

Proof of “If A is a product of elementary matrices, then A is invertible”:

–1 –1 –1 –1 –1 –1 If A = E1E2… En–1En, then A exists and A = En En–1 …E2 E1 because every elementary matrix is invertible.-

Theorem 2.15 Equivalent Conditions for Invertibility

If A is an n n matrix, then the following statements are equivalent.

1) A is invertible. 2) Ax = b has a unique solution for every n1 column matrix b (namely, x = A–1b). 3) Ax = 0 has only the trivial solution. 4) A is row-equivalent to Inn. 5) A can be written as a product of elementary matrices.

Returning to our example of Gaussian elimination using elementary matrices, we found the  0 1  3 2  reduced echelon form of the augmented matrix    1  2 1  5  3 6 9 3 

1  2 1  5 1 0 0  1 0 0 0 1 0   =       0 1  3 2  0 1 0  0 1 0 1 0 0 0 0 1  2 0 0 1/ 6 3 0 1 0 0 1

Look just at the 33 coefficient matrix instead of the 34 augmented matrix. We have

row-echelon form mRows mRowAdds row swaps 1  2 1 1 0 0  1 0 0 0 1 0  0 1  3   0 1  3 = 0 1 0  0 1 0 1 0 0            1  2 1 0 0 1  0 0 1/ 6 3 0 1 0 0 1  3 6 9          U A

The row-echelon form is upper triangular. In general, we may have more than one mRow (multiply a row by a nonzero constant) elementary matrix, more than one mRowAdd (add a multiple of one row to another row) elementary matrix, and more than one row swap matrix. The product of and arbitrary number of row swap matrices is called a permutation matrix P.

mRows mRowAdds

U= Fn F2 F1 Em E2 E1 PA

p. 34 Chapter 2: Matrices. 2.4 Elementary Matrices. mRowAdds mRows E 1E 1 E 1 F 1F 1 F 1 U= PA 12m 1 2n lower triangular diagonal

Lemma The product of diagonal matrices is diagonal. The product of lower triangular matrices is lower triangular.

LDU = PA E 1E 1E 1 F 1F 1F 1 12m 12 n row-echelon form row swaps 1 0 0 1 0 0  1  2 1 0 1 0  0 1  3       0 1 0 0 1 0 0 1  3 = 1 0 0            1  2 1 3 0 1 0 0 1/ 6 0 0 1  0 0 1  3 6 9       L D U P A

Theorem LU-Factorization

Every square matrix can be factored as PA = LDU, where P is a permutation matrix, L is lower triangular with all ones on the main diagonal, D is diagonal, and U is upper triangular with all ones on the main diagonal.

A variation on this is PA = LU, where this L equals the LD from above, and does not necessarily have ones on the diagonal.

The PA = LU factorization is the usual method used by computers for solving systems of linear equations, finding inverse matrices, and calculating determinants (Chapter 3). It is also useful in proofs.

p. 35 Chapter 2: Matrices. 2.4 Elementary Matrices.

0 1 1  Example: Find the PA = LDU factorization of A =   1 0 1 2  3 6

0 1 1 1 0 1 1 0 1 Solution: A =   R1R2    2R1R3R3    3R2R3R3  1 0 1 0 1 1 0 1 1 2  3 6 2  3 6 0  3 4 1 0 1 1 0 1    R2 R2    = U 0 1 1 0 1 1 0 0 1 0 0 1 

1 0 1  1 0 0 1 0 0  1 0 0 0 1 0 0 1 1 *   =   0 1 0  0 1 0     0 1 1 0 1 0     1 0 0 1 0 1 0 0 1  0 0 1 0  3 1  2 0 1 0 0 1 2  3 6        U F E2 E1 P A

1 0 0 1 0 0 1 0 0 1 0 1  0 1 0 0 1 0 0 1 0   =       0 1 1 2 0 1 0 3 1 0 0 1 0 0 1      1 1 1 U E1 E2 F

1 0 0 1 0 0     = 0 1 0 0 1 0 2 3 1 0 0 1   L D

p. 36 Chapter 2: Matrices. 2.5 Applications of Matrix Operations.

2.5 Applications of Matrix Operations.  Objective: Write and use a stochastic (Markov) matrix.  Objective: Use matrix multiplication to encode and decode messages.  Objective: Use matrix algebra to analyze and economic system (Leontief input-output model).

Consider a situation in which members of a population occupy a finite number of states {S1, S2, …, Sn}. For example, a multinational company has $4 trillion in assets (the population). Some of the money is in the Americas, some in Asia, and the rest is in Europe (the three states). In a Markov process, at each discrete step in time, members of the population may move from one state to another, subject to the following rules:

1) The total number of individuals stays the same. 2) The numbers in each state never become negative. 3) The new state depends only on the current state (history is disregarded). from  The behavior of a Markov process is described by a matrix  SSS of transition probabilities (or stochastic matrix or Markov 21 n  1211  ppp 1n S1  matrix).    2221  ppp 2n S2  P     to pij is the probability that a member of the population will     th th  change from the j state to the i state. The rules above    ppp S  become  nn 21 nn n 

1) Each column of the transition matrix adds up to one. 2) Every probability entry is 0 ≤ pij ≤ 1.

1  Example: Stochastic Matrix. A chemistry course is taught in two sections. Every week, 4 of the 1 1 students in Section A and 3 of the students in Section B drop, and 6 of each section transfer to the other section. Write the transition matrix. At the beginning of the semester, each section has 144 students. Find the number of students in each section and the number of students who have dropped after one week and after two weeks. 1 7 6 1 Solution: 12 A B 2

* 1 1 1 (A) 7 1 1 4  6 6 0 12 6 0  1 1 1  (B)  1 1  P = 6 6  3 01 = 6 2 0 .     1  1 1 1 (d)  1 1 1 4 0 1  4 3   4 3  0 3 144 108 (A)  79  P   =   (B) . P =   144  96   66  drop  0   84  (d) 143

1

p. 37 Chapter 2: Matrices. 2.5 Applications of Matrix Operations. Matrix multiplication can be used to encode and decode messages. The encoded messages are called cryptograms.

To begin, assign a number to each letter of the alphabet (and assign 0 to a space).

0 1 2 3 4 5 6 7 8 9 10 11 12 13 _ A B C D E F G H I J K L M

14 15 16 17 18 19 20 21 22 23 24 25 26 N O P Q R S T U V W X Y Z

Use these numbers to convert a message in to a row matrix, including spaces but ignoring punctuation. Then partition the row matrix into 13 uncoded row matrices.

Example:

M A K E _ I T _ S O _ _ [13 1 11] [5 0 9] [20 0 19] [15 0 0]

 1  4 1  Example: Use the invertible matrix A =   to encode the message MAKE IT SO.  2 7 6   0 1 3  Solution:

*--[ 13 1 11] = [11 –56 26] [20 0 19] = [20 –99 37]

[5 0 9] = [5 –29 22] [15 0 0] = [15 –60 –15]

The sequence of encoded matrices is [11 –56 26] [5 –29 22] [20 –99 37] [15 –60 –15]

Removing the brackets yields the cryptogram 11 –56 26 5 –29 22 20 –99 37 15 –60 –15

In order to decrypt a message, we need to know the encryption matrix A.

p. 38 Chapter 2: Matrices. 2.5 Applications of Matrix Operations.

 1  4 1  Example: Use the invertible matrix A =   to decode  2 7 6   0 1 3  –8 26 31 13 –73 50 19 –97 44 16 –64 –16. Solution:

27 13 17 A–1 =    6 3  4   2 1 1  *-–-  

[–8 26 31] = [2 5 1] [19 –97 44] = [19 0 21]

[13 –73 50] = [13 0 21] [16 –64 –16] = [16 0 0]

[2 5 1] [13 0 21] [19 0 21] [16 0 0] B E A M _ U S _ U P _ _

User (Output) In economics, an input-output model (developed by 

Leontief) consists of n different industries In, each of I1 I 2  I n

which needs inputs (e.g. steel, food, labor, …) and has d11 d12  d1n  I1  and output. To produce a unit (e.g. $1 million) of output,    d 21 d 22  d 2n I 2  Supplier an industry may use the outputs of other industries and of D     itself. For example, production of steel may use steel,         (Input)    food, and labor. d n1 d n2  d nn I n 

Let dij be the amount of output the industry j needs from industry i to produce one unit of output per year. (We assume the dij are constant, i.e. fixed prices.) The matrix of these coefficients is called the input-output matrix or consumption matrix D. A column represents all of the inputs to a given industry. For this model to work, 0 ≤ dij ≤ 1 and the sum of the entries in each column must be less than or equal to 1. (Otherwise, it costs more than one unit to produce a unit in that industry.)

Let xi be the total output matrix of industry i, and X = [xi]. If the economic system is closed (self- sustaining: total output = “intermediate demand,” i.e. what is needed to produce it), then X = DX. If the system is open with external demand matrix E (e.g. exports) then X = DX + E.

To find what output matrix is needed to produce a given external demand matrix, we solve

p. 39 Chapter 2: Matrices. 2.5 Applications of Matrix Operations. X = DX + E X – DX = E (I – D)X = E X = (I – D)–1E

 Example: Input-Output Economic Model

Production of one unit of steel requires 0.4 units of steel, no food, and 0.5 units of labor. Production of one unit of food requires no steel, 0.1 units of food, and 0.7 units of labor. Production of one unit of labor requires 0.1 units of steel, 0.8 units of food, and 0.1 units of *-–labor.- Find the output matrix when the external demands are 300 units of steel, 200 units of food, and no labor.

Solution:

1 0.4 0 0.1 300  1 0 0 0.4 0 0.1  848   D =   . E =   . X = 0 1 0   0 0.1 0.8     0 0.1 0.8 200      2076   0.5 0.7 0.1  0   0 0 1 0.5 0.7 0.1 2086

We need 848 units of steel, 2076 units of food, and 2086 units of labor.

p. 40 Chapter 3: Determinants. 3.1 The Determinant of a Matrix.

Chapter 3: Determinants.

3.1 The Determinant of a Matrix.  Objective: Find the determinant of a 22 matrix.  Objective: Find the minors and cofactors of a matrix.  Objective: Use expansion by cofactors to find the determinant of a matrix.  Objective: Find the determinant of a triangular matrix.

Every square matrix can be associated with a scalar called its determinant. Historically, determinants were recognized as a pattern of nn systems of linear equations. The system

11 1 12 bxaxa12 1  1222baab2 11  12abba21 has the solution x1  and x2  . 21 1 22 bxaxa22  aaaa21122211  aaaa21122211

 The determinant of a 11 matrix A = a11 is det(A) = |A| = a11. The |…| symbols mean determinant, not absolute value.

 aa 1211  aa 1211  The determinant of a 22 matrix A =   is det(A) = |A| = = a11a22 – a12a21.  aa 2221  aa 2221

Geometrically, the signed area of a parallelogram with vertices at (0, 0), (x1, y1), (x2, y2), and (x1 + x2, y1 + y2) is yx A = 11 yx 22 (x2, y2)

(The area is positive if the angle from (x1, y1) (x1, y1) to (x2, y2) is counterclockwise; otherwise, the area is negative.)

Proof:

The area A of the parallelogram is A = area of large rectangle  areas of four triangles  areas of two small rectangles 1 1 = (x1 + x2)(y1 + y2)  2 x1y1  2 x1y1  x2y2  x2y2  2x2y1

= x1y1 + x1y2 + x2y1 + x2y2  x1y1  x2y2  2x2y1

= x1y2  x2y1 =

p. 41 Chapter 3: Determinants. 3.1 The Determinant of a Matrix. To define the determinant of a square matrix A of order (dimensions) higher than 2, we define minors and cofactors. The minor Mij of the entry aij is the determinant of the matrix obtained by i+j deleting row i and column j of A. The cofactor Cij of the entry aij is Cij = (–1) Mij. Notice that             (–1)i+j is a “checkerboard” pattern: (–1)i+j =                  

 Examples: Finding Cofactors. Let A =

a12 a13 Find C21. Solution:  C21 = – = –a12a33 + a13a32 a32 a33

a11 a13 Find C22. Solution:  C22 = + = a11a33 – a13a31 a31 a33

The determinant of an nn matrix A (n ≥ 2) is the sum of the entries in the first row of A multiplied by their respective cofactors.

n det(A) = a1 j C1 j = a11C11 + a12C12 + … + a1nC1n j1

3 3 4 8 7  6 7  6 8 Example: 6 8 7 = 3 + 3  + 4 5  9   3  9   3 5  3 5  9   = 3(–72 – 35) +3(54 – 21) + 4(30 + 24) = 3(–107) +3(33) +4(54) = –6

Theorem 3.1 Expansion by Cofactors

Let A be a square matrix of order n. Then the determinant of A is given by an expansion in any row i

n det(A) =  Ca ijij = ai1Ci1 + ai2Ci2 + … + ainCin j1

and also by an expansion in any column j

n det(A) = aijCij = a1jC1j + a2jC2j + … + anjCnj i1

p. 42 Chapter 3: Determinants. 3.1 The Determinant of a Matrix.

When expanding, you don’t need to find the cofactors of zero entries, because aijCij = (0)Cij = 0.

The definition of the determinant is inductive, because it uses the determinant of a matrix of order n – 1 to define the determinant of a matrix of order n.

 Example: Expanding by Cofactors to Find a Determinant

= = –1 + 3(– )

= –1(2 +7 ) +3(–2 ) = –1(2(3) + 7(–6)) + 3(–2)(0) = –(6 – 42) = 36

To find a determinant using Mathematica, type Det[a] (also on Basic Math Assistant, More drop-down menu)

To find a determinant on the TI-89, type Matrixdet(A

To find a determinant of a 33 matrix, you can also use the following shortcut. Copy Columns 1 and 2 into Columns 4 and 5. To calculate the determinant, add and subtract the indicated products. subtract

a11 a12 a13 a11 a12 a13 a11 a12

a21 a22 a23  a21 a22 a23 a21 a22

a31 a32 a33 a31 a32 a33 a31 a32 add

= a11a22a33 + a12a23a31 + a13a21a32 – a31a22a13 – a32a23a11 – a33a21a12

12 30 0 1 1 3 1 1 3 1 1 Example: 0 4 6  0 4 6 0 4 so = 32 + 6 + 0 – 12 – 30 – 0 = –4 1 5 8 1 5 8 1 5 32 6 0

p. 43 Chapter 3: Determinants. 3.1 The Determinant of a Matrix. Theorem 3.2 Determinant of a Triangular Matrix

The determinant of a triangular matrix A of order n is the product of the entries on the main diagonal. det(A) = a11a22… ann

Proof by Induction for upper triangular matrices:

 When k = 1, A = [a11] so |A| = a11

 Assume that the theorem holds for all upper triangular matrices of order k. Let A be an upper triangular matrix of order k + 1. Then expanding in the last row,

a11 a12  a1k a1,k 1 a11 a12  a1k 0 a22  a2k a2,k 1 0 a22  a2k |A| =      = ak+1,k+1     0 0  akk ak,k 1 0 0  akk 0 0  0 ak1,k 1

= ak+1,k+1(a11a22…akk) = a11a22…akk ak+1,k+1

The proof for lower triangular matrices is similar.

Optional application to multivariable calculus:

u2 x2 du Remember integration by substitution: f (u)du = f (u(x)) dx   dx u1 x1 cos(x) optional For example, cot(x)dx = dx   sin(x) du cos(x) 1 du 1 Let u = sin(x), = cos(x) so dx = dx = du = ln|u| + C = ln|sin(x)| + C dx  sin(x)  u dx  u

In multivariable calculus, u u u x y z v v v  f (u,v, w)dudvdw =  f (u(x, y, z),v(x, y, z), w(x, y, z)) dxdydz V V x y z w w w x y z

The determinant is called the Jacobian.

p. 44 Chapter 3: Determinants. 3.2 Determinants and Elementary Operations.

3.2 Determinants and Elementary Operations.  Objective: Use elementary row operations to evaluate a determinant.  Objective: Use elementary column operations to evaluate a determinant.  Recognize conditions that yield zero determinants.

In practice, we rarely evaluate determinants using expansion by cofactors. The properties of determinants under elementary operations provide a much quicker way to evaluate determinants.

Theorem 3.9 det(AT) = det(A). [Proof is in Section 3.4]

Theorem 3.3 Elementary Row (Column) Operations and Determinants.

Let A and B be nn square matrices.

a) When B is obtained from A by swapping two rows (two columns) of A, det(B) = –det(A). b) When B is obtained from A by adding a multiple of one row of A to another row of A (or one column of A to another column of A), det(B) = det(A). c) When B is obtained from A by multiplying of a row (column) of A by a nonzero constant c, det(B) = c det(A).

Theorem 3.4 Conditions that Yield a Zero Determinant.

If A is an nn square matrix and any one of the following conditions is true, then det(A) = 0

a) An entire row (or an entire column) consists of zeros. b) Two rows (or two columns) are equal. c) One row is a multiple of another row (or one column is a multiple of another column).

Proof by Induction of 3.3a (for rows):

 aa 1211   aa 2221   When k = 2, A =   and B =   so  aa 2221   aa 1211  det(B) = a21a12 – a22a11 = –(a11a22 – a12a21) = –det(A)

 Assume that the theorem holds for all matrices of order k. Let A be a matrix of order k + 1 and B be a matrix obtained by swapping two rows of A. To find det(A) and det(B), expand in any row other than the swapped rows. The respective cofactors are opposites, because they come from kk matrices that have two rows swapped. Thus, det(B) = –det(A).

Proof of 3.4a (for rows): Suppose that row i of A is all zeroes. Expand by cofactors in row i. n n det(A) =  Ca ijij = 0 Cij = 0 j1 j1

p. 45 Chapter 3: Determinants. 3.2 Determinants and Elementary Operations. Proof of 3.4b: Let B be the matrix obtained from A by swapping the two identical rows (columns) of A, so det(B) = –det(A). But B = A, so det(A) = –det(A) so det(A) = 0.

Proof of 3.3b (for rows): Suppose B is obtained from A by adding c times row k to row i. Expand by cofactors in row i. Note that the cofactors of Cij are the same for matrices A and B, because the matrices are the same everywhere except row i. n n n n det(B) =  Cbijij =  ca kj  )(Caijij =  kj Cacij +  Caijij = c·0 + det(A) = det(A). j1 j1 j1 j1 n because  kj Caij is the determinant of a matrix with two identical rows (row k and row i). j1 See Theorem 3.4b.

Another way of writing this is

a11  a1n 11  aa 1n 11  aa 1n    

ak1  akn k1  aa kn k1  aa kn   = c  +  = c·0 + det(A) = det(A).

ca  aik 11  ca kn  ain )()( k1  aa kn i1  aa in    

an1  ann n1  aa nn n1  aa nn

Proof of 3.3c (for rows): Suppose B is obtained from A by multiplying row i by a nonzero scalar c. Expand by cofactors in row i. n n n det(B) =  Cb ijij = ca Cijij = c  Ca ijij = c det(A) j1 j1 j1

Proof of 3.4c: Suppose B is a matrix with two equal rows (or two equal columns), and A is obtained from B by multiplying one of those rows (or columns) by a nonzero scalar c. Using 3.3c on that row (or column), det(A) = c det(B). Using 3.4b, det(B) = 0. Thus, det(A) = 0.

Geometrically, the signed area of a parallelogram with edges from (0, 0) to (x1, y1) and from (0, 0) to (x2, y2) has the same properties yx as 11 when you perform an elementary row operation. Also, yx 22 the signed area of a parallelepiped with edges from (0, 0) to (x1, y1, z1), from (0, 0) to (x2, y2, z2), and from (0, 0) to (x3, y3, z3) has the

zyx 111 same properties as zyx 222 when you perform an elementary

zyx 333 row operation.

p. 46 Chapter 3: Determinants. 3.2 Determinants and Elementary Operations. Row Swapping (Theorem 3.3a)

yx x y If we swap two rows, e.g. 11  2 2 yx 22 x1 y1

zyx111 x2 y2 z2

or zyx222  x1 y1 z1

zyx333 x3 y3 z3

then the sign of the area/volume changes because we change from a right-hand orientation to a left-hand orientation.

Adding a multiple of one row to another row (Theorem 3.3b)

x y If we add a multiple of one row to another row, e.g.  1 1 , then x2  0.5x1 y2  0.5y1 the A = bh is unchanged.

p. 47 Chapter 3: Determinants. 3.2 Determinants and Elementary Operations. Adding a multiple of one row to another row (Theorem 3.3b)

yx cx cy If we multiply a row by a nonzero constant c, e.g. 11  1 1 , then the A = bh  yx 22 x2 y2 (cb)h = cA is also multiplies by the constant c.

 Example: Finding a Determinant Using Elementary Row Operations

4 3  2 14 11 0 5 4 1 = 5 4 1 R + 2R  R = –1[14(–13) – 11(–22)] 1 2 1

 2 3 4  2 3 4 = –1(60) 14 11 0 = –60 = 5 4 1 R3 – 4R2  R3

 22 13 0 14 11 = –1  22 13

p. 48 Chapter 3: Determinants. 3.2 Determinants and Elementary Operations.  Example: Finding a Determinant Using Elementary Column Operations

 C2  3C1  C2 1 3 3 4 1 0 0 0    2  4 6 5  2 ______ C3  ____  C3 7 5 1 3 = 7 ______ 9 5 1 2 9 ______ C4  ____  C4 1 0 0 0  2 1 __ __

= (__) 7 ______C2  C2

9 ______1 0 0 0 ______ C3  2 1 0 0   = (__) 7 ____________ C 9 ______ 4 1 0 0 ______= (__)(__) = (__)(__)(__) ______

=

p. 49

Chapter 3: Determinants. 3.3 Properties of Determinants.

3.3 Properties of Determinants.  Objective: Find the determinant of a matrix product and of a scalar multiple of a matrix.  Find the determinant of an inverse matrix and recognize equivalent conditions for a nonsingular matrix.  Find the determinant of the transpose of a matrix.

Theorem 3.5 Determinant of a Matrix Product

If A and B are square matrices of the same order, then det(AB) = det(A) det(B).

Proof: To begin, let E be an elementary matrix. By Thm 2.12, EB is the matrix obtained from applying the corresponding row operation to B. By Thm. 3.3,  B)det(  exchanging two rows    det(EB) =  B)det( if the row operation is  ofmultiple one a adding another torow     Bc)det(  constantgmultiplyin a c row nonzero aby Also by Thm 3.3, 1   det(E) = det(EI) =  1  if the row operation is    c  Thus, det(EB) = det(E) det(B). This can be generalized by induction to conclude that |Ek…E2E1B| = |Ek| |…| |E2| |E1| |B| where the Ei are elementary matrices. If A is nonsingular, then by Thm. 2.14, it can be written as the product A = Ek…E2E1 so |AB| = |A| |B|.

If A is singular, then A is row-equivalent to a matrix with an entire row of zeroes (for example, the reduced row echelon form). From Thm 3.4, we know |A| = 0. Moreover, because A is singular, it follows that AB must be singular. (Proof by contradiction: if AB were nonsingular, then A[B(AB)-1] = I would show that A is not singular, because A–1 = B(AB)-1.) Therefore, |AB| = 0 = |A| |B|.

Comment on Proof by Contradiction: “P implies Q” is equivalent to “not Q implies not P.”

Theorem 3.6 Determinant of a Scalar Multiple of a Matrix

If A is a square matrix of order n and c is a scalar, then det(cA) = cndet(A).

Proof: Apply Property (c) of Thm. 3.3 to each of the n rows of A to obtain n factors of c.

Theorem 3.7 Determinant of an Invertible Matrix

A square matrix A is invertible (nonsingular) if and only if det(A)  0.

Proof: On the one hand, if A is invertible, then AA–1 = I, so . |A| |A–1| = | I | = 1. Therefore, |A|  0. On the other hand, assume det(A)  0. Then use Gauss-Jordan elimination to find the reduced row-echelon form R. Since R is in reduced row-echelon form, it is either the identity matrix or

p. 51 Chapter 3: Determinants. 3.3 Properties of Determinants. it must have at least one row of all zeroes. The second case is not possible: if R had a row of all zeroes, then det(R) = 0, but then det(A) = 0 (which contradicts the assumption). Therefore, A is row-equivalent to R = I, so A is invertible.

Theorem 3.8 Determinant of an Inverse Matrix

1 If A is an invertible matrix, then det (A–1) = det(A)

1 Proof: AA–1 = I, so . |A| |A–1| = | I | = 1 and |A|  0, so |A–1| = . | A |

 Equivalent Conditions for a Nonsingular nn Matrix (Summary)

1) A is invertible. 2) Ax = b has a unique solution for every n1 column matrix b. 3) Ax = 0 has only the trivial solution for the n1 column matrix 0. 4) A is row-equivalent to I. 5) A can be written as a product of elementary matrices. 6) det(A)  0.

Theorem 3.9 Determinant of a the Transpose of a Matrix

If A is a square matrix, then det(AT) = det(A).

Proof: Let A be a square matrix of order n. From Section 2.4, we know that A can be factored as PA = LDU, where P is a permutation matrix, L is lower triangular with all ones on the main diagonal, D is diagonal, and U is upper triangular with all ones on the main diagonal. L is obtained from I by adding a multiple of the rows containing the diagonal ones to the rows below the diagonal, so |L| = |I | = 1 by Thm. 3.3b. Likewise, U is obtained from I by adding a multiple of the rows containing the diagonal ones to the rows above the diagonal, so |U| = |I | = 1 by Thm. 3.3b. By Thm. 3.2, |D| = d11d22… dnn

i.e. the product of its diagonal elements. P is a product of elementary row-swap matrices, each of which has determinant –1. So |P| is the product of some number of –1’s.

|P| = 1 if the number of row swaps is even; |P| = –1 if the number of row swaps is odd.

p. 52 Chapter 3: Determinants. 3.3 Properties of Determinants.

T 1 0 0 e i   1  0 1 0 eT  i2  Let e1 =   , e2 =   , …, en =   be n1 matrices. Then P = where i1, i2, …, in is               0 0 1 eT        in  some permutation of 1, 2, …, n. Now PT = e e  e , so i1 i2 in eT e eT e  eT e  1 0  0 i1 i1 i1 i2 i1 in  T T T    T ei ei ei ei  ei ei 0 1  0 PP =  2 1 2 2 2 n  =   = I,               eT e eT e  eT e 0 0  1  in i1 in i2 in in   

and by Thm. 3.5, det(P) det(PT ) = det(PPT ) = det(I) = 1.

Then either det(P) = 1 so det(PT) = 1, or det(P) = –1 so det(PT) = –1. In both cases,

det(P) = det(PT).

So we have PA = LDU which gives us |P| |A| = |L| |D| |U| = (1) |D| (1) =|D|, so | D | |A| = . | P | Taking the transpose, we have ATPT = UTDTLT which gives us |AT| |PT| = |UT| |DT| |LT|. Now |PT| = |P|; |DT| = |D| because DT = D since D is diagonal; |LT| = 1

because LT is upper triangular with all ones on the main diagonal; and

|UT| = 1

because UT is lower triangular with all ones on the main diagonal. Thus, |AT| |PT| = |UT| |DT| |LT| becomes |AT| |PT| = (1) |D| (1), so

| D | | D | |AT| = = = |A|. | PT | | P |

p. 53

Chapter 3: Determinants. 3.4 Applications of Determinants.

3.4 Applications of Determinants.  Objective: Find the adjoint of a matrix and use it to find the inverse of a matrix.  Objective: Use Cramer’s Rule to solve a system of n linear equations in n unknowns.  Objective: Use determinants to find area, volume, and the equations of lines and planes.

Using the adjoint of a matrix to calculate the inverse is time-consuming and inefficient. In practice, Gauss-Jordan elimination is used for 33 matrices and larger. However, “adjoint” is vocabulary you may be expected to know in future classes.

i+j Recall from Section 3.1 that the cofactor Cij of a matrix A is (–1) times the determinant of the matrix obtained by deleting row i and column j of A.

 11 12  CCC1n    CCC The matrix of cofactors of A is  21 22 2n  .      nn21  CCCnn 

 11 21  CCC n1    CCC   The adjoint of A is the transpose of matrix of cofactors: adj(A) =  12 22 n2        21 nn  CCC nn 

 Theorem 3.10 The Inverse of a Matrix Given by Its Adjoint

1 If A is an invertible matrix, then A–1 = adj(A). A)det(

 1211  aaa 1n    2221  aaa 2n  11 21  j1  CCCC n1         12 22  j2  CCCC n2 Proof: Consider A[adj(A)] =         ii 21  aaa ki        21 nn  jn  CCCC nn      optional  nn 21  aaa nn 

The ij entry of this product is ai1Cj1 + ai2Cj2 + … + ainCjn. If i = j, this is det(A) (expanded by cofactors in row i). If i  j, this is the determinant of the matrix B, which is the same as A except that row j has been replaced with row i.

p. 55 Chapter 3: Determinants. 3.4 Applications of Determinants.

a11 a12  a1n  a11 a12  a1n  a a  a  a a  a   21 22 2n   21 22 2n             11 21  j1  CCCCn1        a a  a a a  a   CCCC A =  i1 i2 ki  , B[adj(A)] =  i1 i2 ki   12 22 j2 n2                      CCCC a j1 a j2  a jn  ai1 ai2  ain   21nn jn nn                an1 an2  ann  an1 an2  ann  The j column of the cofactor matrix is unchanged, because it does not depend on the j row of A or B. Since two rows of B are the same, the cofactor expansion for i  j is zero.

det(A) 0  0   0 det(A)  0  Thus, A[adj(A)] =   = det(A)I.          0 0  det(A)

1 det(A)  1 because A is invertible, so we can write A[ adj(A)] = I, so A–1 = adj(A). A)det(

For 33 matrices and larger, Gauss-Jordan elimination is much more efficient than the adjoint method for finding the inverse of a matrix.

a b  d  b –1 1 a b However, for a 22 matrix A =   , adj(A) =   so A =   . c d  c a ad  bc c d      

Cramer’s Rule to solve n linear equation in n variables is time-consuming and inefficient. In practice, Gaussian elimination is used to solve linear systems. However, Cramer’s Rule is vocabulary you may be expected to know in future classes.

 Theorem 3.11 Cramer’s Rule

If an nn system Ax = b has a coefficient matrix with nonzero determinant |A|  0, then det(A ) det(A ) det(A ) x  1 , x  2 , … x  n 1 det(A) 2 det(A) n det(A)

where Ai is the matrix A but with column i replace by b.

p. 56 Chapter 3: Determinants. 3.4 Applications of Determinants.

 x1   11 21  CCCn1  b1  x  1   CCC b  Proof:  2  x = A–1b = adj(A)b =  12 22 n2   2     A)det(            xn   21nn  CCCnn  bn 

so xi = (b1C1i + b2C2i + … + bnCni). The sum in parentheses is the cofactor expansion of

det(Ai ) det(Ai), so xi = det(A)

1x  2y  10 Example: Solve using Cramer’s Rule. 3x  4y  15

10 2 1 10 15 4 10 3 15 15 15 Solution: x = = = –5; y = = = 1 2  2 1 2  2 2 3 4 3 4

 Area, Volume and Equations of Lines and Planes:

We already know that the signed area of a parallelogram yx is given by a 22 determinant (Section 3.1). A = 11 yx 22

x1 y1 1

The signed area of a triangle with vertices (x1, y1), (x2, y2), and (x3, y3) is A = x2 y2 1

x3 y3 1

Proof: (x3, y3) The area of a triangle is half of the area of a parallelogram. So the area of the triangle we want is

x1 y1 1 x y x y x y 1 1 2 2 + 3 3 + 1 1 = x y 1 2 x y x y x y 2 2 3 3 1 1 2 2 2 (x2, y2)    x3 y3 1 pos neg neg (x1, y1)

x1 y1 1 1 For a triangle the signed area is A = x2 y2 1 2 x3 y3 1

If the vertices (x1, y1), (x2, y2), and (x3, y3) are ordered clockwise then the area is positive; otherwise, it is negative. (The homework asks for the absolute value of the area.)

p. 57 Chapter 3: Determinants. 3.4 Applications of Determinants. The area of the triangle is zero if and only if the three points are collinear.

x1 y1 1

 (x1, y1), (x2, y2), and (x3, y3) are collinear if and only if x2 y2 1 = 0

x3 y3 1

x y 1  The equation of a line through distinct points (x1, y1) and (x2, y2) is x1 y1 1 = 0

x2 y2 1

Similarly to the two-dimensional case of a parallelogram, the signed volume of a parallelepiped is given

zyx111

by a 33 determinant (Section 3.1). V = zyx222

zyx333

Let’s find the volume of the tetrahedron (pyramid with four triangular faces) with vertices at (x1, y1, z1), (x2, y2, z2), (x3, y3, z3), and (x4, y4, z4).

1 The volume of the tetrahedron with vertices at (0, 0, 0), (x1, y1, z1), (x2, y2, z2), and (x3, y3, z3) is 6

x1 y1 z1 1 times the volume of the parallelepiped, i.e. x y z . For the triangle above 6 2 2 2 x3 y3 z3

x1 y1 1 1 ( 2 x2 y2 1 ), we had 3 sides and the areas of 3 triangles to add/subtract. Now we have 4faces

x3 y3 1 and the volumes of 4 tetrahedrons to add/subtract. The signed volume of the tetrahedron with vertices at (0, 0, 0), (x1, y1, z1), (x2, y2, z2), and (x3, y3, z3) is (imagine for the moment that (x4, y4, z4) in front of the triangle (x1, y1, z1), (x2, y2, z2), (x3, y3, z3)

x1 y1 z1 x4 y4 z4 x2 y2 z2 x3 y3 z3 1 1 1 1 V = x y z + x y z + x y z + x y z 6 4 4 4 6 1 1 2 6 3 3 3 6 2 2 2 x3 y3 z3 x2 y2 z2 x4 y4 z4 x1 y1 z1

For a tetrahedron, the signed volume is x y z 1 1 1 1 1 x2 y2 z2 1 V = – 6 x3 y3 z3 1

x4 y4 z4 1

p. 58 Chapter 3: Determinants. 3.4 Applications of Determinants. If when you wrap the fingers of your right from (x1, y1, z1) to (x2, y2, z2) to (x3, y3, z3), your thumb points toward (x4, y4, z4), then the signed volume is positive. If when you wrap the fingers of your left from (x1, y1, z1) to (x2, y2, z2) to (x3, y3, z3), your thumb points toward (x4, y4, z4), then the signed volume is negative. (The homework asks for the absolute value of the volume.)

 Four points (x1, y1, z1), (x2, y2, z2), (x3, y3, z3), and (x4, y4, z4) are coplanar if and only if x y z 1 1 1 1 x y z 1 2 2 2 = 0 because that is when the tetrahedron has zero volume.

x3 y3 z3 1

x4 y4 z4 1

 The equation of a plane through distinct points (x1, y1, z1), (x2, y2, z2), and (x3, y3, z3) is x y z 1

x1 y1 z1 1 = 0 x2 y2 z2 1

x3 y3 z3 1

p. 59

Chapter 4: Vector Spaces. 8.1 Complex Numbers (Optional).

Chapter 4: Vector Spaces.

8.1 Complex Numbers (Optional).  Objective: Use the to find all zeroes of a quadratic polynomial with real coefficients.  Objective: Add and subtract two complex numbers and multiply a by a real scalar.  Objective: Graphically represent complex numbers in the complex plane as directed line segments.  Objective: Multiply two complex numbers.  Objective: Multiply and find determinants of matrices with complex entries.  Objective: Perform Gaussian and Gauss-Jordan elimination on matrices with complex entries.

So far, the scalars we have been using have been real numbers R. However, any mathematical field can be used as the scalars.

Properties of a Field F

Let a, b, and c be any elements of F. Then F is a field if it has two operations, addition and multiplication, and it contains two distinct elements 0 and 1, such that the following properties are true.

1) a + b is an element of F. Closure under addition 2) a + b = b + a Commutative property of addition 3) (a + b) + c = a + (b + c) Associative property of addition 4) a + 0 = a Additive identity property 5) There exists a –a such that a + (–a) = 0 Additive inverse property 6) ab is an element of F. Closure under multiplication 7) ab = ba Commutative property of multiplication 8) (ab)c = a(bc) Associative property of multiplication 9) 1a = a Multiplicative identity property 10) Except for a = 0, there exists an a–1 such that a(a–1) = 1 property 11) a(b + c) = ab + ac Distributive property

Three familiar examples of fields are the rational numbers Q = , the real numbers R, and the complex numbers C.

A familiar set that is not a field is the integers Z = {…, –2, –1, 0, 1, 2, …}. Why not?

We are going to need to use the field of complex numbers as our scalars, because we will need to solve polynomial equations. All can be solved using complex numbers (complex numbers are an “algebraically closed field”), but the same is not true of real numbers.

p. 61 Chapter 4: Vector Spaces. 8.1 Complex Numbers (Optional).

def The is i  1 , so i2 = –1.

(Electrical engineers often write j  1 because they use i for electric current.)

 Example: Solve 5x2 – 6x + 5 = 0

Aside: the complex roots of a polynomial with real coefficients are complex conjugates (Section 8.2) of each other (a + bi and a – bi).

Notice that we have assumed a definition of multiplication by a : 1 1 10 (6 + 8i) = ( 10 6) + ( 8)i

 Example: Solving the polynomial equation 2x3 +3x2 +50x +75 = 0 using software.

Solution:

Mathematica: inputting Clear[x];Solve[2x^3+3x^2+50x+75==0,x]

yields output

You can also find Solve in the Palettes Menu::Basic Math Assistant::y = x menu.

TI-89: ComplexcSolve(2x^3+3x^2+50x+75=0,x)

A complex number is a number of the form a + bi, where a and b are real numbers. a is the real part and bi is the imaginary part of a + bi. The form a + bi is the standard form of a complex number, for example 1 + 2i, 0 + 3i, and –4 + 0i.

Geometrically, a complex number a + bi is represented in the complex plane by a directed line segment from the origin to (a, b), where a and b are Cartesian coordinates. In other words, the horizontal axis is the real axis and the vertical axis is the imaginary axis.

Operations in the Set of Complex Numbers C def  Addition: (a + bi) + (c + di)  (a + c) + (b + d)i def  Multiplication by a real number: c(a + bi)  ca + cbi  Negative: –(a + bi) –a + –bi. Notice that –(a + bi) = (–1)(a + bi).  Subtraction: (a + bi) + (c + di) (a + bi) + –(c + di) = (a – c) + (b – d)i

p. 62 Chapter 4: Vector Spaces. 8.1 Complex Numbers (Optional).  Examples

Let z = 2 + 4i and w = –4 + 1 i. Illustrate the following graphically. z, w, z + w, 1.5z, – w, w – z p. 63 Chapter 4: Vector Spaces. 8.1 Complex Numbers (Optional).  Multiplication of complex numbers is defined using the distributive property and using i 2 = –1. def (a + bi)(c + di)  (ac – bd) + (ad + bc)i

because (a + bi)(c + di) = ac + adi + bci + bdi 2 = ac + adi + bci + bdi 2 = ac + adi + bci – bd

Warning: When you multiply square roots of negative numbers, convert into standard form a + bi (not using the of a negative number) before you multiply. For example, 1 1 = i·i = –1, not 1 = 1()()1 = 1 = 1.

Application (Electrical Engineering): V(t)= I(t)Z, I(t)= current = I0 cos(t) + i I0 sin(t) 1 V(t) = volatge, Z = impedance = R + + Li, Ci R = resistance, C = capacitance, L = inductance

3 4 Example: Use software to check that  i is a zero of the polynomial 5x2 – 6x + 5. 5 5

Solution:

Mathematica: to input the imaginary unit i, denoted by i in Mathematica, type ii (four separate keystrokes). After the first three keystrokes, you will see ii. After the fourth keystroke, ii will change to i.

Type x=3/5+4ii/5 5x^2-6x+5

TI-89: (3/5+4/5). 5x^2-6x+5

Complex matrices   413 i  Example: Find the inverse of A =   by hand. Check your answer by multiplying.   i 541 

–1 1   415 i Solution: A =   . A)det(   i 341  det(A) = 5(3) – (–1 + 4i)(–1 – 4i) = 15 – (1 + 4i – 4i + 16) = –2 1 A–1 =  2

p. 64 Chapter 4: Vector Spaces. 8.1 Complex Numbers (Optional).

–1   413i  1   415i Check: AA =         i 541  2   i 341   1  15 441 ii 16 3 12 3  12ii   02   01 =     =   =    2   5 20 5 20 441 iiii  1516   20  10  Example: By hand, find the determinant of

 Example: Perform Gauss-Jordan elimination using row operations to solve

)2( xiw iy  z  22 iw  31)53(3)42(2 iziyxi ()62()36(3 16  3)7 iziyixiw

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392 Chapter 8 Complex Vector Spaces

8.1 Complex Numbers

Use the imaginary unit i to write complex numbers. Graphically represent complex numbers in the complex plane as points and as vectors. Add and subtract two complex numbers, and multiply a complex number by a real scalar. Multiply two complex numbers, and use the Quadratic Formula to find all zeros of a quadratic polynomial. Perform operations with complex matrices, and find the determinant of a complex matrix.

REMARK COMPLEX NUMBERS When working with products So far in the text, the scalar quantities used have been real numbers. In this chapter, you involving square roots of will expand the set of scalars to include complex numbers. negative numbers, be sure to In algebra it is often necessary to solve quadratic equations such as convert to a multiple of i before x2 3x 2 0. The general is ax 2 bx c 0, and its multiplying. For instance, consider the following solutions are given by the Quadratic Formula operations. b Ίb2 4ac x Ί1Ί1 i i 2a i 2 where the quantity under the radical,b2 4ac, is called the . If 2 1 Correct b 4ac 0, then the solutions are ordinary real numbers. But what can you conclude about the solutions of a quadratic equation whose discriminant is negative? For Ί Ί Ί͑ ͒͑ ͒ 1 1 1 1 instance, the equation x2 4 0 has a discriminant of b2 4ac 16, but there is Ί1 no real number whose square is 16. To overcome this deficiency, mathematicians 1 Incorrect invented the imaginary unit i, defined as i Ί1 where i 2 1. In terms of this imaginary unit, Ί16 4Ί1 4i. With this single addition of the imaginary unit i to the real number system, the system of complex numbers can be developed.

Definition of a Complex Number If and ba are real numbers, then the number a bi is a complex number, where a is the real part and bi is the imaginary part of the number. The form a bi is the standard form of a complex number.

Some examples of complex numbers written in standard form are 2 2 0i, 4 3i, and 6i 0 6i. The set of real numbers is a subset of the set of complex numbers. To see this, note that every real number a can be written as a complex number using b 0. That is, for every real number, a a 0i. A complex number is uniquely determined by its real and imaginary parts. So, two complex numbers are equal if and only if their real and imaginary parts are equal. That is, if a bi and c di are two complex numbers written in standard form, then a bi c di if and only if a c and b d. 9781133110873_0801.qxp 3/10/12 6:52 AM Page 393

8.1 Complex Numbers 393

Imaginary THE COMPLEX PLANE axis Because a complex number is uniquely determined by its real and imaginary parts, (a, b) or a + bi it is natural to associate the number a bi with the ordered pair ͑a, b͒. With this association, complex numbers can be represented graphically as points in a coordinate b plane called the complex plane. This plane is an adaptation of the rectangular Real axis coordinate plane. Specifically, the horizontal axis is the real axis and the vertical axis a is the imaginary axis. The point that corresponds to the complex number a bi is ͑a, b͒, as shown in Figure 8.1. The Complex Plane Figure 8.1 Plotting Numbers in the Complex Plane

Plot each number in the complex plane. a.4 3i b.2 i c.3i d. 5 SOLUTION Figure 8.2 shows the numbers plotted in the complex plane.

a.Imaginary b. Imaginary axis axis 4 2 3 4 + 3i 1 or (4, 3) Real 2 axis −31−2 −1 2 3 1 Real −2 − i axis or (−2, −1) −2−1 1234 −3 −2 −4

c.Imaginary d. Imaginary axis axis 2 4 1 3 Real axis 2 −3123−2 −1 1 5 or (5, 0) Real −2 axis −1 12345 −3 −3i or (0, −3) −4 −2

Figure 8.2

Another way to represent the complex number a bi is as a vector whose horizontal component is a and whose vertical component is b. (See Figure 8.3.) (Note that the use of the letter i to represent the imaginary unit is unrelated to the use of i to represent a unit vector.)

Imaginary axis

1 Horizontal component Real axis

−1

−2 Vertical component 4 − 2i −3

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394 Chapter 8 Complex Vector Spaces

ADDITION, SUBTRACTION, AND SCALAR MULTIPLICATION OF COMPLEX NUMBERS Because a complex number consists of a real part added to a multiple of i, the operations of addition and multiplication are defined in a manner consistent with the rules for operating with real numbers. For instance, to add (or subtract) two complex numbers, add (or subtract) the real and imaginary parts separately.

Definition of Addition and Subtraction of Complex Numbers The sum and difference of a bi and c di are defined as follows. ͑a bi͒ ͑c di͒ ͑a c͒ ͑b d͒i Sum ͑ ͒ ͑ ͒ ͑ ͒ ͑ ͒ REMARK a bi c di a c b d i Difference Note in part (a) of Example 2 that the sum of two complex numbers can be a real number. Adding and Subtracting Complex Numbers

a. ͑2 4i͒ ͑3 4i͒ ͑2 3͒ ͑4 4͒i 5 b. ͑1 3i͒ ͑3 i͒ ͑1 3͒ ͑3 1͒i 2 4i

Using the vector representation of complex numbers, you can add or subtract two complex numbers geometrically using the parallelogram rule for vector addition, as shown in Figure 8.4.

Imaginary Imaginary axis axis z = 3 + 4i 4 2 3 w = 3 + i 1 2 Real 1 z w axis + = 5 Real −3 1 23 axis −1 162345

−2 −3 − −3 z = 1 3i −4 w = 2 − 4i z − w = −2 − 4i Addition of Complex Numbers Subtraction of Complex Numbers Figure 8.4 Many of the properties of addition of real numbers are valid for complex numbers as well. For instance, addition of complex numbers is both associative and commutative. Moreover, to find the sum of three or more complex numbers, extend the definition of addition in the natural way. For example, ͑ 2 i͒ ͑3 2i͒ ͑2 4i͒ ͑2 3 2͒ ͑1 2 4͒i 3 3i. 9781133110873_0801.qxp 3/10/12 6:52 AM Page 395

8.1 Complex Numbers 395

Another property of real numbers that is valid for complex numbers is the distributive property of scalar multiplication over addition. To multiply a complex number by a real scalar, use the definition below.

Definition of Scalar Multiplication If c is a real number and a bi is a complex number, then the scalar multiple of c and a bi is defined as c͑a bi͒ ca cbi.

Scalar Multiplication with Complex Numbers

a. 3 ͑2 7i͒ 4͑8 i͒ 6 21i 32 4i 38 17i b. 4͑1 i͒ 2͑3 i͒ 3͑1 4i͒ 4 4i 6 2i 3 12i 1 6i

Geometrically, multiplication of a complex number by a real scalar corresponds to the multiplication of a vector by a scalar, as shown in Figure 8.5.

Imaginary Imaginary axis axis

4 3 3 2 2z = 6 + 2i z = 3 + i 2 1 z = 3 + i Real 1 axis Real 1 23 axis − − − 1 2345 6 z = 3 i −1 −2 −2 −3

Multiplication of a Complex Number by a Real Number Figure 8.5 With addition and scalar multiplication, the set of complex numbers forms a of dimension 2 (where the scalars are the real numbers). You are asked to verify this in Exercise 55.

LINEAR Complex numbers have some useful applications in ALGEBRA electronics. The state of a circuit element is described by APPLIED two quantities: the voltage V across it and the current I flowing through it. To simplify computations, the circuit element’s state can be described by a single complex number z V li, of which the voltage and current are simply the real and imaginary parts. A similar notation can be used to express the circuit element’s capacitance and inductance. When certain elements of a circuit are changing with time, electrical engineers often have to solve differential equations. These can often be simpler to solve using complex numbers because the equations are less complicated.

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396 Chapter 8 Complex Vector Spaces

MULTIPLICATION OF COMPLEX NUMBERS The operations of addition, subtraction, and scalar multiplication of complex numbers have exact counterparts with the corresponding vector operations. By contrast, there is no direct vector counterpart for the multiplication of two complex numbers.

Definition of Multiplication of Complex Numbers The product of the complex numbers a bi and c di is defined as ͑a bi͒͑c di͒ ͑ac bd͒ ͑ad bc͒i.

Rather than try to memorize this definition of the product of two complex numbers, simply apply the distributive property, as follows. TECHNOLOGY ͑a bi͒͑c di͒ a͑c di͒ bi͑c di͒ Distributive property Many graphing utilities and 2 software programs can ac ͑ad͒i ͑bc͒i ͑bd͒i Distributive property calculate with complex ac ͑ad͒i ͑bc͒i ͑bd͒͑1͒ Use i2 1. numbers. For example, on ͑ ͒ ͑ ͒ some graphing utilities, you ac bd ad i bc i Commutative property can express a complex number ͑ac bd͒ ͑ad bc͒i Distributive property a bi as an ordered pair ͑a, b͒. Try verifying the result of Example 4(b) by multiplying Multiplying Complex Numbers ͑2, 1͒ and ͑4, 3͒.You should obtain the ordered pair ͑11, 2͒. a. ͑2͒͑1 3i͒ 2 6i b. ͑2 i͒͑4 3i͒ 8 6i 4i 3i 2 8 6i 4i 3͑1͒ 8 3 6i 4i 11 2i

Complex Zeros of a Polynomial

Use the Quadratic Formula to find the zeros of the polynomial p͑x͒ x2 6x 13 and verify that p͑x͒ 0 for each zero. SOLUTION Using the Quadratic Formula, b Ίb2 4ac 6 Ί16 6 4i x 3 2i. 2a 2 2 Substitute each value of x into the polynomial p͑x͒ to verify that p͑x͒ 0. p͑3 2i͒ ͑3 2i͒2 6͑3 2i͒ 13 ͑3 2i͒͑3 2i͒ 6͑3 2i͒ 13 REMARK 9 6i 6i 4 18 12i 13 0 A well-known result from p͑3 2i͒ ͑3 2i͒2 6͑3 2i͒ 13 algebra states that the complex ͑ ͒͑ ͒ ͑ ͒ zeros of a polynomial with real 3 2i 3 2i 6 3 2i 13 coefficients must occur in 9 6i 6i 4 18 12i 13 0 conjugate pairs. (See Review Exercise 81.) In Example 5, the two complex numbers 3 2i and 3 2i are complex conjugates of each other (together they form a conjugate pair). More will be said about complex conjugates in Section 8.2. 9781133110873_0801.qxp 3/10/12 6:52 AM Page 397

8.1 Complex Numbers 397

COMPLEX MATRICES Now that you are able to add, subtract, and multiply complex numbers, you can apply these operations to matrices whose entries are complex numbers. Such a matrix is called complex.

Definition of a Complex Matrix A matrix whose entries are complex numbers is called a complex matrix.

All of the ordinary operations with matrices also work with complex matrices, as demonstrated in the next two examples.

Operations with Complex Matrices

Let A and B be the complex matrices i 1 i 2i 0 A ΄ ΅ and B ΄ ΅ 2 3i 4 i 1 2i and determine each of the following. a.3A b.͑2 i͒B c.A B d. BA SOLUTION i 1 i 3i 3 3i a. 3A 3΄ ΅ ΄ ΅ 2 3i 4 6 9i 12 2i 0 2 4i 0 b. ͑2 i͒B ͑2 i͒΄ ΅ ΄ ΅ i 1 2i 1 2i 4 3i i 1 i 2i 0 3i 1 i c. A B ΄ ΅ ΄ ΅ ΄ ΅ 2 3i 4 i 1 2i 2 2i 5 2i 2i 0 i 1 i d. BA ΄ ΅΄ ΅ i 1 2i 2 3i 4 2 0 2i 2 0 ΄ ΅ 1 2 3i 4i 6 i 1 4 8i 2 2 2i ΄ ΅ 7 i 3 9i

Finding the Determinant of a Complex Matrix

Find the determinant of the matrix TECHNOLOGY 2 4i 2 A ΄ ΅. Many graphing utilities and 3 5 3i software programs can perform matrix operations on complex SOLUTION matrices. Try verifying the 2 4i 2 calculation of the determinant det͑A͒ Խ 3 5 3iԽ of the matrix in Example 7. ͑ ͒͑ ͒ ͑ ͒͑ ͒ You should obtain the same 2 4i 5 3i 2 3 answer, ͑8, 26͒. 10 20i 6i 12 6 8 26i 9781133110873_0801.qxp 3/10/12 6:52 AM Page 398

398 Chapter 8 Complex Vector Spaces

8.1 Exercises

Simplifying an Expression In Exercises 1Ð6, determine 42. p͑x͒ x3 2x2 11x 52 Zero: x 4 the value of the expression. 43. p͑x͒ 2x3 3x2 50x 75 Zero: x 5i Ί Ί Ί Ί Ί Ί 1.2 3 2.8 8 3. 4 4 44. p͑x͒ x3 x2 9x 9 Zero: x 3i 4.i 3 5.i 4 6. ͑i͒7 Operations with Complex Matrices In Exercises Equality of Complex Numbers In Exercises 7Ð10, 45Ð54, perform the indicated matrix operation using the determine x such that the complex numbers in each pair complex matrices A and B. are equal. 1 ؉ i 1 1 ؊ i 3i ΅ ΄ ؍ and B ΅ ΄ ؍ A x 3i, 6 3i 2 ؊ 2i ؊3i ؊3 ؊i .7 8. ͑2x 8͒ ͑x 1͒i, 2 4i 45.A B 46. B A 9. ͑x 2 6͒ ͑2x͒i, 15 6i 1 47.2A 48. 2B 10. ͑x 4͒ ͑x 1͒i, x 3i 1 49.2iA 50. 4iB ͑ ͒ ͑ ͒ Plotting Complex Numbers In Exercises 11Ð16, plot 51. det A B 52. det B the number in the complex plane. 53.5AB 54. BA 11.z 6 2i 12.z 3i 13. z 5 5i 55. Proof Prove that the set of complex numbers, with the 14.z 7 15.z 1 5i 16. z 1 5i operations of addition and scalar multiplication (with real scalars), is a vector space of dimension 2. Adding and Subtracting Complex Numbers In Exercises 17Ð24, find the sum or difference of the complex numbers. Use vectors to illustrate your answer. 56. Consider the functions ͑ ͒ 2 ͑ ͒ 2 17.͑2 6i͒ ͑3 3i͒ 18. ͑1 Ί2i͒ ͑2 Ί2i͒ p x x 6x 10 and q x x 6x 10. 19.͑5 i͒ ͑5 i͒ 20. i ͑3 i͒ (a) Without graphing either function, determine whether the graphs of p and q have x -intercepts. ͑ ͒ ͑ ͒ ͑ ͒ 21.6 2i 22. 12 7i 3 4i Explain your reasoning. ͑ ͒ ͑ ͒ ͑ ͒ ͑ ͒ 23.2 i 2 i 24. 2 i 2 i (b) For which of the given functions is x 3 i a Scalar Multiplication In Exercises 25 and 26, use zero? Without using the Quadratic Formula, find the vectors to illustrate the operations geometrically. Be sure other zero of this function and verify your answer. to graph the original vector. 25. u and 2u, where u 3 i 57. (a) Evaluate in for n 1, 2, 3, 4, and 5. 3 2010 26. 3u and 2u, where u 2 i (b) Calculate i . (c) Find a general formula for in for any positive integer n. Multiplying Complex Numbers In Exercises 27Ð34, 0 i find the product. 58. Let A ΄ ΅. ͑ ͒͑ ͒ ͑ ͒͑ 2 ͒ i 0 27.5 5i 1 3i 28. 3 i 3 i (a) Calculate An for n 1, 2, 3, 4, and 5. 29.͑Ί7 i͒͑Ί7 i͒ 30. ͑4 Ί2i͒͑4 Ί2 i ͒ (b) Calculate A2010. 31.͑a bi͒2 32. ͑a bi͒͑a bi͒ (c) Find a general formula for An for any positive 33.͑1 i͒3 34. ͑2 i͒͑2 2i͒͑4 i͒ integer n. Finding Zeros In Exercises 35Ð40, determine all the True or False? In Exercises 59 and 60, determine zeros of the polynomial function. whether each statement is true or false. If a statement is ͑ ͒ 2 ͑ ͒ 2 35.p x 2x 2x 5 36. p x x x 1 true, give a reason or cite an appropriate statement from 37.p͑x͒ x 2 5x 6 38. p͑x͒ x 2 4x 5 the text. If a statement is false, provide an example that 39.p͑x͒ x4 16 40. p͑x͒ x4 10x 2 9 shows the statement is not true in all cases or cite an appropriate statement from the text. Finding Zeros In Exercises 41Ð44, use the given zero 59.Ί2Ί2 Ί4 2 60. ͑Ί10͒2 Ί100 10 to find all zeros of the polynomial function. 41. p͑x͒ x3 3x2 4x 2 Zero: x 1 61. Proof Prove that if the product of two complex numbers is zero, then at least one of the numbers must be zero. 9781133110873_08_ANS.qxd 3/10/12 6:58 AM Page A1

Answer Key

Ϫ5 Ϫ15 ϩ 10i Section 8.1 51. Ϫ5 Ϫ 3i 53. ΄ ΅ 55. Proof 25i 15 ϩ 30i 57. (a) i1 ϭ i (b) i2010 ϭϪ1 1.ϪΊ6 3.Ϫ4 5. 1 7.x ϭ 6 9. x ϭ 3 i2 ϭϪ1 11. Imaginary 13. Imaginary i3 ϭϪi axis axis 4 ϭ z = −5 + 5i i 1 5 2 5 i ϭ i Real 4 1 n ϭ 4k axis 3 i n ϭ 4k ϩ 1 246 (c) in ϭ , where k is an integer. −2 2 ΆϪ1 n ϭ 4k ϩ 2 − z = 6 2i Ϫ ϭ ϩ −4 1 i n 4k 3 Real 59. False. See the Remark, page 392. 61. Proof axis −5 −4 −3 −2 −1 15. Imaginary axis

5 z = 1 + 5i 4 3 2 1 Real axis 12345 17. 5 ϩ 3i 19. 2i Imaginary Imaginary axis axis

10 6 8 − u = 2 + 6i 4 u v = 2i 6 2 u = 5 + i 4 u + v = 5 + 3i Real 2 axis Real −2 8 axis v = 5 − i −2 4 6810 −4 −4 v = 3 − 3i 21. 6 ϩ 2i 23. 4 ϩ 2i Imaginary Imaginary axis axis

4 4 u − v = 6 + 2i 3 2 u + v = 4 + 2i Real 2 axis v = 2 + i 24u = 6 1 u = 2 + i −2 v = −2i Real axis 1234 25. Imaginary axis

4 −u = −3 + i Real axis −4 −2 6 −2 2u = 6 − 2i − 4 u = 3 − i 27. 20 ϩ 10i 29. 8 31. ͑a2 Ϫ b2͒ ϩ 2abi Ϫ ϩ Ϫ1 ± 3 ± ± 33. 2 2i 35. 2 2i 37. 2, 3 39. 2, 2 2 1 ϩ 3i 41. 1, 1 ± i 43. Ϫ3, ±5i 45. ΄ ΅ 2 Ϫ1 Ϫ 2i Ϫ4i 2 ϩ 2i 2 Ϫ2 ϩ 2i 2i 47. ΄ ΅ 49. ΄ ΅ 4 Ϫ 4i Ϫ6i 4 ϩ 4i 6 Chapter 4: Vector Spaces. 8.2 Conjugates and Division of Complex Numbers (Optional).

8.2 Conjugates and Division of Complex Numbers (Optional).  Objective: Find the conjugate of a complex number.  Objective: Find the modulus of a complex number.  Objective: Divide complex numbers.  Objective: Perform Gaussian on and find the inverses of matrices with complex entries.

 The conjugate of the complex number z = a + bi is denoted by 푧̅ or z* and is given by z* = a – bi

Theorem 8.1 Properties of Complex Conjugates

For a complex numbers z = a + bi,  1) zz* = 푧푧̅ = a2 + b2 2) zz* = 푧푧̅  0 3) zz* = 푧푧̅ = 0 if and only if z = 0 4) (z*)* = (푧̅) = z

 The modulus of the complex number z = a + bi is denoted by |z| and is given by |z| =  ba 22

 Theorem 8.2 The modulus of a complex number. |z| = zz* = √푧푧̅

 Example: Find z* and |z| if z = 4 – i. Solution: z* = 4 + i |z| = 2  )1(4 2 = 17

The quotient of two complex numbers z = a + bi and w = c + di, w  0, is  z z w* zw* (a bi)(c  di)  adbcbdac )( i = = = = w w w* w || 2  dc 22  dc 22

 72 i  Example: Find .  34 i  72 i 72( )(  ii )34 68 ii  2128 13 34 Solution: = = =  i  34 i 34( )(  ii )34  22 )34( 25 25

p. 67 Chapter 4: Vector Spaces. 8.2 Conjugates and Division of Complex Numbers (Optional). Theorem 8.3 Properties of Complex Conjugates

For complex numbers z and w (w  0),  1) (z + w)* = z* + w* i.e. 푧̅̅̅+̅̅̅푤̅̅ = 푧̅ + 푤̅ 2) (z – w)* = z* – w* i.e. 푧̅̅̅−̅̅̅푤̅̅ = 푧̅ − 푤̅ 3) (zw)* = z*w* i.e. 푧푤̅̅̅̅ = 푧̅ 푤̅ 4) (z/w)* = z*/w* i.e. 푧̅̅/̅̅푤̅ = 푧̅/푤̅

 5  5i 4  i   Example: Find the inverse of A =   by hand.  3  4i 1 3i Solution

p. 68 Chapter 4: Vector Spaces. 8.2 Conjugates and Division of Complex Numbers (Optional).  Example: Perform Gaussian elimination using row operations to solve

(3  i)x  (8  4i)y  (9  3i)z  52  4i 2ix  7iy  (12 10i)z  67  57i (5  5i)x  20iy  (15 16i)z  57 100i

Solution:

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8.2 Conjugates and Division of Complex Numbers 399

8.2 Conjugates and Division of Complex Numbers

Find the conjugate of a complex number. Find the modulus of a complex number. Divide complex numbers, and find the inverse of a complex matrix.

COMPLEX CONJUGATES In Section 8.1, it was mentioned that the complex zeros of a polynomial with real coefficients occur in conjugate pairs. For instance, in Example 5 you saw that the zeros of p͑x͒ ϭ x2 Ϫ 6x ϩ 13 are 3 ϩ 2i and 3 Ϫ 2i. In this section, you will examine some additional properties of complex conjugates. You will begin with the definition of the conjugate of a complex number.

Definition of the Conjugate of a Complex Number The conjugate of the complex number z ϭ a ϩ bi is denoted by z and is given by z ϭ a Ϫ bi.

REMARK In part (d) of Example 1, note Finding the Conjugate of a Complex Number that 5 is its own complex conjugate. In general, it can Complex Number Conjugate be shown that a number is its a. z ϭϪ2 ϩ 3i z ϭϪ2 Ϫ 3i own complex conjugate if and only if the number is real. (See b. z ϭ 4 Ϫ 5i z ϭ 4 ϩ 5i Exercise 39.) c. z ϭϪ2i z ϭ 2i d. z ϭ 5 z ϭ 5 Imaginary axis Geometrically, two points in the complex plane are conjugates if and only if they z = −2 + 3i 3 are reflections in the real (horizontal) axis, as shown in Figure 8.6. Complex conjugates have many useful properties. Some of these are shown in Theorem 8.1. 2

Real THEOREM 8.1 Properties of Complex Conjugates axis −4 −3 −112 For a complex number z ϭ a ϩ bi, the following properties are true. −2 1. zz ϭ a2 ϩ b2 2. zz Ն 0 −3 3. zz ϭ 0 if and only if z ϭ 0. 4. ͑z͒ ϭ z z = −2 − 3i

Imaginary axis PROOF z = 4 + 5i 5 To prove the first property, let z ϭ a ϩ bi. Then z ϭ a Ϫ bi and 4 3 zz ϭ ͑a ϩ bi͒͑a Ϫ bi͒ ϭ a2 ϩ abi Ϫ abi Ϫ b2i 2 ϭ a 2 ϩ b2. 2 1 Real The second and third properties follow directly from the first. Finally, the fourth axis −3567−2 23 property follows from the definition of the complex conjugate. That is, −2 ͑z͒ ϭ ͑ ϩ ͒ ϭ Ϫ ϭ ϩ ϭ −3 a bi a bi a bi z. −4 −5 − z = 4 5i Finding the Product of Complex Conjugates Conjugate of a Complex Number Figure 8.6 When z ϭ 1 ϩ 2i, you have zz ϭ ͑1 Ϫ 2i͒͑1 ϩ 2i͒ ϭ 12 ϩ 22 ϭ 1 ϩ 4 ϭ 5. 9781133110873_0802.qxp 3/10/12 6:57 AM Page 400

400 Chapter 8 Complex Vector Spaces

THE MODULUS OF A COMPLEX NUMBER REMARK Because a complex number can be represented by a vector in the complex plane, it The modulus of a complex makes sense to talk about the length of a complex number. This length is called the number is also called the modulus of the complex number. absolute value of the number. In fact, when z is a real number, ԽzԽ Ίa2 02 ԽaԽ. Definition of the Modulus of a Complex Number The modulus of the complex number z a bi is denoted byԽzԽ and is given by ԽzԽ Ίa2 b2.

Finding the Modulus of a Complex Number

For z 2 3i and w 6 i, determine the value of each modulus. a.zԽ b.ԽwԽԽ c. ԽzwԽ SOLUTION a. ԽzԽ Ί22 32 Ί13 b. ԽwԽ Ί62 ͑1͒2 Ί37 c. Because zw ͑2 3i͒͑6 i͒ 15 16i, you have ԽzwԽ Ί152 162 Ί481.

Note that in Example 3,ԽzwԽ ԽzԽ ԽwԽ. In Exercise 41, you are asked to prove that this multiplicative property of the modulus always holds. Theorem 8.2 states that the modulus of a complex number is related to its conjugate.

THEOREM 8.2 The Modulus of a Complex Number For a complex number z, ԽzԽ2 zz.

PROOF Let z a bi, then z a bi and zz ͑a bi͒͑a bi͒ a2 b2 ԽzԽ2.

LINEAR Fractals appear in almost every part of the universe. They ALGEBRA have been used to study a wide variety of applications such APPLIED as bacteria cultures, the human lungs, the economy, and galaxies. The most famous fractal is called the Mandelbrot Set, named after the Polish-born mathematician Benoit Mandelbrot (1924–2010). The Mandelbrot Set is based on the following sequence of complex numbers. ͑ ͒2 zn zn1 c, z1 c The behavior of this sequence depends on the value of the complex number c. For some values of c, the modulus of

each term zn in the sequence is less than some fixed number N, and the sequence is bounded. This means that c is in the Mandelbrot Set, and its point is colored black. For other values of c, the moduli of the terms of the sequence become infinitely large, and the sequence is unbounded. This means that c is not in the Mandelbrot Set, and its point is assigned a color based on “how quickly” the sequence diverges. Andrew Park/Shutterstock.com 9781133110873_0802.qxp 3/10/12 6:57 AM Page 401

8.2 Conjugates and Division of Complex Numbers 401

DIVISION OF COMPLEX NUMBERS One of the most important uses of the conjugate of a complex number is in performing division in the complex number system. To define division of complex numbers, consider z a bi and w c di and assume that c and d are not both 0. For the quotient z x yi w to make sense, it has to be true that z w͑x yi͒ ͑c di͒͑x yi͒ ͑cx dy͒ ͑dx cy͒i. But, because z a bi, you can form the linear system below. cx dy a dx cy b Solving this system of linear equations for x and y yields ac bd bc ad x and y . ww ww Now, because zw ͑a bi͒͑c di͒ ͑ac bd͒ ͑bc ad͒i, the following definition is obtained.

Definition of Division of Complex Numbers The quotient of the complex numbers z a bi and w c di is defined as z a bi REMARK w c di If c2 d 2 0, then c d 0, ac bd bc ad i and w 0. In other words, as c2 d 2 c2 d 2 is the case with real numbers, 1 division of complex numbers ͑ zw͒ by zero is not defined. ԽwԽ2 provided c2 d 2 0.

In practice, the quotient of two complex numbers can be found by multiplying the numerator and the denominator by the conjugate of the denominator, as follows. a bi a bi c di ΂ ΃ c di c di c di ͑a bi͒͑c di͒ ͑c di͒͑c di͒ ͑ac bd͒ ͑bc ad͒i c2 d 2 ac bd bc ad i c2 d 2 c2 d 2

Division of Complex Numbers

1 1 1 i 1 i 1 i 1 1 a. ΂ ΃ i 1 i 1 i 1 i 12 i 2 2 2 2 2 i 2 i 3 4i 2 11i 2 11 b. ΂ ΃ i 3 4i 3 4i 3 4i 9 16 25 25 9781133110873_0802.qxp 3/10/12 6:57 AM Page 402

402 Chapter 8 Complex Vector Spaces

Now that you can divide complex numbers, you can find the (multiplicative) inverse of a complex matrix, as demonstrated in Example 5.

Finding the Inverse of a Complex Matrix

Find the inverse of the matrix 2 i 5 2i A ΄ ΅ 3 i 6 2i 1 and verify your solution by showing that AA I2. SOLUTION Using the formula for the inverse of a 2 2 matrix from Section 2.3, 1 6 2i 5 2i A1 ΄ ΅. ԽAԽ 3 i 2 i Furthermore, because ԽAԽ ͑2 i͒͑6 2i͒ ͑5 2i͒͑3 i͒ ͑12 6i 4i 2͒ ͑15 6i 5i 2͒ 3 i it follows that 1 6 2i 5 2i A1 ΄ ΅ TECHNOLOGY 3 i 3 i 2 i If your graphing utility or 1 1 ͑6 2i͒͑3 i͒ ͑5 2i͒͑3 i͒ ΂ ΃΄ ΅ software program can perform 3 i 3 i ͑3 i͒͑3 i͒ ͑2 i͒͑3 i͒ operations with complex 1 20 17 i matrices, then you can verify ΄ ΅. the result of Example 5. If you 10 10 7 i have matrix A stored on a To verify your solution, multiply A and A1 as follows. graphing utility, evaluate A1. 2 i 5 2i 1 20 17 i 1 10 0 1 0 AA1 ΄ ΅ ΄ ΅ ΄ ΅ ΄ ΅ 3 i 6 2i 10 10 7 i 10 0 10 0 1 The last theorem in this section summarizes some useful properties of complex conjugates.

THEOREM 8.3 Properties of Complex Conjugates For the complex numbers z and w, the following properties are true. 1. z w z w 2. z w z w 3. zw z w 4. z͞w z͞w

PROOF To prove the first property, let z a bi and w c di. Then z w ͑a c͒ ͑b d͒i ͑a c͒ ͑b d͒i ͑a bi͒ ͑c di͒ z w. The proof of the second property is similar. The proofs of the other two properties are left to you. 9781133110873_0802.qxp 3/10/12 6:57 AM Page 403

8.2 Exercises 403

8.2 Exercises

Finding the Conjugate In Exercises 1Ð6, find the Finding the Inverse of a Complex Matrix In Exercises complex conjugate z and geometrically represent both 31Ð36, determine whether the complex matrix A has an z and z. inverse. If A is invertible, find its inverse and verify that ؍ ؊1 1.z 6 3i 2. z 2 5i AA I. 6 3i 2i 2 i 3.z 8i 4. z 2i 31.A ΄ ΅ 32. A ΄ ΅ i i i 5.z 4 6. z 3 2 3 3 1 i 2 1 i 2 33.A ΄ ΅ 34. A ΄ ΅ Finding the Modulus In Exercises 7Ð12, find the 1 1 i 0 1 i ؊3 ؉ 2i, and؍ ؉ i, w 2 ؍ indicated modulus, where z i 0 0 1 0 0 .؊5i؍ v 35.A 0 i 0 36. A 0 1 i 0 2 ΄ ΅ ΄ ΅ 7.ԽzԽ 8. Խz Խ 0 0 i 0 0 1 i 9.ԽzwԽ 10. ԽwzԽ 11.ԽvԽ 12. Խzv2Խ Singular Matrices In Exercises 37 and 38, determine all values of the complex number z for which A is (.and solve for z 0 ؍ Verify that ԽwzԽ ԽwԽԽzԽ ԽzwԽ, where z 1 i and singular. (Hint: Set detͧAͨ .13 w 1 2i. 2 2i 1 i 5 z 14. Verify that Խzv2Խ ԽzԽԽv2Խ ԽzԽԽvԽ2, where z 1 2i 37.A ΄ ΅ 38. A 1 i 1 i z 3i 2 i ΄ ΅ and v 2 3i. 1 0 0

Dividing Complex Numbers In Exercises 15Ð20, 39. Proof Prove that z z if and only if z is real. perform the indicated operations. 2 i 1 1 i 15. 16. 40. Consider the quotient . i 6 3i 6 2i 3 Ί2i 5 i (a) Without performing any calculations, describe 17. 18. 3 Ί2i 4 i how to find the quotient. ͑2 i͒͑3 i͒ 3 i (b) Explain why the process described in part (a) 19. 20. 4 2i ͑2 i͒͑5 2i͒ results in a complex number of the form a bi. (c) Find the quotient. Operations with Complex Rational Expressions In Exercises 21Ð24, perform the operation and write the result in standard form. 41. Proof Prove that for any two complex numbers z and w, each of the statements below is true. 2 3 2i 5 21. 22. (a) ԽzwԽ ԽzԽԽwԽ 1 i 1 i 2 i 2 i (b) If w 0, then Խz͞wԽ ԽzԽ͞ԽwԽ. i 2i 1 i 3 23. 24. 42. Graphical Interpretation Describe the set of 3 i 3 i i 4 i points in the complex plane that satisfies each of the Finding Zeros In Exercises 25Ð28, use the given zero to statements below. find all zeros of the polynomial function. (a) ԽzԽ 3 (b) Խz 1 iԽ 5 25. p͑x͒ 3x3 4x2 8x 8 Zero: 1 Ί3i (c) Խz iԽ 2 (d) 2 ԽzԽ 5 26. p͑x͒ 4x3 23x2 34x 10 Zero: 3 i 43. (a) Evaluate ͑1͞i͒n for n 1,2, 3, 4, and 5. 27. p͑x͒ x4 3x3 5x2 21x 22 Zero: 3 Ί2i (b) Calculate ͑1͞i͒2000 and ͑1͞i͒2010. 28. p͑x͒ x3 4x2 14x 20 Zero: 1 3i (c) Find a general formula for ͑1͞i͒n for any positive integer n. Powers of Complex Numbers In Exercises 29 and 30, 1 i 2 find each power of the complex number z. 44. (a) Verify that ΂ ΃ i. Ί2 a)z 2 (b)z 3 (c)z؊1 (d) z؊2) (b) Find the two square roots of i. 29.z 2 i 30. z 1 i (c) Find all zeros of the polynomial x 4 1. Answer Key

Section 8.2 1. 6 ϩ 3i 3. 8i Imaginary Imaginary axis axis

4 z = 6 + 3i 8 z = 8i 2 4 Real axis 246 −8 −4 84 −2 −4 −4 − z = 6 3i −8 z = −8i

5. 4 Imaginary axis

3 2 1 z and z = 4 Real axis −1 31245 −2

7.Ί5 9.Ί65 11. 5 13. ԽwzԽ ϭ ԽϪ3 ϩ iԽ ϭ Ί10 ԽwԽԽzԽ ϭ Ί5Ί2 ϭ Ί10 ԽzwԽ ϭ ԽϪ3 ϩ iԽ ϭ Ί10 7 6Ί2 13 9 15. 1 Ϫ 2i 17. Ϫ i 19. ϩ i 11 11 10 10 Ϫ1 Ϫ 5 1 ϩ 9 Ϫ2 ± Ί 21. 2 2i 23. 10 10i 25. 3, 1 3i 27. 1, 2, Ϫ3 ± Ί2i Ϫ Ϫ 2 ϩ 1 3 ϩ 4 29. (a)3 4i (b) 2 11i (c)5 5i (d) 25 25i i Ϫ3i 31. AϪ1 ϭϪ1΄ ΅ 33. Not invertible 3 Ϫ2 ϩ i 6 Ϫi 0 0 Ϫ1 ϭ Ϫ ϪϪ5 10 35. A ΄ 0 i 0΅ 37. 3 3 i 0 0 Ϫ i 39. Proof 41. (a) and (b) Proofs 43. (a) ͑1͞i͒1 ϭ 1͞i ϭϪi, ͑1͞i͒2 ϭϪ1, ͑1͞i͒3 ϭ i, ͑1͞i͒4 ϭ 1, ͑1͞i͒5 ϭϪi (b) ͑1͞i͒2000 ϭ 1, ͑1͞i͒2010 ϭϪ1 1 n ϭ 4k Ϫi n ϭ 4k ϩ 1 (c) ͑1͞i͒n ϭ , where k is an integer ΆϪ1 n ϭ 4k ϩ 2 i n ϭ 4k ϩ 3 Chapter 4: Vector Spaces. 4.1 Vectors in Rn.

4.1 Vectors in Rn.  Objective: Represent a vector in the plane as a directed line segment.  Objective: Perform basic vector operations in R2 and represent them graphically.  Objective: Perform basic vector operations in Rn.  Prove basic properties about vectors and their operations in Rn.

In physics and engineering, a vector is an object with magnitude and direction and represented graphically by a directed line segment. In mathematics we have a much more general definition of a vector.

Geometrically, a vector in the plane is represented by a directed line segment with its initial point at the origin and its terminal (final) point at (x1, x2). The same ordered pair used to represent the terminal point is used to represent the vector. That is, x = (x1, x2). The coordinates x1 and x2 are called the components of the vector x. Two vectors u = (u1, u2) and v = (v1, v2) are equal iff u1 = v1 and u2 = v2.

Vector operations in R2

def  Vector Addition: u + v = (u1, u2) + (v1, v2)  (u1 + v1, u2 + v2). def  Scalar Multiplication: cu = c(u1, u2)  (cu1, cu2).

 Negative: –u (–u1, –u2). Notice that –u = (–1)u.

 Subtraction: u – v u + (–v) = (u1, u2) + (–v1, –v2) = (u1 – v1, u2 – v2).

The zero vector in R2 is 0 = (0, 0).

p. 71 Chapter 4: Vector Spaces. 4.1 Vectors in Rn.  Examples

Let u = (2, 4) and v = (–4, 1). Illustrate the following graphically. u, v, u + v, 1.5u, – v, v – u p. 72 Chapter 4: Vector Spaces. 4.1 Vectors in Rn. Theorem 4.1 Properties of Vector Addition and Scalar Multiplication in the Plane (R2)

Let u, v, and w be vectors in R2, and let c and d be scalars.

1) u + v is a vector in R2. Closure under addition 2) u + v = v + u Commutative property of addition 3) (u + v) + w = u + (v + w) Associative property of addition 4) u + 0 = u Existence of additive identity 5) u + (–u) = 0 Existence of additive inverses 6) cv is a vector in R2. Closure under scalar multiplication 7) c(u + v) = cu + cv Distributive property over vector addition 8) (c + d)u = cu + du Distributive property over scalar addition 9) c(du) = (cd)u Associative property 10) 1u = u Multiplicative identity property

 Proof of (3): Associative property of addition

(u + v) + w = [(u1, u2) + (v1, v2)] + (w1, w2)

=

=

= Assoc. prop. of addition of real numbers

=

=

= u + (v + w)

p. 73 Chapter 4: Vector Spaces. 4.1 Vectors in Rn.  Proof of (8): Distributive property of scalar multiplication over real number addition

(c + d)u = (c + d)(u1, u2)

=

= Distributive property of real numbers

=

=

= cu + du

To add (1, 4) + (2, –2) in Mathematica, type {1,4}+{2,-2} You can also assign a variable by typing u={1,4} You can perform scalar multiplication by 3u or 3*u

To add (1, 4) + (2, –2) on the TI-89, you can type 1,4+2,-2 or 14+2-2 You can also assign a variable by typing 1,4.U You can perform scalar multiplication by 3u or 3u

Vector operations in Rn

We can generalize from the 2-dimensional plane R2 to an n-space Rn of ordered n-tuples. For example, R1 = R = set of all real numbers; R2 = 2-space = set of all ordered pairs of real numbers; R3 = 3-space = set of all ordered triples of real numbers.\

n An n-tuple (x1, x2, …, xn) can be viewed as a point in R with the xi as its coordinates, or as a vector with the xi as its components.

The standard vector operations in Rn are

def  Vector Addition: u + v = (u1, u2, …, un) + (v1, v2, …, vn)  (u1 + v1, u2 + v2, …, un + vn). def  Scalar Multiplication: cu = c(u1, u2, …, un)  (cu1, cu2, …, cun).

 Negative: –u (–u1, –u2, …, –un). Notice that –u = (–1)u.

 Subtraction: u – v u + (–v) = (u1, u2, …, un) + (–v1, –v2, …, –vn) = (u1 – v1, u2 – v2, …, un – vn).

The zero vector in Rn is 0 = (0, 0, …, 0).

p. 74 Chapter 4: Vector Spaces. 4.1 Vectors in Rn. Theorem 4.2 Properties of Vector Addition and Scalar Multiplication in the Plane (Rn)

Let u, v, and w be vectors in Rn, and let c and d be scalars.

1) u + v is a vector in Rn. Closure under addition 2) u + v = v + u Commutative property of addition 3) (u + v) + w = u + (v + w) Associative property of addition 4) u + 0 = u Existence of additive identity 5) u + (–u) = 0 Existence of additive inverses 6) cv is a vector in Rn. Closure under scalar multiplication 7) c(u + v) = cu + cv Distributive property over vector addition 8) (c + d)u = cu + du Distributive property over scalar addition 9) c(du) = (cd)u Associative property 10) 1u = u Multiplicative identity property

The vector 0 is called the additive identity in Rn and –v is the additive inverse of v.

Theorem 4.3 Properties of Vector Addition and Scalar Multiplication in Rn

Let v be a vector in Rn and let c be a scalar. Then

1) The additive identity is unique. That is, if v + u = v, then u = 0. 2) The additive inverse of v is unique. That is, if v + u = 0, then u = –v. 3) 0v = 0 4) c0 = 0 5) If cv = 0, then c = 0 or v = 0. 6) –(–v) = v

 Proof of (1): Uniqueness of the additive identity

v + u = v Given

(v + u) + (–v) = v + (–v) Add –v to both sides

= v + (–v)

= v + (–v)

= 0

u = 0

p. 75 Chapter 4: Vector Spaces. 4.1 Vectors in Rn.  Proof of (2): Uniqueness of the additive inverse

v + u = 0 Given

(–v) + (v + u) = (–v) + 0

= (–v) + 0

= (–v) + 0

= (–v) + 0

u = –v

p. 76 Chapter 4: Vector Spaces. 4.2 Vector Spaces.

4.2 Vector Spaces.  Objective: Define a vector space and recognize some important examples of vector spaces.  Objective: Show that a given set is not a vector space. (Optional)

Theorem 4.2 listed ten properties of vector addition and scalar multiplication in Rn. However, there are many other sets (Cn, sets of matrices, polynomials, functions) besides Rn that can be given suitable definitions of vector addition and scalar multiplication so that they too satisfy the same ten properties. Hence, one branch of mathematics, linear algebra, can study all of these.

 Definition of a Vector Space

Let V be a set on which two operations (vector addition and scalar multiplication) are defined. If the axioms listed below are satisfied for every u, v, and w in V and every scalar c and d in a given field F (usually, F = R or F = C), then V is called a vector space over F.

1) u + v is in V. Closure under addition 2) u + v = v + u Commutative property * 3) (u + v) + w = u + (v + w) Associative property 4) V has a zero vector 0 such that Existence of additive identity for every u in V, u + 0 = u 5) For every u in V, there is a vector Existence of additive inverses (opposites) denoted by –u such that u + (–u) = 0

6) cv is a vector in V. Closure under scalar multiplication 7) c(u + v) = cu + cv Distributive property over vector addition 8) (c + d)u = cu + du Distributive property over scalar addition 9) c(du) = (cd)u Associative property 10) 1u = u Scalar identity

Notice that a vector space actually consists of four entities: a set V of vectors, a field F of scalars, and two defined operations (vector addition and scalar multiplication). Be sure all four entities are clearly understood. (For example, I could keep the set V of vectors, the field F of scalars, the same definition of scalar multiplication, but change the definition of how to add vectors and end up with a different vector space, or end up with something that is no longer a vector space.)

 Examples of Vector Spaces. (Unless otherwise stated, assume the field is R.)

 R2 with the standard operations * def def u + v = (u1, u2) + (v1, v2)  (u1 + v1, u2 + v2). cu = c(u1, u2)  (cu1, cu2). 0 = (0, 0) –u = (–u1, –u2)  Rn with the standard operations. Note that this includes R2, which is just R with the usual addition and multiplication.

 Cn over the field C with the standard operations

p. 77 Chapter 4: Vector Spaces. 4.2 Vector Spaces.  More Examples of Vector Spaces. (Unless otherwise stated, assume the field is R.)

 The vector space M2,3of all 23 real matrices with the standard operations

a11 a12 a13  b11 b12 b13  a11  b11 a12  b12 a13  b13  A + B =   +   =   a21 a22 a23 b21 b22 b23 a21  b21 a22  b22 a23  b23

ca11 ca12 ca13  and cA = c =   ca 21 ca 22 ca 23

 The vector space Mm,n of all mn real matrices with the standard operations.

 The vector space P2 of all polynomials of degree 2 or less with the usual operations. 2 2 Let p(x) = a0 + a1x + a2x and q(x) = b0 + b1x + b2x . def Define the usual operations (p + q)(x)  p(x) + q(x) and (cp)(x) c[p(x)]

We can verify closure under addition: 2 2 (p + q)(x) = p(x) + q(x) = a0 + a1x + a2x + b0 + b1x + b2x 2 = (a0 + b0) + (a1 + b1)x + (a2 + b2)x which is a polynomial of degree 2 or less (less if a2 + b2 = 0). Notice that we have used the commutative and distributive properties of real numbers. The other axioms can be verified in a similar manner. . Note that 0(x) = 0 + 0x + 0x2.

 The vector space Pn of all polynomials of degree n or less with the usual operations.

 The vector space P of all polynomials with the usual operations.

 The vector space C(–,) of continuous real- valued functions on the domain (–,) For 8 example, x2 + 1, , sin(x), and ex are vectors 1 x 2 in this space. Addition and scalar multiplication are defined in the usual way. (f + g)(x) f(x) + g(x) and (cf )(x) c[f (x)]

f, g, and f + g are vectors in C(–,), just as u, v, and u + v and are vectors in Rn. f (x) can be

thought of as a component of f, just as ui is a component of u. u has n components: u1, u2, …, 2 un. f has an infinite number of components: …, f (–2), …, f (  3 ), …, f (0), …, f ( ), ….

The additive identity (zero function) is f0(x) = 0 (the x-axis), and given f (x), the additive inverse of f is [–f ](x) = –[ f (x)].

p. 78 Chapter 4: Vector Spaces. 4.2 Vector Spaces.  Another Example of a Vector Spaces .  The vector space C[a, b] of continuous real-valued functions on the domain [a, b] over the field R.

The most important reason for defining an abstract vector space using the ten axioms above is that we can make general statements about all vector spaces. I.e. the same proof can be used for Rn and for C[a, b].

Theorem 4.4 Properties of Vector Addition and Scalar Multiplication

Let v be a vector in Vn and let c be a scalar. Then

1) 0v = 0 2) c0 = 0 3) If cv = 0, then c = 0 or v = 0. 4) –1v = –v

Proof of (2): c0 = 0

c0 = 0 Given

c0 = c(0 + 0) Additive identity

c0 =

c0 + –( c0)

c0 + –( c0) =

0 = c0 Additive inverse

Proof of (3): If cv = 0, then c = 0 or c  0. If c  0, then

cv = c0 Given

Multiply both sides by Multiplicative c–1cv = c–10 inverse in R

p. 79 Chapter 4: Vector Spaces. 4.2 Vector Spaces.

c(c–1)v = c–10 Commutative property of multiplication

1v = c–10 Multiplicative inverse in R

v c–10 Scalar identity

v = 0 Theorem 4.3(2) – just proved

Thus, either c = 0 or v = 0.

 Examples that are not Vector Spaces

2 2

 Z (ordered pair of integers) over the field R. Z is not closed under scalar multiplication, for 1 1 2 example 2 (1, 2) = ( 2 , 1)  Z .

Aside on notation: 1  Z means “1 is an element of (is a member) of the set of integers. optional  Z means “1 is not an element of the set of integers. Although Z2 satisfied Axioms 1–5 and 10 of a vector space, it is not a vector space because not all axioms are satisfied. .() V =  The set of second-degree polynomials is not a vector space because it is not closed under addition. For example, let p(x) = x2 and q(x) = –x2 + x + 1. Then p(x) + q(x) = x + 1 is a first degree polynomial.

 Let V = R2 with the standard vector addition but nonstandard scalar multiplication defined by

c(u1, u2) = (cu1, 0). Show that V is not a vector space.

It turns out that the only axiom that is not satisfied in this case is (10) Scalar identity. For example, 1(2, 3) = (2, 0)  (2, 3).

p. 80 Chapter 4: Vector Spaces. 4.2 Vector Spaces.

 Another Example that is not a Vector Spaces

 Rotations in three dimensions represented as arrows using the right-hand rule. The direction of the arrow represents the direction of the rotation, via

the right-hand rule, while the length of the arrow represents the optional magnitude of the direction in degrees. Scalar multiplication is the standard operation (stretching the arrow, or reversing the direction if

the scalar is negative). “Vector addition” (e.g. ) is the first rotation followed by

the second. This is not a vector space because “vector addition” is not commutative.

.

(In Chapter 6, we will see that rotations can be represented not as vectors, but as matrices.)

p. 81

Chapter 4: Vector Spaces. 4.3 Subspaces of Vector Spaces.

4.3 Subspaces of Vector Spaces.  Objective: Determine whether a subset W of a vector space V is a subspace of V.  Objective: Determine subspaces of Rn.

Many vector spaces are subspaces of larger spaces.

.() A V nonempty = subset W of a vector space V is a called a subspace of V when is a vector space under the operations of vector addition and scalar multiplication defined in V.

 Theorem 4.5 Test for a Subspace

If W is a nonempty subset of V, then W is a subspace of V if and only if the following .() conditionsV = hold.

0) W is not empty. 1) If u and v are in W, then u + v is in W. Closure under addition 2) If u is in W and c is a scalar, then cu is in W. Closure under scalar multiplication

Proof: If W is a subspace of V, then W is a vector space satisfying the closure axioms, so u + v is in W and cu is in W. On the other hand, assume W is closed under vector addition and scalar multiplication. By assumption, two axioms are satisfied. 1) u + v is in W. Closure under addition 6) cv is a vector in W. Closure under scalar multiplication Then if u, v, and w are in W then they are also in V, so the following axioms are automatically satisfied. 2) u + v = v + u Commutative property 3) (u + v) + w = u + (v + w) Associative property 7) c(u + v) = cu + cv Distributive property over vector addition 8) (c + d)u = cu + du Distributive property over scalar addition 9) c(du) = (cd)u Associative property 10) 1u = u Scalar identity Because W is closed under scalar multiplication, we know that for any v in W, 0v and (–1)v are also in W. From Thm. 4.4, we know that 0v = 0 and (–1)v = –v so the remaining axioms are also satisfied. 4) W contains the zero vector 0. Additive identity 5) For every v in W, W contains –v. Additive inverse (opposite)

p. 82 Chapter 4: Vector Spaces. 4.3 Subspaces of Vector Spaces.  Examples of Subspaces and Sets that are not Subspaces.

 Show that the set W = {(v1, 0, v3): v1 and v3 are real numbers} is a subspace of R3 with the standard operations.

Graphically, W is the x-z plane in R3. W is nonempty, because it contains (0, 0, 0). W is closed under addition because if u, v  W, then u + v = (u1, 0, u3) + (v1, 0, v3) = (u3 + v1, 0, u3 + v3)  W. W is closed under scalar multiplication because if c is a scalar and v  W, then cv = c(v1, 0, v3) = (cv1, 0, cv3)  W.

 Is Z2 (ordered pair of integers) with the standard operations a subspace of R2?

Z2 is closed under addition. But as we saw in 4.2, Z2 is not closed under 1 1 2 2 scalar multiplication, for example 2 (1, 2) = ( 2 , 1)  Z . So Z is not a subspace of R2.

 Show that the set W = {(v1, v2): v1 = 0 or v2 = 0} with the standard operations is not a subspace of R2.

W is closed under scalar multiplication, but W is not closed under addition. * For example, (1, 0) + (0, 1) = (1, 1)  W.

 Is that the set W = {(v1, v2): v1  0 and v2  0} with the standard operations a subspace of R2?

W is closed under addition, but W is not closed under scalar multiplication when c < 0 and v  0. For example, (–1)(2, 3) = (–2, –3)  W.

In general, a subspace is “straight” (line) or “flat” (plane or higher-dimensional object), is “infinite” in all directions, has no “holes,” and contains the origin.

 Let W be the set of all symmetric 22 matrices. Show that W is a subspace of M2,2 with the standard operations.

T 0 0 “A is symmetric” means that A = A . W is nonempty, because it contains   . W is closed 0 0 under addition because if A, B  W, then (A + B)T = AT + BT = A + B, so A + B is symmetric. W is closed under scalar multiplication because if c is a scalar and A  W, then (cA)T = c(AT) cA, so cA is symmetric.

p. 83 Chapter 4: Vector Spaces. 4.3 Subspaces of Vector Spaces.  More Examples of Subspaces and Sets that are not Subspaces.

 Let W be the set of all singular matrices of order 2. Show that W is not a subspace of M2,2 with the standard operations.

1 0 0 0 1 0 0 0 1 0   and   are singular matrices (det = 0), but   +   =   is not 0 0 0 1 0 0 0 1 0 1 singular, so W is not closed under addition and is therefore not a subspace. (Notice that a b W = {   : ad – bc = 0}, and ad – bc = 0 is not a linear equation.) c d

2 2 2  Let W be the unit in R , i. e. W = {(v1, v2): v1 + v2 = 1}, with the standard operations. Is W a subspace of R2

No. W is not closed under addition: (1, 0) + (0, 1) = (1, 1)  W. Another reason is that W is not closed under scalar multiplication: 0(1, 0) = (0, 0)  W.

 Is W = {(0, 0)} with the standard operations a subspace of R2

Yes, the only addition to check is 0 + 0 = 0  W. Scalar multiplication is also easy to check: c0 = 0  W. Is W = {0} and W = V are called the trivial subspaces of c.

 Which of these two subsets is a subspace of R3 with the standard operations? 3 a) U = {(v1, v2, v3)  R : 2v1 + 3v2 + 4v3 = 12} This is a the equation of a plane through the points (6, 0, 0), (0, 4, 0), and (0, 0, 3). U is not closed under addition, because (6, 0, 0) + (0, 4, 0) = (6, 4, 0), but 2(6) + 3(4) + 4(0) = 24  12. Moreover, U is not closed under scalar multiplication, because 0(6, 0, 0) = (0, 0, 0), but 2(0) + 3(0) + 4(0) = 0  12. Either one of these reasons is enough to show that U is not a subspace.

3 b) W= {(v1, v2, v3)  R : 2v1 –3 v2 + 4v3 = 0} This is a the equation of a plane parallel to U, but W contains 0. If v and w  W, then

v + w = (v1, v2, v3) + (w1, w2, w3) = (v1 + w1, v2 + w2, v3 + w3), and 2(v1 + w1) –3(v2 + w2) + 4(v3 + w3) = 2v1 + 2w1 –3v2 –3w2 + 4v3 + 4w3

= (2v1 –3v2 + 4v3) + (2w1 –3w2 + 4w3) = 0 + 0 = 0, so v + w  W.

Also, if c is a scalar and v  W, then cv = (cv1, cv2, cv3) and 2(cv1) –3(cv2) + 4(cv3)

= c(2v1 –3v2 + 4v3) = c·0 = 0, so cv  W. Therefore, W is a subspace.

p. 84 Chapter 4: Vector Spaces. 4.3 Subspaces of Vector Spaces.  More Example of Subspaces.

 P[0, 1] is the set of all real-valued polynomial funtions on the domain [0, 1]. C1[0, 1] is the set of all real-valued, continuously differentiable functions on the domain [0, 1]. C0[0, 1], also written as C[0, 1], is the set of all real-valued, continuous functions on the domain [0, 1]. Let W be the set of all real-valued, integrable functions on the interval [0, 1]. Let V be the set of all real-valued functions on the interval [0, 1].

If we take the usual definitions of vector addition and scalar multiplication, and we use R as the field of scalars, then V is a vector space, and the other four sets are subspaces of V. In fact, P[0, 1] is a subspace of C1[0, 1], which is is a subspace of C0[0, 1], which is is a subspace of W, which is is a subspace of V.

 Theorem 4.6 The Intersection of Two Subspaces is a Subspace.

If V and W are both subspaces of a vector space U, then the interaction of V and W (denoted V  W) is also a subspace of U. (Note: V  W is the set of all vectors that are in both V  W.)

Proof: Because V and W are both subspaces of U, we know that they both .() V =contain 0, so V  W contains 0 and is not empty. To show that V  W is closed under addition, let u1 and u2 be two vectors in V V  W. Because u1 and u2 are both in V, and—being a subspace—V is closed, u1 + u2 is also in V. Likewise, u1 + u2 is also in W. Since V  W U u + u is in both V and W, u + u is in V  W, so V  W is closed 1 2 1 2 W under vector addition. A similar argument shows that V  W is closed under scalar multiplication, so V  W is a subspace of U.

p. 85 Chapter 4: Vector Spaces. 4.4 Spanning Sets and Linear Independence.

4.4 Spanning Sets and Linear Independence.  Objective: Write a vector as a linear combination of other vectors in a vector space V.  Objective: Determine whether a spanning set S of vectors in a vector space is a spanning set of V.  Objective: Determine whether a set of vectors in a vector space V is linearly independent.  Objective: Prove results about spanning sets and linear independence.

 A vector v in a vector space V is called a linear combination of the vectors u1, u2, …, uk in V if and only if v can be written in the form

v = c1u1 + c2u2 + … + ckuk where c1, c2, …, ck are scalars.

 Example: Write the vector v = (–1, –2, 2) as a linear combination of the vectors in the set S = {(2, –1, 3), (5, 0, 4)} (if possible).

Solution: We need to solve v = c1u1 + c2u2 for the scalars ci. Substituting in the given vectors, we have

(–1, –2, 2) = c1(2, –1, 3) + c2(5, 0, 4)

(–1, –2, 2) = (2c1, –1c1, 3c1) + (5c2, 0c2, 4c2)

(–1, –2, 2) = (2c1 + 5c2, –1c1 + 0c2, 3c1 + 4c2) cc  152  62  1 21 c  which gives the system cc  201 or  01  1   2 21      c2  cc 21  243  43   2  Solve this system by finding the reduced row-echelon form (using software)  162   201   01   2       c1      201    110  or  10    1 so c1 = 2 and c2 = –1. c2   243   000   00   0 

Answer: (–1, –2, 2) = 2(2, –1, 3) – 1(5, 0, 4).

1 4 1 Example: Write the vectors v = (–3, 15, 18) and w = ( 3 , 3 , 2 ) as linear combinations of the vectors in the set S = {(2, 0, 7), (2, 4, 5), (2, –12, 13)} (if possible).

Solution: We need to solve

p. 86 Chapter 4: Vector Spaces. 4.4 Spanning Sets and Linear Independence.

.

24  48 Example: Write the vector v =   as linear combinations of the vectors in the set 57 20   1 2 2  7  4  9 6  5 S = {   ,   ,   ,   } (if possible).  5 4 6 2  11 12  4  5 Solution:

p. 87 Chapter 4: Vector Spaces. 4.4 Spanning Sets and Linear Independence.

 Let S = {v1, v2, …, vk} be a subset of a vector space V. Then S is called a spanning set of V if and only if every vector in V can be written as a linear combination of vector in S. In such cases, we say that S spans V.

Examples of spanning sets: 3 3 The set {(1, 0, 0), (0, 1, 0), (0, 0, 1)} spans R because any vector v = (v1, v2, v3) in R can be written as v = v1(1, 0, 0) + v2(0, 1, 0) + v3(0, 0, 1).

2 2 The set {1, x, x } spans P2 because any vector (polynomial) p(x) = ax + bx + c in P2 can be written as p(x) = c(1) + b(x) + a(x2).

3  Example: Determine whether the set S1 = {(5, 7, 6), (4, 2, 4), (1, –3, 2)} spans R .

3 R consists of all the vectors of the form (v1, v2, v3), where v1, v2, and v3 are real numbers. S1 3 spans R if we can always solve for the scalars c1, c2, and c3 in the equation

(v1, v2, v3) = c1(5, 7, 6) + c2(4, 2, 4) + c3(1, –3, 2)

(v1, v2, v3) = (5c1, 7c1, 6c1) + (4c2, 2c2, 4c2) + (1c3, –3c3, 2c3)

(v1, v2, v3) = (5c1 + 4c2 + 1c3, 7c1 + 2c2 – 3c3, 6c1 4c2 + 2c3)

v1  5c1  4c2 1c3  This vector equation is equivalent to the system v2  7c1  2c2  3c3  v3  6c1  4c2  2c3

5 4 1 c1  v1  or the matrix equation 7 2  3c   v  .   2   2  6 4 2 c3  v3 

p. 88 Chapter 4: Vector Spaces. 4.4 Spanning Sets and Linear Independence.

5 4 1

This equation can always be solved for c1, c2, and c3, because 7 2  3  0 , so the matrix is 6 4 2 3 invertible. (To be precise, the determinant is –32.) Therefore, S1 spans R .

3 Example: Determine whether the set S2 = {(5, 7, 6), (3, 2, 4), (1, –3, 2)} spans R .  only this number has changed.

Solution:

Geometrically, the vectors in S2 all lie in the same plane (a “2-dimensional” object), while the vectors in S1 do not. We need a “3-dimensional” space to contain S1.

S1 S2.

p. 89 Chapter 4: Vector Spaces. 4.4 Spanning Sets and Linear Independence.

 If S = {v1, v2, …, vk} is a set of vectors in a vector space V, then the span of S is the set of all linear combinations of the vectors in S.

span(S) = {c1v1 + c2v2 + …+ ckvk: c1, c2, … ck, are scalars}

We sometimes write span{v1, v2, …, vk} instead of span(S).

Another notation for span(S)—which we will avoid because it will be confusing later—is v1, v2, …, vk.

Theorem 4.7 Span(S) is a Subspace of V

* If S = {v1, v2, …, vk} is a set of vectors in a vector space V, then span(S) is a subspace of V. Moreover, span(S) is the smallest subspace of V that contains (S), in the sense that every subspace that contains S must also contain span(S).

 Proof First, we want to show that span(S) is a subspace of V.

So we need to show that

Let c be a scalar and let u and w be any vectors in span(S). Then

Next, we want to show that every subspace that contains S must also contain span(S). This is Lab Problem 4.4.55.

p. 90 Chapter 4: Vector Spaces. 4.4 Spanning Sets and Linear Independence.

Sometimes, one vector can be written “in terms of” other vectors. For example, in S2 = {(5, 7, 6), (5,7,6)  (1,3,2) (3, 2, 4), (1, –3, 2)} from above, (3, 2, 4) = . We could say that (3, 2, 4) “is 2 dependent” upon (5, 7, 6) and (1, –3, 2). But we could just as easily solve for (5, 7, 6); there is no good reason to treat (3, 2, 4) as special. A more equitable equation is

1(5, 7, 6) – 2(3, 2, 4) + 1(1, –3, 2) = (0, 0, 0).

In fact, there are an infinite number of solutions to c1(5, 7, 6) + c2(3, 2, 4) + c3(1, –3, 2) = 0.

c1 = t, c2 = –2t, c3 = t. (This includes c1 = c2 = c3 = 0.)

On the other hand, for S1 = {(5, 7, 6), (4, 2, 4), (1, –3, 2)}, the only solution to

5 4 1 c1  0 c (5, 7, 6) + c (4, 2, 4) + c (1, –3, 2) = (0, 0, 0) or 7 2  3c   0 1 2 3   2    6 4 2 c3  0 1 c1  5 4 1  0 0 is c   7 2  3 0  0 . This is called the trivial solution.  2        c3  6 4 2  0 0

 If S = {v1, v2, …, vk} is a set of vectors in a vector space V, then S is called linearly independent if and only if the equation

c1v1 + c2v2 + …+ ckvk = 0

has only the trivial solution c1 = c2 = …= ck = 0. S is called linearly dependent if and only if there are also nontrivial solutions.

Examples #1-4: Let w1 = (7, 3, –5), w2 = (1, 4, 6), w3 = (9, 11, 7), w4 = (12, –5, –4), and w5 = (–6, 5, 8).

3 Example #1: Is {w1, w2} linearly independent? Does {w1, w2} span R ? Solution: To decide linear independence, we want to solve c1w1 + c2w2 = 0. 3 To decide spanning R , we want to solve c1w1 + c2w2 = (v1, v2, v3) for arbitrary v1, v2, and v3.

Both equations look like c1(7, 3, –5) + c2(1, 4, 6) = (v1, v2, v3). (For linear independence, the right-hand side is v1 = v2 = v3 = 0.)

p. 91 Chapter 4: Vector Spaces. 4.4 Spanning Sets and Linear Independence.

So we have (7c1, 3c1, –5c1) + (1c2, 4c2, 6c2) = (v1, v2, v3)

(7c1 + 1c2, 3c1 + 4c2, –5c1 + 6c2) = (v1, v2, v3)

 7 1 v  c  1  3 4 1  v      2  c2   5 6 v3 

 7 1    3 4 is not square, so we cannot take the determinant as we did before.  5 6

 7 1 v1  Instead, find the reduced row-echelon form:  3 4 v    2   5 6 v3 

(Don’t worry about reproducing this result; I’ll explain soon why Mathematica and the TI-89 give a different answer.)

Writing this as a system of equations, we have

When the right-hand side is v1 = v2 = v3 = 0, we have only the trivial solution c1 = c2 = 0, because each variable ci corresponds to a pivot. Therefore, {w1, w2} is linearly independent.

On the other hand, there are many choices of v1, v2, and v3 47 25 for which the last equation 0 = v1 – 38 v2 + 38 v3 can be solved. This happens whenever the (reduced) row- echelon form has a row of all zeroes. Therefore,{w1, w2} does not span R3.

Notice that to answer these questions we didn’t need to pay attention to the coefficients on the right-hand side of the line in the augmented matrix. All that we needed to know was the coefficient matrix on the left-hand side of the line.

1 0 0   (Mathematica and the TI-89 both give  0 1 0 because they assume 0 0 1

you can divide Row 3 by v1 – v2 + v3, and they do not know that we intend the last

p. 92 Chapter 4: Vector Spaces. 4.4 Spanning Sets and Linear Independence. column to be on the right-hand side of the equation. However, the right-hand column is not important to deciding linear independence and spanning.)

3 Example #2: Is {w1, w2, w3} linearly independent? Does {w1, w2, w3} span R ? Solution: To decide linear independence, we want to solve c1w1 + c2w2 + c3w3 = 0. 3 To decide spanning R , we want to solve c1w1 + c2w2 + c3w3 = v for arbitrary v.

c1  v1  Using block multiplication notation, we write [ w | w | w ] c  = v  1 2 3  2   2  c3  v3 

 7 1 9  c1  v1  so  3 4 11 c  = v  .    2   2   5 6 7  c3  v3 

is singular (not invertible) because its

determinant (found using software) is zero. This tells us that sometimes we cannot solve c1w1 + c2w2 + c3w3 = v, so {w1, 3 w2, w3} does not span R . This also tells us that c1w1 + c2w2 + c3w3 = 0 has an infinite number of solutions, so {w1, w2, w3} is linearly dependent.

 7 1 9  1 0 1     For another perspective, the reduced row-echelon form of  3 4 11 is 0 1 2 .  5 6 7  0 0 0 To investigate linear independence, we set the right-hand side equal to zero. The third column in the matrix, which doesn’t have a pivot, gives us a free parameter. (Fewer pivots than variables.)

 1c1  0c2 1c3  0 c3  t       0c1 1c2  2c3  0 has the solutions c2  2t where t is a free parameter.     0c1  0c2  0c3  0 c1  t  Since we have non-trivial solutions, {w1, w2, w3} is linearly dependent.

To investigate spanning, we set the right-hand side equal to arbitrary numbers. The third row of all zeroes in the matrix gives us 0c1 + 0c2 + 0c3 = #, which does not have a solution when the right-hand side is not zero. (Fewer pivots than equations.) Therefore, {w1, w2, w3} does not span R3.

3 Example #3: Is {w1, w2, w4} linearly independent? Does {w1, w2, w4} span R ? Solution: To decide linear independence, we want to solve c1w1 + c2w2 + c4w4 = 0. 3 To decide spanning R , we want to solve c1w1 + c2w2 + c4w4 = v for arbitrary v.

p. 93 Chapter 4: Vector Spaces. 4.4 Spanning Sets and Linear Independence.

c1  v1  Using block multiplication notation, we write [ w | w | w ] c  = v  1 2 4  2   2  c4  v3 

 7 1 12  c1  v1  so  3 4  5 c  = v  .    2   2   5 6  4 c4  v3 

is invertible (non-singular) because its

determinant (found using software) is 591. This tells us that we can solve c1w1 + c2w2 + c4w4 = v, so {w1, w2, 3 w4} spans R . This also tells us that c1w1 + c2w2 + c4w4 = 0 has a unique of solution, so {w1, w2, w4} is linearly independent.

3 Example #4: Is {w1, w2, w4, w5} linearly independent? Does {w1, w2, w4, w5} span R ? Solution: To decide linear independence, we want to solve c1w1 + c2w2 + c4w4 + c5w5 = 0. 3 To decide spanning R , we want to solve c1w1 + c2w2 + c4w4 + c5w5 = v for arbitrary v.

c1    c2 Using block multiplication notation, we write [ w1 | w2 | w4 | w5]   = c4    c5  c   7 1 12  6 1 c     2  so  3 4  5 5  = . c4   5 6  4 8    c5 

We cannot take the determinant of a non-square matrix. The reduced row-echelon form of 128  7 1 12  6 1 0 0  591    3 4  5 5  is 0 1 0 506 .    591     263  5 6  4 8  0 0 1  591

To investigate linear independence, we see that the fourth column in the matrix (which doesn’t have a pivot) gives us a free parameter, so {w1, w2, w4, w5} is linearly dependent.

To investigate spanning, we see that there is no row of all zeroes (every row has a pivot), so 3 we will never have an inconsistent equation 0 = #. Therefore, {w1, w2, w4, w5} spans R .

p. 94 Chapter 4: Vector Spaces. 4.4 Spanning Sets and Linear Independence. Summary:

n  {w1, …, wk} in R is linearly independent if and* only if the reduced row-echelon form of the matrix [w1 | … | wk ] has as many pivots as columns (variables). n  {w1, …, wk} spans R if and only if the reduced row-echelon form of the matrix [w1 | … | wk ] has as many pivots as rows (equations), i.e. no rows of all zeroes.

Theorem 4.8 A Property of Linearly Dependent Sets

A set S = {v1, v2, …, vk}, k  2, is linearly dependent if and only if at least one of the vectors vj can be written as a linear combination of the other vectors in S.  Proof

To prove “only if (),” assume set S is linearly dependent and k  2.

Then

Because one of the coefficients must be nonzero, we can assume without loss of generality (WOLOG) that c1  0. Then *

Conversely (which means changing  to ), suppose v1 is a linear combination of the other vectors in S. Then

v1 =

0 =

Thus, {v1, v2, …, vk} is linearly dependent because

Example: Given that {w1, w2, w4, w5} is linearly dependent, and that

c5  t   263  c4  509t  c1w1 + c2w2 + c4w4 + c5w5 = 0 has the solutions  , write w2 as a linear c   506t  2 509   128  c2  509t  combination of the other vectors in the set.

p. 95 Chapter 4: Vector Spaces. 4.4 Spanning Sets and Linear Independence. Solution:

128 506 263 509 tw1 – 509 tw2 + 509 tw4 + tw5 = 0

so 128w1 + 263w4 + 509w5 = 506w2

128 263 509 so w2 = 506 w1 + 506 w4 + 506 w5

Theorem 4.9 Corollary

Two vectors v1 and v2 in a vector space V are linearly dependent if and only if one is a scalar multiple of the other. *

 Proof: From the proof of Theorem 4.8,

Theorem

Any set S = {0, v1, v2, …, vk}, containing the zero vector is linearly dependent. *

 Proof: 10 + 0v1 + 0v2 + … + 0vk = 0.

Since not all coefficients are zero, S is linearly dependent.

p. 96

Chapter 4: Vector Spaces. 4.5 Basis and Dimension.

4.5 Basis and Dimension. n  Objective: Recognize bases in the vector spaces R , Pn, and Mm,n.  Objective: Find the dimension of a vector space.  Objective: Prove results about spanning sets and linear independence.

 A set of vectors S = {v1, v2, …, vk} in a vector space V is called a basis for V when

1) S spans V, and 2) S is linearly independent.

“S spans V” says that the set S is not too small to be a basis; “S is linearly independent” says that the set S is not too big to be a basis. (The plural of “basis” is “bases.”)

3 In the diagram on the left, you can see that S1 = {u1, u2} is too small to span R . For example, v is outside of span(S1). In the figure on the right, you can see that S1 = {u1, u2, u3} is large enough to span R3. For example, you can see the linear combination that gives v. Finally, In the figure on the right, you can see that S3 = {u1, u2, u3, v} is large to be linearly independent, because v is a linear combination of u1, u2, and u3.

p. 97 Chapter 4: Vector Spaces. 4.5 Basis and Dimension.  Example. The standard basis for R3: Show that {(1, 0, 0), (0, 1, 0), (0 0, 1)} is a basis for R3.

Solution:

To decide linear independence, we want to solve c1(1, 0, 0) + c2(0, 1, 0) + c3(0 0, 1) = 0. 3 To decide spanning R , we want to solve c1(1, 0, 0) + c2(0, 1, 0) + c3(0 0, 1) = v for arbitrary v. * 1 0 0 c1  0 1 0 0 c1  v1  These equations are 0 1 0 c  = 0 and 0 1 0 c  = v     2       2   2  0 0 1 c3  0 0 0 1 c3  v3  The first equation has only the trivial solution, and the second equation always has a solution, so {(1, 0, 0), (0, 1, 0), (0 0, 1)}is linearly independent and it spans R3. Therefore, {(1, 0, 0), (0, 1, 0), (0 0, 1)} is a basis for R3.

e1  1, 0, , 0  n e 2  0, 1, , 0 The standard basis in R is     e n  0, 0, , 1

 Example: Show that {(7, –2, 5), (–3, –9, –1), (1, 7, –7)} is a basis for R3.

Solution: To decide linear independence, we want to solve

.

To decide spanning R3, we want to solve

* As matrix equations, we obtain

The matrix is invertible (its determinant is 474). Because of this, the first equation has only the trivial solution, and the second equation always has a solution, so {(7, –2, 5), (–3, –9, –1), (1, 7, –7)} is linearly independent and it spans R3. Therefore, {(7, –2, 5), (–3, –9, –1), (1, 7, –7)} is a basis for R3.

p. 98 Chapter 4: Vector Spaces. 4.5 Basis and Dimension. 2 n The standard basis for Pn is {1, x, x , …, x }.

1 0 0 1 0 0 0 0 0 0 0 0             The standard basis for M2,3 is { 0 0 , 0 0 , 1 0 , 0 1 , 0 0 , 0 0 }. 0 0 0 0 0 0 0 0 1 0 0 1

Theorem 4.9 Uniqueness of a Basis Representation

If S = {v1, v2, …, vk} is a basis for a vector space V, then every vector in V, can be written in one and only one way as a linear combination of vectors in S.

 Proof: Let u be an arbitrary vector in V,

u can be written in at least one way as a linear combination of vectors in S because

Now suppose that u can be written as a linear combination of vectors in S in two ways:

u = b1v1 + b2v2 + …+ bkvk and u = c1v1 + c2v2 + …+ ckvk * Subtracting these two equations gives

Since S is linearly independent,

Thus, u can be written as a linear combination of vectors in S in only one way.

 Example: Using the basis {(7, –2, 5), (–3, –9, –1), (1, 7, –7)} for R3 from the previous example,

find the unique representation of u = (u1, u2, u3) using this basis. In other words, find the unique solution for the constants c1, c2, c3 in the equation

u = c1(7, –2, 5), + c2(–3, –9, –1), + c3(1, 7, –7)}

p. 99 Chapter 4: Vector Spaces. 4.5 Basis and Dimension.

Solution: As a matrix equation, we obtain

Lemma 4.9½ Dimension of Spanning Sets and Linearly Independent Sets

If S1 = {v1, v2, …, vn} spans a vector space V and S2 = {u1, u2, …, um} is a set of m linearly independent vectors in V, then m  n.

Proof by contradiction:

Suppose m > n. To show that S2 is linearly dependent (a contradiction), we need to find scalars k1, k1, …, km (not all zero) such that

u1  k u + k u + … + k u = k  k    = 0 1 1 2 2 m m  1 m  1m   u   m  m1

Because S1 spans V, each ui is a linear combination of vectors in S1:

u1  v1  u1  C11v1  ...  C1n v n     C    i.e.      mn   u  v  u  C v  ...  C v  m m1  n n1 m m1 1 mn n

Substituting into the first equation gives

v1  k  k C    = 0  1 m  1m mn   v   n  n1

 k1  0 Now consider [C T ]      nm     k  0  m  m1   n1

This is n equations for m unknowns (ki), with n < m, so we have (infinitely many) nontrivial (non-zero) solutions (k1, k2, …, km). Taking the transpose, we have a non-zero solution to

p. 100 Chapter 4: Vector Spaces. 4.5 Basis and Dimension.

0  01n =k1  kn 1m Cmn Therefore, we have a non-zero solution to

v1  u1  0 = k  k C    = k  k     1 m  1m mn    1 m  1m   v  u   n  n1  m  m1

This says that S2 = {u1, u2, …, um} is linearly dependent, which contradicts the premise, thus completing the proof by contradiction.

Theorem 4.10 Bases and Linear Dependence * If S = {v1, v2, …, vn} is a basis for a vector space V, then every set containing more than n vectors in linearly dependent.

Proof:

S = {v1, v2, …, vn} spans V so by Lemma 4.9½, any set of m linearly independent vectors in V has m  n. Therefore, any set of m vectors in V where m > n must be linearly dependent.

Note: the last step is true because “P  Q” is logically equivalent to its contrapositive “(not Q)  (not P)”.

Theorem 4.11 Bases and Linear Dependence * If a vector space V has one basis with n vectors, then every basis for V has n vectors.

Proof:

Let S1 = {u1, u2, …, un} be one basis for Vand let S2 = {v1, v2, …, vm} be another basis for V. Because S1 is a basis, and S2 is linearly independent, Thm. 4.10 tells us that m < n. Similarly, because S2 is a basis, and S1 is linearly independent, Thm. 4.10 tells us that n < m. Therefore, m = n.

If a vector space V has basis consisting of n vectors,* then the number n is called the dimension of V, denoted by dim(V) = n. We define the dimension of the trivial vector space {0} to be n = 0.

The dimension of a vector space is well-defined (unambiguous) because of Thm. 4.11.

p. 101 Chapter 4: Vector Spaces. 4.5 Basis and Dimension. Examples:

 dim(Rn) = n

 dim(Pn) = n + 1

 dim(Mm,n) = mn

 Example: Find the dimension of the subspace of R3 given by W= {(2a, a –3b, a + b): a and b are real numbers}.

Solution:We can write (2a, a –3b, a + b) = a(2, 1, 1) + b(0, –3, 1), so W is spanned by S = . {(2, 1, 1), (0, –3, 1)}. Moreover, S is linearly independent, because the reduced row echelon 2 0  1 0 0     a   form of 1  3 is 0 1 (every column has a pivot). The only solution to   = 0 b 1 1  0 0 0 is a = b = 0. So S is a basis, and dim(W) = 2.

 Example: Find the dimension of the subspace W of P4 spanned by S = {–3 – 5x + x2 + 2x4, –5x – 4x2 + 4x3, –12 – 10x + 11x2 – 12x3 + 11x4, 3 + 10x + 4x2 – 5x4}

Solution: S spans W, but it might not be linearly independent. To decide linear independence, we solve

p. 102 Chapter 4: Vector Spaces. 4.5 Basis and Dimension. Theorem 4.12 Basis Tests in an n-Dimensional Space

Let V be a vector space of dimension n. *

1) If S = {v1, v2, …, vn}is a linearly independent set of vectors in V, then S is a basis for V. 2) If S = {v1, v2, …, vn}spans V, then S is a basis for V.

 Proof by Contradiction of Part(1) Assume that S is not a basis for V. Since S is linearly independent, it must not span V. Choose a vector u in V that is not in span(S) Then the set {v1, v2, …, vn, u} is also linearly independent.

To see this, note that c1v1 + c2v2 + …+ cnvn = u has no solution.

c1v1 + c2v2 + …+ cnvn = –cn+1u

has a solution only when cn+1 = 0, and in that case, c1 = c2 = … = cn = 0 because {v1, v2, …, vn} is linearly independent.

{v1, v2, …, vn, u} being linearly independent contradicts Thm. 4.10—we have n + 1 linearly independent vectors in an n-dimensional vector space. So S must be a basis for V

 Proof by Contradiction of Part(2) Assume that S is not a basis for V. Since S spans V, it must not be linearly independent. So c1v1 + c2v2 + …+ cnvn = 0 has a solution where not all of the ci are zero. Without loss of generality, we can assume that cn  0. Then the set {v1, v2, …, vn–1}also spans V.

c1 c2 cn1 To see this, first observe that vn = – v1 – v2 – … – vn–1. cn cn cn Then

p. 103

Chapter 4: Vector Spaces. 4.6 Rank of a Matrix and Systems of Linear Equations.

4.6 Rank of a Matrix and Systems of Linear Equations.  Objective: Find a basis for the row space, a basis for the column space, and the rank of a matrix.  Find the nullspace of a matrix.  Find the solution of a consistent system Ax = b in the form xp + xh.  Objective: Prove results about subspaces associated with matrices.

Given an mn matrix A, in this section we will speak of its row vectors, which are in Rn

 11 12  aaa1n   11 12,,  aaa1n    aaa  ,,  aaa A =  21 22 2n  21 22 2n       mm31  aaamn   m m21,,  aaamn 

and its column vectors, which are in Rm

 11 12  aaa 1n  a11   a12  a1n    aaa  a  a  a  A =  21 22 2n   21   22    2n                      mm 31  aaa mn  a 1  amm 2  amn 

 Given an mn matrix A.

The row space of A is the subspace of Rn spanned by the row vectors of A. The column space of A is the subspace of Rm spanned by the column vectors of A.

 Theorem 4.13 Row-Equivalent Matrices Have the Same Row Spaces .( If an mn matrix A is row-equivalent to an mn matrix U, then the row space of A is equal to the row space of U.

 Proof Because the rows of U can be obtained from the rows of A by elementary row operations (scalar multiplication and addition), it follows that the row vectors of U are linear combinations of the row vectors of A. The row vectors of U lie in the row space of A, and the subspace .( spanned by the row vectors of U is contained in the row space of A: span(U)  span(A). Since the rows of A can also be obtained from the rows of U by elementary row operations, we also have that subspace spanned by the row vectors of A is contained in the row space of U: span(A)  span(U). Therefore, the two row spaces are equal: span(U) = span(A).

p. 105 Chapter 4: Vector Spaces. 4.6 Rank of a Matrix and Systems of Linear Equations.  Theorem 4.14 Basis for the Row Space of a Matrix .( If a matrix A is row-equivalent to a matrix U in row-echelon form, the then the nonzero row vectors of U form a basis for the row space of A.

  34291     26410 Example: Show that the nonzero rows of  31000 (a matrix in row-echelon form)    10000  00000 are linearly independent.

c1  34291   c   26410  Solution: We want to solve  00000 = 2  c3  31000 

 c4  10000 

The first column of the right-hand side sum is c1, so c1 = 0. The second column of the right- hand side sum is –9c1 + c2 = 0 + c2, so c2 = 0. The fourth column of the right-hand side sum is 0 + 0 + c3 (because c1 = c2 = 0), so c3 = 0. Finally, the fifth column of the right-hand side sum is 0 + 0 + 0 + c4 (because c1 = c2 = c3 = 0), so c4 = 0. Therefore the nonzero rows are linearly independent.

Alternate Solution: The transpose of =

0  1   0  0 0  0001              c1  0  9  1  0 0  0019    c2 is 0 = c1  2 1 + c2  4  + c3 0 + c4 0 =  0042    .             c3  0  4  6 1 0   0164    c4  0  3   2  3 1  1323 

0 0 Solving this by “forward substitution” (first row down to the last row) yields =   . 0   0

p. 106 Chapter 4: Vector Spaces. 4.6 Rank of a Matrix and Systems of Linear Equations.

  4981    4  224433 Example: Find a basis for the row space of  4  457637    9  5211376  40  3032

1  8  9 4    0 1 8  6 Solution: The row-echelon form is 0 0 0 1  , so a basis for the row space is   0 0 0 0  0 0 0 0 

{[1, –8, –9, 4], [0, 1, 8, –6], [0, 0, 0, 1]}. 1 0 55 0   0 1 8 0 You could also take the reduced row-echelon form 0 0 0 1 to obtain the basis   0 0 0 0 0 0 0 0 {[1, 0, 55, 0], [0, 1, 8, 0], [0, 0, 0, 1]}.

Example: Find a basis for the subspace of R3 spanned by {[1, –6, 3], [1, –5, 7], [–7, 38, –37]}.

Solution:

Lemma: The pivot columns of a matrix U in row-echelon form (or reduced row-echelon form) are linearly independent.

p. 107 Chapter 4: Vector Spaces. 4.6 Rank of a Matrix and Systems of Linear Equations.

1 xxxxxx    100 xxxx Example: U =  1000 xxx . If we look only at the pivot columns and solve    100000x  0000000 1 xxx 0 c  0  10 xx 1 0   c    0  100x  2  = 0 , we see by back substitution that the unique solution is =     c3    0  1000   0   c4  0  0000 0 , so the pivot columns of U are linearly independent.

Lemma: The pivot columns of a matrix U in row-echelon form (or reduced row-echelon form) span the column space of U.

Example: U = . The column space of U is the subset of R5 consisting

of vectors whose fifth component is zero (because we are taking linear combinations of vectors whose fifth component is zero).

v1  v   2 

If we look only at the pivot columns, we can always solve = v3  , by back   v4   0  substitution so the pivot columns of U span the column space of U.

 Theorem Basis for the Column Space of a Matrix .( Given an mn matrix A and its row-echelon (or reduced row-echelon) form U. Then the columns of A that correspond to the pivot columns of U form a basis for the column space of A.

Proof: by combining the two preceding lemmas, we know that the pivot columns of U form a basis for the column space of U. Let ai be the column vectors of A and ui be the column vectors of U, so A = [a1| …| an ] and U = [u1| …| un ]. We can factor PA = LU, so Pai = Lui for each column vector, –1 –1 ui = L Pai, and ai = P Lui. (essential idea)

p. 108 Chapter 4: Vector Spaces. 4.6 Rank of a Matrix and Systems of Linear Equations. To show that the pivot columns of A are linearly independent, consider the equation 0 = , where the terms in the sum include only the pivot columns. Then c aii ipivots L–1P0 = L–1P , so 0 = 1 = c aii  PLcaii c uii ipivots ipivots ipivots

Since the pivot columns of U are linearly independent, all of the ci (i  pivots) are zero, so the pivot columns of A are also linearly independent.

To show that the pivot columns of A span the column space of A, consider any vector v in the n column space of A. Then v can be written as v = c aii so i1 n n n –1 –1 1 L Pv = L P c aii =  PLcaii = c uii i1 i1 i1 –1 Since L Pv is a linear combination of the ui, it can be written as a linear combination of the pivot columns (which are a basis for the column space of U): –1 –1 –1 –1 1 L Pv = d uii so v = P L(L Pv) = P L =  i LPdui = d aii ipivots ipivots ipivots Therefore, the pivot columns of A span the column space of A.

Taken together, the two parts of this proof show that the pivot columns of A form a basis for the column space of A.

 Example: Find a basis for the column space of   4981     4  224433  A =  4  457637  .    9  5211376   40  3032 

.(

p. 109 Chapter 4: Vector Spaces. 4.6 Rank of a Matrix and Systems of Linear Equations.

 Theorem 4.15 Row Space and Column Space have the Same Dimension. .( If A is an mn matrix, then the row space and column space of A have the same dimension.

Proof: The dimension of the row space and the dimension of the column space are both equal to the number of pivots in the row-echelon form (or in the reduced row echelon form) of A.

.( The rank r = rank(A) of a matrix is the dimension of its row space, the dimension of its column space, and the number of pivots in the row-echelon form (or in the reduced row echelon form).

 Theorem 4.16 Solutions to a Homogeneous System Ax = 0

If A is an mn matrix, then the set of all solutions to the homogeneous system of linear equations Ax = 0 is a subspace of Rn.

Proof: Because A is mn, x is n1, so the set of all solutions to the homogeneous system of linear equations Ax = 0 is a subset of Rn. We will call this subset N. N is not empty because it contains 0: A0 = 0. Now we must show …

.(

 If A is an mn matrix, the subspace of Rn consisting of all solutions to the homogeneous system Ax = 0 is a called the nullspace of A and is denoted by N(A). We sometimes call N(A) the solution space of Ax = 0. The dimension of N(A) is called the nullity of A, and is sometimes written as  (the Greek letter nu).

p. 110 Chapter 4: Vector Spaces. 4.6 Rank of a Matrix and Systems of Linear Equations.  Theorem 4.17+ Solutions of a Homogeneous System Ax = 0

If A is an mn matrix of rank r and nullity , then the dimension of the solution space of Ax = 0, i.e. N(A), is  = n – r. That is, n = rank(A) + nullity(A), or the number of variables = number of pivot variables + number of free variables.

Moreover, the solution set to the homogeneous system Ax = 0 is the same as the solution set to the reduced row-echelon form Rx = 0. This solution has the form

xh = t1x1 + t2x2 + … + tx

where the ti are free parameters, one for each free (non-pivot) variable, and {x1, x2, …, x} is a basis for N(A).

 Proof Ax = 0 and Rx = 0 have the same solution set, i.e. N(A) = N(R), because A and R are row- equivalent. .( Ax = 0 iff En…E2E1Ax = En…E2E10 iff Rx = 0

Each column of R is multiplied by a component of x. R has r pivots, corresponding to r pivot variables in x. The remaining n – r variables are free variables, which we can replace with free parameters t1, t2, …, tn–r. Now we can solve r equations (on for each row of R that has a pivot, and is therefore not all zeroes) for r pivot variables by back-substitution. The set of these solutions xh is N(A). If we collect like terms in ti, and factor out the ti, we have

xh = t1x1 + t2x2 + … + tx

so N(A) is spanned by {x1, x2, …, x}. Moreover, {x1, x2, …, x} is linearly independent, because the solution to 0 = t1x1 + t2x2 + … + tx = xh is 0 = xfree(i) = ti and for each free variable and 0 = xpivot(j) = (some linear combination of the ti) for each pivot variable. Since all of the ti are 0, {x1, x2, …, x} is linearly independent, so {x1, x2, …, x} is a basis for N(A), and N(A) has dimension  = nullity(A) = n – r. That is, the number of variables = number of pivot variable. + number of free variables.

 Example: Find a basis for nullspace of 1 2 1 0 0  2 5 1 1 0  A =   . 3 7 2 2  2   4 9 3 1 4 

p. 111 Chapter 4: Vector Spaces. 4.6 Rank of a Matrix and Systems of Linear Equations. Solution: 1 0 3 0  4 0 0 1 1 0 2 0 The reduced row-echelon form is [ R | 0 ] =   . 0 0 0 1  2 0   0 0 0 0 0 0

Now we need to solve the system pivots

 x1  1 0 3 0  4 0 1x  3x  4x  0 x  1 3 5 0 1 1 0 2   2  0  1x 1x  2x  0      2 3 5 x3  = or the equivalent form      0 0 0 1  2   0  1x4  2x5  0   x4   0 0 0 0 0   0  0  0        x5 

The two free (non-pivot) variables are x3 and x5. We can choose parameters x3 = s and x5 = t, and solve for the pivot variables x1, x2, and x4 using back substitution.

1x1  3s  4t  0   1x2 1s  2t  0  x4 = 2t, x2 = s – 2t, and x1 = –3s + 4t   1x4  2t  0

 3s  4t  3  4         s  2t   1   2 That means that N(A) is the set of all vectors of the form =  s  = s  1  + t  0         2t   0   2   t   0   1 

So { , } is a basis for N(A).

p. 112 Chapter 4: Vector Spaces. 4.6 Rank of a Matrix and Systems of Linear Equations. Example: Find a basis for nullspace of  1 4 2 1  A =   .  0 1 1 1  2  8  4  2

Solution:

Now that we have solved Ax = 0, let us consider Ax = b, where b  0. We have already seen (Ch. 1.2) that Ax = b has an infinite number of solutions if there are more variables than equations (n > m). And since you can’t have more pivots than rows, m  r, so n > r, or  > 0.

Theorem 4.18 Solutions of a Nonhomogeneous System Ax = b

If xp is a particular solution to the nonhomogeneous equation Ax = b, then every solution of this system can be written in the form x = xp + xh, whereS* x h is a solution of the corresponding homogeneous system Ax = 0, i.e. xh is in N(A). The general solution of Ax = b, parameterized by the free parameters t1, t2, …, t is

x = xp + t1x1 + t2x2 + … + t x

 Proof

Let x be any solution of Ax = b. Then (x – xp) is a solution of the homogeneous system Ax = 0, because

A(x – xp) = Ax – Axp = b – b = 0 .( If we let xp = x – xp, then x = xp + xh.

(For ease of calculation, we usually find xp by solving Ax = b when all of the free variables are set to zero.)

From Thm. 4.17+ we know that xh = t1x1 + t2x2 + … + tx.

p. 113 Chapter 4: Vector Spaces. 4.6 Rank of a Matrix and Systems of Linear Equations.  Example: Find the general solution (the set of all solutions) of the system of linear system.

x1  2x2  x3  x4  5x5  0  5x 10x  3x  3x  55x  8 1 2 3 4 5 x1  2x2  2x3  3x4  5x5  14

1x1  2x2  x3  x4 15x5  2

 1 2 1 1 5 0  1 2 0 0  5 1   5 10 3 3 55  8 0 0 1 0 6 2  Solution:   rref    1 2 2  3  5 14  0 0 0 1 4  3     1  2 1 1 15  2 0 0 0 0 0 0 

1x1  2(0)  5(0)  1

To find xp, set the free variables x2 = 0, x5 = 0. 1x3  6(0)  2

1x4  4(0)  3

 1     0  .( So x4 = –3, x3 = 2, x1 = 1. xp =  2     3  0 

1x1  2s  5t  0

To find xh, set x2 = s, x5 = t. Ax = 0  1x3  6t  0

1x4  4t  0

 2s  5t  2  5         s   1   0  So x4 = –4t, x3 = –6t, x1 = –2s + 5t. xh =   6t  = s  0  + t  6         4t   0   4  t   0   1 

 5     0  x = + s + t  6    4  1 

p. 114 Chapter 4: Vector Spaces. 4.6 Rank of a Matrix and Systems of Linear Equations.  Example: Find the general solution (the set of all solutions) of the system of linear system. x  3y 10z  18  2x  7y  32z  29

 x  3y 14z  12 x  y  2z  8

Solution:

 Example: Find the general solution (the set of all solutions) of the system of linear system. 3x  8y  4z  19  6y  2z  4w  5

5x  22z  w  29 x  2y  2z  8

Solution:

p. 115 Chapter 4: Vector Spaces. 4.6 Rank of a Matrix and Systems of Linear Equations.  Theorem 4.19 Consistency of a System of Linear Equations

The system Ax = b is consistent if and only if b is in the column space of A.  Proof Suppose A is mn, so x is n1 and b is m1. a a  a   x  a   a  a  11 12 1n 1 11 S* 12 1n a a  a  x  a  a  a  Then Ax =  21 22 2n   2  =  21 x   22 x  2n x (see Section 2.1)            1    2    n           am1 am2  amn  xn  am1  am2  amn 

So Ax = b is if and only if b is a linear combination of the columns of A, i.e. b is in the column space of A.

Summary of Equivalent Conditions for Square Matrices

If A is an nn matrix, then the following conditions are equivalent.

1) A is invertible. 2) Ax = b has a unique solution for any n1 matrix b. 3) Ax = 0 has only the trivial solution. 4) A is row-equivalent to In. 5) det(A)  0 6) rank(A) = n 7) The n row vectors of A are linearly independent. 8) The n column vectors of A are linearly independent.

p. 116 Chapter 4: Vector Spaces. 4.7 Coordinates and Change of Basis.

4.7 Coordinates and Change of Basis.  Objective: Represent coordinates in general n-dimensional spaces. Objective: Find a coordinate matrix relative to a basis in Rn.  Objective: Find the transition matrix from the basis B to the basis B in Rn.   Let B = {v1, v2, …, vn} be an ordered basis for a vector space V and let x be a vector in V such that

x = c1v1 + c2v2 + …+ cnvn

The scalars c1, c2, …, cn are called the coordinates of the vector x relative to (or with respect to) the basis B. The coordinate vector (or matrix) of x relative B is the column matrix in Rn whose components are the coordinates of x. S*

c1    c2 [x]B =        cn 

Example: The coordinate matrix of x = (x1, x2, x3) relative to the standard basis S = {e1, e2, e3} = {(1, 0, 0), (0, 1, 0), (0 0, 1)} in R3 is simply

 x1  [x] = x  because x = x e + x e + x e S  2  1 1 2 2 3 3 x3 

Example: Find the coordinate matrix of p(x) = 2x2 –5x + 6 relative to the standard ordered basis 2 S = {1, x, x } in P2

 6  Solution: p(x) = 6(1) – 5(x) + 2(x2) so [p] =   S  5  6 

 28  Example: Find the coordinate matrix of A =   relative to the standard ordered basis  71   01   10   00   00  S = {   ,   ,   ,   } in M2,2  00   00   01   10 

8 2 Solution: A = 8 + 2 + 1 + 7 so [A]S =   1   7

p. 117 Chapter 4: Vector Spaces. 4.7 Coordinates and Change of Basis. Example 1: Given that the coordinate matrix of x in R2 relative to the nonstandard ordered basis B = {v1, v2} = {(1, 0), (1, 2)} is 3 [x]B =   2 Find the coordinate matrix of x relative to the standard basis B = {u1, u2} = {(1, 0), (0, 1)} = S.

Solution: [x]B = means that x = 3v1 + 2v2 = 3(1, 0) + 2(1, 2) = (5, 4).

5 Now (5, 4) = 5(1, 0) + 4(0, 1) = 5u1 + 4u2 so [x]B=S =   . 4

Another method is to take the coordinate matrices relative to B of      11  3 x = 3v1 + 2v2 =  vv 21  to obtain [x]B=S =  B vv ][][ B'2'1  =     = .      20  2

This procedure is called a change of basis from B to B. Given [x]B, we found [x]B using an equation of the form

1 –1   [x]B = P [x]B where P =  B vv ][][ B'2'1  .  

The matrix P–1 is called the transition matrix from B to B.

Example 2: Given that the coordinate matrix of x in R3 relative to the standard ordered basis B = S = {e1, e2, e3} is  1  [x] =   . B  2  1

Find the coordinate matrix of x relative to the nonstandard ordered basis B = {u1, u2, u3} = {(1, 0, 1), (0, –1, 2), (2, 3, –5)}.

p. 118 Chapter 4: Vector Spaces. 4.7 Coordinates and Change of Basis. Solution: Write x as a linear combination of the vectors in B:

  c1    x = c u + c u + c u =  uuu c  =  [x] =  uuu][][][ 1 1 2 2 3 3  321  2  B  1 B 2 B 3 B    c3    1  1   201 c1   201  5    Since B = S,  2  =   310 so c  =  310 =  8 .      2      1   521 c3    521  2

Thus we have [x]B=S = = .

Moreover, we found that [x]B = [x]B

1 Comparing to [x]B = P [x]B after Example 1above, we must have

[x]B = P[x]B where P = .

The matrix P is called the transition matrix from B to B.

 Given nonstandard bases B = {v1, v2, …, vn}, B = {u1, u2, …, un}, and the standard basis S = n {e1, e2, …, en} of R . Define matrices     B =  vv  v ][][][  and (B) =  uu  u ][][][  . BS  1 S 2 S Sn  BS  1 S 2 S Sn      S* (B and B are square, invertible matrices.) B is the transition matrix from B to S; B is the –1 transition matrix from B to S. To get P , the transition matrix from B to B, first apply BBS, –1 then apply (B) SB. –1 –1 –1 –1 P = (B) B or (P )BB = (B) SB BBS The inverse is –1 –1 P = B B or PBB = (B )SB (B)BS

p. 119 Chapter 4: Vector Spaces. 4.7 Coordinates and Change of Basis.

2  Example: Given B = {v , v , v } = {(1, 0, 1), (1, 1, 0), (0, 1, 1)} and [x] =   . Find [x] 1 2 3 B 3 S 1

Solution:

 Example: Given B = {u1, u2, u3} = {(8, 11, 0), (7, 0, 10), (1, 4, 6)} and x =(3, 19, 2). Find [x]B

Solution:

 Example: Given B = {v1, v2, v3} = {(1, 0, 2), (0, 1, 3), (1, 1, 1)},  1  B = {u , u , u } = {(2, 1, 1), (1, 0, 0), (0, 2, 1)}, and [x] =   . 1 2 3 B  2  1 –1 Find the transition matrix P from B to B the transition matrix P from B to B, and [x]B.

Solution:

p. 120 Chapter 5: Inner Product Spaces. 5.1 Length and Dot Product in Rn.

Chapter 5: Inner Product Spaces.

5.1 Length and Dot Product in Rn.  Objective: Find the length of a vector and find a unit vector.  Objective: Find the distance between two vectors.  Objective: Find a dot product and the angle between two vectors, determine orthogonality, and verify the Cauchy-Schwarz Inequality, the triangle inequality, and the Pythagorean Theorem.  Objective: Use a matrix product to represent a dot product.

From the Pythagorean Theorem, we know that the length, or norm, of a vector v in R2 is 2 2 3 ||v|| = 1  vv2 . Also from the Pythagorean Theorem, in R we have 2 ||v|| =  2  2   vvv2 = 2 2 vvv2 .  1 2  3 1 2 3

n S* 2 2 2  In R we define the length or norm of v to be ||v|| = 1 2  vvv n

Notice that  |v||  0  ||v|| = 0 if and only if v = 0.

A unit vector is a vector whose length is one. In R2 and R3 alternate notation is sometimes used for the standard unit vectors i = (1, 0), j = (0, 1) in R2 i = (1, 0, 0), j = (0, 1, 0) , k = (0, 0, 1) in R3

Two nonzero vectors u and v are parallel when one of them is a scalar multiple of the other: u = cv. If c > 0, then u and v have the same direction. If c < 0, then u and v have opposite directions.

p. 121 Chapter 5: Inner Product Spaces. 5.1 Length and Dot Product in Rn. Theorem 5.1 Length of a scalar multiple

Let v be a vector in Rn and let c be a scalar. Then ||cv|| = |c| ||v||, where |c| is the absolute value of c.

2 2 2 2 2 2 2 Proof: ||cv|| = 1 cvcv2 cv n )()()( = 1 2 vvvcn = |c| ||v||

Theorem 5.2 Unit Vector in the Direction of v

v If v is a nonzero vector in Rn, then the vector u = has length one and has the same direction v as v. u is called the unit vector in the direction of v.

  1  1  Proof: Because v  0, we know that ||v|| > 0 so > 0. Then u =   v so u has the same v  v  v direction as v. Moreover, ||u|| = = ||v|| = 1, so u is a unit vector. v

The process of finding the unit vector in the same direction as v is called normalizing v.

The distance between two vectors u and v is d(u, v) = ||u – v||. In Rn, 2 2 2 d(u, v) = 11 vuvu 22   vu nn )()()(

Notice that  d(u, v)  0  d(u, v) = 0 if and only if u = v.  d(u, v) = d(v, u)

 Example: Let u = (–2, 3, 4), v = (1, 2, –2). Find the length of u, normalize v, and find the distance between u and v.

Solution by hand: ||u|| =  43)2( 222 = 29 or  5.38516

 )2,2,1(  1 2 2  = =  , ,  22  )2(21 2  3 3 3 

d(u, v) = 2 2  2(4()23()12( ))2 =

p. 122 Chapter 5: Inner Product Spaces. 5.1 Length and Dot Product in Rn.

61)3(222 = 46 = 6.78233

Solution using Mathematica:

u={-2,3,4} Out = {-2,3,4}

Norm[u] Out =

v={1,2,-2} Out = {1,2,-2}

Normalize[v] Out =

Norm[u-v] Out =

Solution using TI-89: [-2,3,4].U

[1,2,-2].V

MatrixNorms norm(U)

MatrixVector ops unitV(V)

MatrixNorms norm(U-V)

To find the angle between to vectors in Rn, we start with the Law of Cosines in R2. u – v ||u – v||2 = ||u||2 + ||v||2 – 2||u|| ||v|| cos( ) v 2 2 2 2 2 2  (u1 – v1) + (u2 – v2) = u1 + u2 + v1 + v2 – 2||u|| ||v|| cos( ) u

2 2 2 2 2 2 2 2 u1 –2u1v1 + v1 + u2 –2u2v2 + v2 = u1 + u2 + v1 + v2 – 2||u|| ||v|| cos( )

–2u1v1 –2u2v2 = – 2||u|| ||v|| cos( )

u1v1 + u2v2 = ||u|| ||v|| cos( )

This equation leads us to the following definitions.

n  The dot product between two vectors u = (u1, u2, …, un) and v = (v1, v2, …, vn) in R

u•v = u1v1 + u2v2 + … + unvn

S*   n   vu   The angle between two nonzero vectors u and v in R is  = arccos  , 0    .  vu 

p. 123 Chapter 5: Inner Product Spaces. 5.1 Length and Dot Product in Rn.  Two vectors u and v in Rn are orthogonal (perpendicular) when u•v = 0. (Notice that the 0 vector is orthogonal to every vector.)

n  We often represent a vectors u = (u1, u2, …, un) and v = (v1, v2, …, vn) in R by their coordinate matrices relative to the standard basis {e1, e2, …, en}

u1  v1  u  v  u =  2  and v =  2            u n  vn  T T We then write u v for u•v because u v = [u1v1 + u2v2 + … + unvn].  Warning: In our formulas, we will be sloppy and pretend that uTv is equal to the scalar u•v. However, Mathematica and the TI-89 know that uTv is a 11 matrix, and will not let you use it as a scalar.

 Example: Using u = (–2, 3, 4), v = (1, 2, –2) from the previous example, find u•v and the angle between u and v.

Solution by hand: u•v = (–2)(1) + (3)(2) + (4)(–2) = –4 ||u|| =  43)2( 222 = 29 or  5.38516 ||v|| = = 22  )2(21 2 = 3   4   = arccos   1.821 or  104.3  3 29 

Solution using Mathematica:

u.v Out = 4 period, not asterisk

ArcCos[u.v/Norm[u]/Norm[v]] Out =

Out = 1.82099

Solution using TI-89: MatrixVector ops  U V dotP( , )

  Matrix    Vector ops dotP(u, v) MatrixNorms norm(u) MatrixNorms norm(v))

  Matrix    Vector ops dotP(u, v)

p. 124 Chapter 5: Inner Product Spaces. 5.1 Length and Dot Product in Rn. MatrixNorms norm(u) MatrixNorms norm(v))

To obtain a decimal approximation instead of an exact formula, use N[...] in Mathematica or  instead of  on the TI-89. You can change modes between degrees and radians on the TI-89 using  Angle……… DEGREE

Theorem 5.3 Properties of the Dot Product

If u, v, and w are vectors in Rn and c is a scalar, then

1) u•v = v•u 2) u•(v + w) = u•v + u•w 3) c(u•v) = (cu)•v = u•(cv) 4) v•v = ||v||2 5) v•v  0 and v•v = 0 if and only if v = 0.

Proof of 5.3.2:

u•(v + w) = optional

Rn combined with the usual operations of vector addition, scalar multiplication, vector length, and the dot product is called Euclidean n-space.

(Optional aside: non-Euclidean spaces include elliptic space, which is curved, cannot be smoothly mapped Rn and a has a different definition of distance; hyperbolic space, which is curved and has different definition of distance; and Minkowski space, used in Einstein’s special optional relativity, which replaces the non-negative dot product with an “interval” that can be negative.)

p. 125 Chapter 5: Inner Product Spaces. 5.1 Length and Dot Product in Rn.

 Theorem 5.4 The Cauchy-Schwarz Inequality

If u and v are vectors in Rn, then |u•v|  ||u|| ||v||, where |u•v| is the absolute value of u•v.  Proof If u = 0, then equality holds, because

|u•v| = |0•v| = |0| = 0, and ||u|| ||v|

= ||0|| ||v|| = 0 ||v|| = 0.

If u  0, then let t be any real number and consider the vector tu + v. Then

0  || tu + v ||2 = S* The left-hand side is a f (t) = at2 + bt + c, where

a = , b = , c =

−푏±√푏2−4푎푐 The quadratic formula is 푡 = 2푎

Because f (t)  0, we have either one or zero roots, so 푏2 − 4푎푐 ≤ 0

 Theorem 5.5 The Triangle Inequality

If u and v are vectors in Rn, then ||u + v||  ||u|| + ||v||.  Proof ||u + v||2 = S*

p. 126 Chapter 5: Inner Product Spaces. 5.1 Length and Dot Product in Rn.

 Theorem 5.6 The Pythagorean Theorem

If u and v are vectors in Rn, then u and v are orthogonal if and only if ||u + v||2 = ||u||2 + ||v||2.  Proof By definition, u and v are orthogonal if and only if S* ||u + v||2 =

The conclusion follows from these two equations.

p. 127

Chapter 5: Inner Product Spaces. 5.2 Inner Product Spaces.

5.2 Inner Product Spaces.  Objective: Determine whether a function defines an inner product. n  Objective: Find the inner product of two vectors in R , Mm,n, Pn, and C[a,b].  Objective: Use the inner product to find angles between two vectors and determine whether two vectors are orthogonal.  Objective: Find an orthogonal projection of a vector onto another vector in an inner product space.

The inner product is an extension of the concept of the dot product from Rn to a general vector space. The standard dot product, also called the Euclidean inner product, on Rn is written as u•v, while the general inner product on a vector space V is written as u, v.

 Definition of Inner Product

Let u, v, and w be vectors in a vector space V, and let c be any scalar. An inner product on V is a function that associates a real number u, v with each pair of vectors u and v. and satisfies the * following axioms.

1) u, v = v, u Symmetric 2) u, v + w = u, v + u, w 3) cu, v = cu, v 4) v, v  0 and v, v = 0 if and only if v = 0. Nonnegative

A vector space V with an inner product is called an inner product space.

2 u1  v1  Example: Consider R . Let u =   and v =   . Instead of the Euclidean inner product u2  v2 

T T  12  u v, let u, v = u   v = 2u1v1 – u1v2 – u2v1 + 3u2v2. Show that u, v is an inner  31  product.

Solution:

T 1) v, u = v u = 2v1u1 – v1u2 – v2u1 + 3v2u2 = u, v

2) Let A = . We know that for matrix multiplication, u, v + w = uTA(v + w) =

(uTA)(v + w) = (uTA)v + (uTA)w = uTAv + uTAw = u, v + u, w.

3) We know that for matrix multiplication, cu, v = c(uT v) = (cuT) v =

= cu, v.

p. 129 Chapter 5: Inner Product Spaces. 5.2 Inner Product Spaces.

2 2 2 2 2 2 2 2 2 4) v, v = 2v1 – 2v1v2 + 3v2 = v1 + 2v2 + v1 – 2v1v2 + v2 = v1 + 2v2 + (v1 – v2)  0, and v, v = if and only if v1 = 0, v2 = 0, and v1 – v2 = 0 if and only if v = 0.

ut  vt  u  v  Example: Consider R4. Let u =  x  and v =  x  . Instead of the Euclidean inner product u y  v y      u z  v z   c 2 000   0010 uTv, let u, v = uT   v, where c is a positive constant (the speed of light), so  0100    1000 2 u, v = –c utvt + uxvx + uyvy + uzvz. Show that u, v is not an inner product. (Optional aside: u, v is called the spacetime interval.)

Solution: 1   2 2 2 2 2 0 4) v, v = –c vt + vx + vy + vz which can be less than zero, for example when v =   . 0   0 Note: Axioms 2 and 3 are satisfied because of the rules of matrix multiplication. Axiom 1 is satisfied because the matrix is symmetric (AT = A).

Theorem 5.7 Properties of Inner Products

If u, v, and w are vectors in an inner product space V “over the real numbers” and c is a scalar, then

1) 0, v = v, 0 = 0 2) u + v, w = u, w + v, w 3) u, cv = cu, v

Proof of 5.7.1: 0, v = 0v, v = 0v, v = 0. By the symmetric property, v, 0 = 0, v = 0

Proof of 5.7.2: u + v, w = w, u + v Symmetry

= w, u + w, v Axiom 2 = u, w + v, w Symmetry optional Proof of 5.7.3: u, cv = cv, u Symmetry = cv, u Axiom 3 = cu, v Symmetry

p. 130 Chapter 5: Inner Product Spaces. 5.2 Inner Product Spaces. Together with the inner product axioms, Thm 5.7 says that the inner product is linear in each of its arguments (we will study linear operators much more in Chapter 6.) Linearity is essentially the same as distributivity: Dot product: (2u + 3v)•(5x + 7y) = 10(u•x) + 14(u•y) + 15(v•x) + 21(v•y) General inner product: 2u + 3v, 5x + 7y = 10u, x + 14u, y + 15v, x + 21v, y

Let u and v be vectors in an inner product space V.

 The length, or norm, of u is ||u|| = ,uu.

* The distance between two vectors u and v is d(u, v) = ||u – v||.

 , vu    The angle between two nonzero vectors u and v is given by  = arccos  , 0    .  vu

 Two vectors u and v in Rn are orthogonal (perpendicular) when u, v = 0. (Notice that the 0 vector is orthogonal to every vector.)

v If ||u|| = 1 then u is called a unit vector. If v is any nonzero vector, then u = is the unit vector v , vu in the direction of v. Notice that the definition of angle presumes that –1   1. This is vu guaranteed by the Cauchy-Schwarz Inequality (Theorem 5.8.1).

Properties of Length

1) ||u||  0 2) ||u|| = 0 if and only if u = 0. 3) ||cu|| = |c| ||u||

These follow from the axioms in the definition of inner product and the definition of norm.

Properties of Distance

1) d(u, v)  0 2) d(u, v) = 0 if and only if u = v. 3) d(u, v) = d(v, u)

These follow from the axioms in the definition of inner product and the definition of distance.

p. 131 Chapter 5: Inner Product Spaces. 5.2 Inner Product Spaces.

a11 a12  b11 b12   Example: Let A =   and B =   be two matrices in M2,2 and define the inner a21 a22 b21 b22 

product A, B = a11b11 + a12b12 + a21b21 + a22b22. 5 3 1 0 0 1  4 4  Let I =   , R =   and L =  3 5  . 0 1 1 0   4 4 

a) Find the norm of I = .

Solution:

b) Find the inner product of and angle between R and L. Are they orthogonal?

Solution: .(

c) Find the inner product of and angle between I and L. Are they orthogonal?

Solution:

p. 132 Chapter 5: Inner Product Spaces. 5.2 Inner Product Spaces.  Example: Let p and q be two polynomials in P and define the inner product 1 p, q =  xqxp)()(dx 1

a) Find the norm of p(x) = 1. (Sometimes we will just say “Find the norm of 1.”)

Solution:

b) Find the inner product of p(x) = 1 and q(x) = x. Are they orthogonal?

Solution: .(

c) Find the angle between 1 and x2. Are they orthogonal?

Solution:

Theorem 5.8

Let u and v be vectors in an inner product space V.

1) Cauchy-Schwarz Inequality: | u, v |  ||u|| ||v|| 2) Triangle inequality: ||u + v||  ||u|| + ||v|| 3) Pythagorean Theorem: u and v are orthogonal if and only if ||u + v||2 = ||u||2 + ||v||2.

The proofs are the same as the proofs of Theorems 5.4, 5.5, and 5.6, except that we replace u•v with u, v. A result such as the triangle inequality, which seems intuitive in R2 and R3, can be generalized through linear algebra to some less obvious statements about functions in C[a, b].

p. 133 Chapter 5: Inner Product Spaces. 5.2 Inner Product Spaces. 4 Example: Consider f (x) = cos(x) and g(x) = (/2 – x)(/2 + x) in C[–/2, /2] with the inner  2  / 2 product  f , g =  f (x)g(x)dx . Verify the triangle inequality by direct calculation.  / 2

Solution: The triangle inequality is || f + g ||  || f || + || g ||.

|| f ||2 =

|| g ||2 =

|| f + g ||2 =

|| f + g || =

|| f || + || g || =

u  Orthogonal Projections: Let u and v be two vectors in R2. If v is nonzero, then u can be orthogonally projected onto v. This projection is denoted by

projv u. Because projv u is a scalar multiple of v, we can write  v projv u v projv u = a v v (Recall that is the unit vector in the direction of v.) v u  v u  v Then a = ||u|| cos = ||u|| = u v v

u  v u  v projv u = v = v v 2 v  v

p. 134 Chapter 5: Inner Product Spaces. 5.2 Inner Product Spaces.

* Let u and v be vectors in a general inner product space V. The orthogonal projection of u onto v is 〈퐮, 퐯〉 proj 퐮 = 퐯 퐯 〈퐯, 퐯〉

 Example: Use the Euclidean inner product find the orthogonal projection of u = (1, 0, 0) onto

v = (1, 1, 1) and of v onto u .

Solution: .( projv u =

projv u =

p. 135 Chapter 5: Inner Product Spaces. 5.2 Inner Product Spaces. Theorem 5.9 Orthogonal Projection and Distance u u Let u and v be two vectors in an inner product d(u, proj u) d(u, cv) space V such that v  0. Then v

, vu projv u v cv v d(u, projv u) < d(u, cv) whenever c  , vv

That is to say, the orthogonal projection projv u is the vector parallel to v that is closest to u.

u Proof: Let b = , c  , p = cv – bv, q = u – bv, and t = u – cv. q t = u – cv The p is orthogonal to q, because bv p, q = cv – bv, u – bv p = (c – b)v

= cv, u – cbv, v – bv, u + b2v, v , vu 2 = cu, v – c v, v – u, v + v, v , vv 2 2 , vu , vu 2 = cv, u – cv, u – + = 0. , vv , vv So the Pythagorean Theorem tells us ||t||2 = ||p||2 + ||q||2 > ||q||2 because p  0 because c  b.

Therefore, d(u, projv u) = ||q|| < ||t|| = d(u, cv).

  Example: Consider C[–, ] with the inner product  f , g =  xgxf )()( dx .  Find the projection of f (x) = x onto g(x) = sin(x).

Solution: .( , gf projg f = xg )( . , gg

p. 136 Chapter 5: Inner Product Spaces. 5.3 Orthogonal Bases: Gram-Schmidt Process.

5.3 Orthogonal Bases: Gram-Schmidt Process.  Objective: Show that a set of vectors is orthogonal and forms an orthonormal basis.  Objective: Represent a vector relative to an orthonormal basis.  Objective: Apply the Gram-Schmidt orthonormalization process.

Although a vector space can have many different bases, some bases may be easier to work with 3 than others. For example, in R , we often use the standard basis S = {e1, e2, e3} = {(1, 0, 0), (0, 1, 0), (0 0, 1)}. Two properties that make the standard basis desirable are that

1) The vectors are normalized, i.e. each basis vector is a unit vector: ||ei|| = 1 for i = 1, 2, 3.

2) The vectors are mutually orthogonal: ei•ej = 0 whenever i  j, i.e. e1•e2 = 0, e1•e3 = 0, e2•e3 = 0

 A set S of vectors in an inner product space V, is called orthogonal when every pair of vectors in S is orthogonal. If, in addition, each vector in the set is a unit vector, then S is called orthonormal. S* If S = {v1, v2, …, vn} then these definitions can be written as

S is orthogonal if and only if vi•vj = 0 whenever i  j.

S is orthonormal if and only if vi•vj = 0 whenever i  j, and ||ei|| = 1 for i = 1, 2, …, n.

 Example: Show that S = {(cos, sin, 0)}, (–sin, cos, 0), (0, 0, 1)} is an orthonormal set.

Solution: S is orthogonal, because

(cos, sin, 0) • (–sin, cos, 0) = – cos sin + sin cos = 0

(cos, sin, 0) • (0, 0, 1) = 0

and (–sin, cos, 0) • (0, 0, 1) = 0 S*

S is normalized, because

||(cos, sin, 0)|| = 2   0sincos 22 = 1

||(–sin, cos, 0)|| = 2   0cossin 22 = 1

p. 137 Chapter 5: Inner Product Spaces. 5.3 Orthogonal Bases: Gram-Schmidt Process.

||(0, 0, 1)|| = 100222 = 1

Theorem 5.10 Orthogonal Sets are Linearly Independent

If S = {v1, v2, …, vn} is an orthogonal set of nonzero vectors in an inner product space V, then S is linearly independent.

Proof: suppose c1v1 + c2v2 + …+ cnvn = 0

Then (c1v1 + c2v2 + …+ cnvn), vi = 0, vi for each of the vi

We can distribute the inner product on the left-hand side to obtain

Because S is orthogonal, when j  i,

so equation becomes

Because vi  0,

2 Example: Consider C[0, 2] with the inner product  f , g =  xgxf )()( dx . 0 Show that the set {1, cos(x), sin(x), cos(2x), sin(2x), …, cos(nx), sin(nx)} is a linearly independent orthogonal set.

Solution: We use software to integrate the following expressions, and we use these identities to

 k  0)2sin( evaluate the results:  whenever k is an integer  k  1)2cos(

2 2 nx)cos( 11 1, sin(nx) =  )sin(1 dxnx =  =  = 0 0 n 0 nn

2 nx)sin( 2 1, cos(nx) =  )cos(1 dxnx = = 0 – 0 = 0 0 n 0

p. 138 Chapter 5: Inner Product Spaces. 5.3 Orthogonal Bases: Gram-Schmidt Process. sin(mx), sin(nx) when m  n is 2 2   xnm ))sin((  xnm ))sin(( mx )sin()sin(dxnx=  = (0 – 0) – (0 – 0) = 0    nm)(2  nm)(2  0   0

cos(mx), cos(nx) when m  n is 2 2   xnm ))sin((  xnm ))sin(( mx )cos()cos(dxnx=  = (0 + 0) – (0 + 0) = 0    nm)(2  nm)(2  0   0

cos(mx), sin(nx) when m  n is 2 2   xnm ))cos((  xnm ))cos(( mx )sin()cos( dxnx =  =    nm)(2  nm)(2  0   0  1 1   1 1         = 0   )(2  )(2    )(2  nmnmnmnm )(2 

2 2 2 nx)(cos 1 1 cos(nx), sin(nx) =  )sin()cos( dxnxnx =  =  = 0 0 2n 0 2 2nn

 Corollary Orthogonal Bases S* If V is an inner product space of dimension n, then any orthogonal set of n nonzero vectors in V is a basis for V.

 Example: Show that S = {(0, 1, 0, 1), (–1, –1, 2, 1), (2, –1, 0, 1), (2, 2, 3, –2)} is an orthogonal basis of R4.

Solution: S is orthogonal, because (0, 1, 0, 1)•(–1, –1, 2, 1) = 0 – 1 + 0 + 1 = 0 (0, 1, 0, 1)•(2, –1, 0, 1) = 0 –1 + 0 – 1 = 0S* (0, 1, 0, 1)•(2, 2, 3, –2) = 0 + 2 + 0 – 2 = 0 (–1, –1, 2, 1)•(2, –1, 0, 1) = –2 + 1 + 0 + 1 = 0 (–1, –1, 2, 1)•(2, 2, 3, –2) = –2 – 2 + 3 – 2 = 0 and (2, –1, 0, 1)•(2, 2, 3, –2) = 4 – 2+ 0 – 2 = 0 By the Corollary to Theorem 5.10, S is an orthogonal basis for R4.

p. 139 Chapter 5: Inner Product Spaces. 5.3 Orthogonal Bases: Gram-Schmidt Process.  Theorem 5.11 Coordinates Relative to an Orthonormal Basis

B V If = {v1, v2, …, vn} is an orthonormal basis for anS* inner product space , then the coordinate representation of a vector w relative to B is

w = w, v1v1 + w, v2 v2 + … +w, vn  vn

Proof: Because B is a basis for V, there are unique scalars c1, c2, …, cn such that

w = c1v1 + c2v2 + …+ cnvn

Taking the inner product with vi of both sides of the equation gives

w, vi = (c1v1 + c2v2 + …+ cnvn), vi

We can distribute the inner product on the left-hand side to obtain

Because S is orthonormal, when

so equation becomes

so

The coordinates of w relative to an orthonormal basis B are called the Fourier coefficients of w relative to B. The corresponding coordinate matrix is

c1   , vw 1    c  2  , vw 2  [w]B =   =           cn   , vw n 

1   Note: contrast Thm 5.11 with [w] =  vv  v ][][][  [w] B  1 S 2 S Sn  S   from Section 4.7 for a general, non-orthonormal basis B.

p. 140 Chapter 5: Inner Product Spaces. 5.3 Orthogonal Bases: Gram-Schmidt Process.  Example: Find the coordinate matrix of w = (–2, 6, 5) relative to the orthonormal basis 1 1 2 2 22 2 2 1 B = {( , , 0), (– , , ), ( , – , )} of R3. 2 2 6 6 3 3 3 3

Solution: S*

so the coordinate matrix is

Given any basis for a vector space, we can construct an orthonormal basis using a procedure called the Gram-Schmidt Orthonormalization.

 Theorem 5.12(Alt) (Alternative) Gram-Schmidt Orthonormalization Process

Let B = {v1, v2, …, vn} be a basis for an inner product space V.

Let B = {u1, u2, …, un}, where ui is given by

w1 w1 = v1, ...... u1 = w1

w 2 w2 = v2 – v2, u1u1, ...... u2 = S* w 2

w 3 w3 = v3 –v3, u1u1 – v3, u2u2,...... u3 = w 3

…

w n wn = vn – vn, u1u1 – vn, u2u2 – … – vn, u n–1un–1, ...... un = w n

Then B is an orthonormal basis for V. Moreover, span{ u1, u2, …, uk} = span{ v1, v2, …, vk} for k = 1, 2, …, n.

Proof: First, B is normalized, since all of the ui are unit vectors: ||ui||= 1. Second, observe that the terms of the form v , u u are projections onto u : v , u u = proj v j i i i j i i ui j

When we subtract from vj, the vector is

orthogonal to ui

p. 141 Chapter 5: Inner Product Spaces. 5.3 Orthogonal Bases: Gram-Schmidt Process. Now we prove by induction that B is also an orthonormal set.

If n =1, then there are no pairs of vectors in B, so there is nothing to prove.

w k If {u1, u2, …, uk–1} is an orthonormal set, then consider uk = where w k

wk = vk – vk, u1u1 – vk, u2u2 – … – vk, ujuj – … – vk, uk–1uk–1. Let j = 1, 2, …, k – 1. Then

1 uk, uj = (vk – vk, u1u1 – vk, u2u2 – … – vk, ujuj – … vk, uk–1uk–1), uj  w n

= (vk, uj – vk, u1u1, uj – vk, u2u2, uj – … – vk, ujuj, uj – …

– vk, uk–1uk–1, uj)

= (vk, uj – vk, u1(0) – vk, u2(0) – … – vk, uj(1) – … – vk, uk–1(0))

= (vk, uj – vk, uj) = 0

So {u1, u2, …, uk} is also an orthonormal set. By mathematical induction, B is orthonormal, and since it has n vectors in an n-dimensional space, B is an orthonormal basis.

 Example: Use the Gram-Schmidt orthonormalization process to construct an orthonormal basis from B = {(1, 2, 2), (–1, –1,0), (1, 0, 0)}.

Solution:

)2,2,1( 1 2 w1 = (1, 2, 2) u1 = = ( , , ) )2,2,1( 3 3

w2 = (–1, –1,0) – ((–1, –1,0)•( , , ))( , , ) = (–1, –1,0) – (–1)(1, 2, 2) S* 2  1 2 ),,( = (– , – , ) u = 3 3 3 = (– , – , ) 2 2 1 2 3  3 3 ),,( w3 =

u3 =

p. 142 Chapter 5: Inner Product Spaces. 5.3 Orthogonal Bases: Gram-Schmidt Process.  Example: Use the Gram-Schmidt orthonormalization process to construct an orthonormal basis 1 2 from B = {1, x, x } in P2 with the inner product p(x), q(x) =  xqxp)()(dx . 1

2 Solution: Let v1 = 1, v2 = x, and v3 = x . w1 = 1 1 u1 = 1 1 1 ||1||2 = 11dx = x = 2  1 1 1 u1 = 2

w2 =

u2 =

p. 143

Chapter 5: Inner Product Spaces. 5.4 Mathematical Models and Least Squares Analysis (Optional).

5.4 Mathematical Models and Least Squares Analysis (Optional).  Objective: Find the orthogonal complement of a subspace.  Objective: Solve a least squares problem.

In this section, we consider inconsistent systems of linear equations, and we find the “best approximation” to a solution.

Example: Given a table of data. We want to find the t y coefficients c0 and c1 of the line y = c0 + c1t that “best fits” these points. 1 0

Solution: The system of linear equations that we 2 1 have to solve comes from plugging the three points 3 3 (ti, yi) into the equation c0 + c1ti = yi

cc 10  01  11  0  11  0 c  c  cc  12 or   0   . Let A =   , b =   , and x = 0 10  21    1  21  1   c1  c1  cc 10  33  31  3  31  3

 001  The system is inconsistent: reduced row-echelon form is   . (We also knew that there  010   100  would be no solution because the three points in the graph are not collinear.) But what is the “best approximation”?

1 1 Recall that Ax = 1  2cc is always in the column space of A. But since Ax = b has no   0   1 1 3 solution, b is not in the column space of A. We want to find an x that gives the Ax that is closest to b.

Least Squares Problem: Given an mn matrix A and a vector b in Rm, find x in Rn, such that ||Ax – b||2 is minimized.

This gives the Ax that is closest to b. It is customary to minimize ||Ax – b||2 instead of ||Ax – b|| because ||Ax – b|| involves a square root. Intuitively, we see that Ax – b is orthogonal to the column space of A.

p. 145 Chapter 5: Inner Product Spaces. 5.4 Mathematical Models and Least Squares Analysis (Optional). To solve the least squares problem, it helps to use the concept of orthogonal subspaces.

Let S1 and S2 be two subspaces of an n-dimensional vector space V.

 S1 and S2 are orthogonal when v1, v2 = 0 for all v1 in S1 and v2 in S2.   The orthogonal complement of S1 is the set S1 = {u  V: u, v= 0 for all vectors v  S1}.  S1 is pronounced “S1 perp.”

 If every vector x  V can be written uniquely as a sum of a vector s1 from S1 and a vector s2

from S2, x = s1 + s2, then V is the direct sum of S1 and S2, and we write V = S1  S2.

Theorem 5.13 Properties of Orthogonal Complements

Let S be a subspace of an n-dimensional vector space V. Then

1) dim(S) + dim(S) = n 2) V = S  S 3) (S) = S

1 0 Examples: Consider R3. {0} = R3. (R3) = {0}. Let S = span(   ), S = span(   ), x 0 y 1 0 0 0 0 S = span(   ), S = span( ,   ), and S = span( , ). Then S is orthogonal to S . z 0 xy 1 yz x y 1 0   Sy is orthogonal to Sz. Sxy is not orthogonal to Syz. Sxy = Sx  Sy. (Sz) = Sxy, and (Sxy) = Sz.

Theorem 5.16 Fundamental Subspaces of a Matrix

Let A be an mn matrix. The notation R(A) means the column space of A (R for range). R(AT ) is the row space of A. N(A) is the nullspace of A.

1) R(AT ) and N(A) are orthogonal complements. That is: all vectors in the row space of A are orthogonal to all vectors in the nullspace of A; dim(R(AT )) + dim(N(A)) = n, and R(AT )  N(A) = Rn. 2) R(A) and N(AT ) are orthogonal complements. That is: all vectors in the column space of A are orthogonal to all vectors in the nullspace of AT; dim(R(A)) + dim(N(AT )) = m, and R(A)  N(AT ) = Rm.

p. 146 Chapter 5: Inner Product Spaces. 5.4 Mathematical Models and Least Squares Analysis (Optional). Sketch of the proof: 1) N(A) ={x  Rn: Ax = 0} shows that the nullspace is orthogonal to the row space, because the null vector x is orthogonal to each row of A.  1 row( of A x 0)   2 row( of A x 0) Ax = 0 is equivalent to     row( of Am x 0) dim(R(AT )) + dim(N(A)) = n is rank + nullity = n, which we already know. The basis vectors of R(AT )) are linear independent of the basis of N(A) because they are orthogonal to the basis of N(A), so the union of a basis of R(AT )) and a basis of N(A) is a basis of Rn, so R(AT )  N(A) = Rn.

2) Replace A with AT in Part (1), and use (AT )T = A. When we replace A with AT, we must interchange m and n.

1 0 2 1  Example: Find the orthogonal complement of S = span(   ,   ). 0 0     0 1  01   12  Solution: S is the column space R(A), where A =   , so S = N(AT).  00     10  00021  02001   rref   To find N(A), solve Ax = 0       01010   01010 

so 1x1 + 0x2 + 0s – 2t = 0 and 0x1 + 1x2 + 0s + 1t = 0 so x2 = –t and x1 = 2t  2t  0  2  0  2   t 0 1 0 1 x =   = s   + t   so S = N(AT) = span(   ,   )  s  1  0  1  0             t  0  1  0  1 

If S is a subspace of an n-dimensional vector space V and v is a vector in V, then v can be written uniquely as the sum of a vector from S and a vector from S  v = s1 + s2, where s1  S and s2  S  because V = S  S . Then s1  S is the orthogonal projection of v onto S, written projS v

p. 147 Chapter 5: Inner Product Spaces. 5.4 Mathematical Models and Least Squares Analysis (Optional).

Theorem: v – projS v is orthogonal to every vector in S.

 Proof: v can be written uniquely as v = + s2, where s2 = v –  S . Since v –  S, v – is orthogonal to every vector in S.

Theorem 5.15 Orthogonal Projection and Distance

Let S be a subspace of an n-dimensional vector space V and v be a vector in V. Then for all u  S, u  , ||v – || < ||v – u||.

In other words, is the vector in S that is closest to v.

Proof: Let all u  S, u  so v – u = (v – ) + ( – u) Now v – is orthogonal to ( – u), so by the Pythagorean Theorem, ||v – u||2 = ||v – ||2 + || – u||2 where || – u|| > 0 because u  . Therefore, ||v – u||2 > ||v – ||2 ||v – u|| > ||v – ||

To solve the least squares problem, we need Ax – b  (R(A)) = N(AT ), so AT(Ax – b) = 0, or

ATAx = ATb “Normal equations”

Solution to Least Squares Problem

p. 148 Chapter 5: Inner Product Spaces. 5.4 Mathematical Models and Least Squares Analysis (Optional). Example: Let’s finish the example from the t y beginning of this section. Given a table of data. We want to find the coefficients c0 and c1 of the line 1 0 y = c + c t that “best fits” these points. 0 1 2 1 Solution: 3 3

cc1001  11  0  11  0 c  c  cc 12 or   0   . Let A =   , b =   , and x = 0 10  21    1  21  1   c1  c1  cc1033  31  3  31  3

1 T T 3 6  c0   4  3 6   5/ 3 So A Ax = A b , i.e.     =   , so =   =   . 6 14 c1  11 6 14  3/ 2 

3 5 Therefore, the least-squares regression line for the data is y = 2 t – 3

 Example: Given the table of data. Find the coefficients 2 t y c0, c1, and c2 of the quadratic y = c0 + c1t + c2t that best fits these points. 0 2

Solution: 1 1.5 2 2.5 1c  0c  02 c  2 1 0 0  2  0 1 2 3 4 2   c0    1c 1c 1 c  1.5 1 1 1 1.5 0 1 2 so   c  =   1c  2c  22 c  2.5 1 2 4  1  2.5 0 1 2   2   c2    1c0  3c1  3 c2  4 1 3 9  4 

 2  1.95  1.5 A = , x = , and b =   . ATAx = ATb so x = (ATA)–1ATb =  0.8 2.5      0.5   4 

y = 1.95 – 0.8t + 0.5t2

p. 149

Chapter 5: Inner Product Spaces. 5.5 Applications of Inner Product Spaces (Optional).

5.5 Applications of Inner Product Spaces (Optional).  Objective: Find the nth-order Fourier approximation of a function.  Objective: Given a subspace with an orthogonal or orthonormal basis, find the projection of a vector onto that subspace.  Objective: Find the cross product of two vectors in R3.

Recall from Thm. 5.15 that projS v is the unique vector in S that is closest to v.

Theorems 5.14 & 5.19 Projection onto a Subspace & Least Squares Approximation

If {u1, u2, …, un} is an orthonormal basis for a subspace S of a vector space V, and v  V, then

= v, u1u1 + v, u2u2 + … + v, unun 

If V is a space of functions, e.g. V = C[a, b] (so the ui are functions), then

projS f =  f, u1u1 +  f, u2u2 + … +  f, unun

is the least-squares approximating function of f with respect to S.

Proof: we will show that = v, u1u1 + v, u2u2 + … + v, unun by showing that v can be written uniquely as the sum of a vector from S and a vector from S  v = s1 + s2, where s1  S and s2  S . Then by definition, = s1.

Let s1 = v, u1u1 + v, u2u2 + … v, unun and s2 = v – v, u1u1 – v, u2u2 – … – v, unun

Then s1  S because it is a linear combination of vectors in S.

For each basis vector ui of S

ui, s2 = ui, (v – v, u1u1 – v, u2u2 – … – v, uiui – … – v, unun) = ui, v – v, u1ui, u1 – v, u2ui, u2 – … – v, uiui, ui – … – v, unui, un = ui, v – 0 – 0 – … – v, ui (1) – … – 0 = 0

Since all vectors in S are linear combinations of the {ui}, s2 is orthogonal to all vectors in S,  so s2  S .

Example: we know from 5.3 that set {1, cos(x), sin(x), cos(2x), sin(2x), …, cos(nx), sin(nx)} is a 2 an orthogonal set using the inner product  f , g =  xgxf )()( dx . Let’s normalize the set. 0

p. 151 Chapter 5: Inner Product Spaces. 5.5 Applications of Inner Product Spaces (Optional).

2 ||1||2 =  (1)(1)dx = 2. 0 2 2 2  x sin(nx)cos(nx) For n  0, ||cos(nx)|| =  cos(nx)cos(nx)dx =    =  0 2 2n  x0 2 2 2  x sin(nx)cos(nx) and ||sin(nx)|| = sin(nx)sin(nx)dx =    =  0 2 2n  x0 So { 1 , 1 cos(x), 1 sin(x), cos(2x), sin(2x), …, cos(nx), sin(nx)} is an 2   orthonormal set.

The nth-order Fourier approximation of a function f on the interval [0, 2] is the projection of f onto span { , cos(x), sin(x), cos(2x), sin(2x), …, cos(nx), sin(nx)}. Fourier approximations are useful in modeling periodic functions such as sound waves, heart rhythms, and electrical signals.

 Example: Find the second-order Fourier approximation to the periodic function

f (x) =  – x for x  [0, 2 ), and f (x + 2) = f (x)

Solution: g(x) =  f,  +  f, cos(x) cos(x) +  f, sin(x) sin(x) +  f, cos(2x) cos(2x) +  f, sin(2x) sin(2x)

Using software to evaluate the integrals, we find 2  f, 1  = 1 1 (  x)dx = 0 2 2    0

2  f, 1 cos(x) = 1 (  x)cos(x)dx = 0    0

2  f, 1 sin(x) = 1 (  x)sin(x)dx = 2    0

2  f, cos(2x) = 1 (  x)cos(2x)dx = 0   0

2  f, sin(2x) = 1 (  x)sin(2x)dx = 1   0 nd So the 2 order Fourier approximation is g2(x) = 2sin(x) + sin(2x)

p. 152 Chapter 5: Inner Product Spaces. 5.5 Applications of Inner Product Spaces (Optional). See http://www.falstad.com/fourier/e-sawtooth.html It is common notation with Fourier series to write the coefficients as 2 2 2 a = 1 xf)(dx a = 1 xf kx)cos()(dx b = 1 xf kx)sin()(dx 0   k   k   0 0 0 so that the nth-order Fourier approximation is

a0 gn(x) = + a1cos(x) + b1sin(x) + a2cos(2x) + b2sin(2x) + … + ancos(nx) + bnsin(nx) 2

2 2 22 2 2 1  Example: Given that S = {(– , , ), ( , – , )} is an orthonormal set in R3. Find 6 6 3 3 3 3

the projection of v = (1, 0, 0) onto span{(– , , ), ( , – , )}.

 Solution:

projS v = ((1, 0, 0)•(– , , ))(– , , )

+ ((1, 0, 0)• ( , – , ))( , – , )

1 1 = – (– , , ) + ( , – , ) = ( , – , 0) 2 2

3 In R , the cross product of two vectors u = (u1, u2, u3) = u1i + u2j + u3k and v = (v1, v2, v3) = v1i + v2j + v3k is kji

uv = (u2v3 – u3v2)i + (u3v1 – u1v3)j + (u1v2 – u2v1)k = = uuu 321

vvv 321 where the right-hand side is a “determinant” containing the vectors i, j, and k. The cross product is undefined for vectors in vector spaces other than R3.

Theorem 5.17 Algebraic Properties of the Cross Product

If u, v, and w are vectors in R3 and c is a scalar, then

p. 153 Chapter 5: Inner Product Spaces. 5.5 Applications of Inner Product Spaces (Optional).

1) uv = –vu uuu321 2) u(v + w) = uv + uw 6) u•(vw) = (uv)•w = vvv 3) c(uv) = (cu)v = u(cv) 321 www 4) u0 = 0u = 0 321 5) uu = 0

p. 154 Chapter 5: Inner Product Spaces. 5.5 Applications of Inner Product Spaces (Optional). Theorem 5.18 Geometric Properties of the Cross Product

If u, v, and w are nonzero vectors in R3, then

1) uv is orthogonal both u and v  2) The angle  between u and v is given by ||uv|| = ||u|| ||v|| sin( ) 3) u and v are parallel if and only if uv = 0 4) The parallelogram having u and v as adjacent sides has an area of ||uv|| 5) The parallelepiped having u, v, and w as edges has a volume of ||u•(vw)||

 Example: Find the area of the parallelogram with vertices at (5, 2, 0), (3, –6, 7), (7, –2, 8), and (5, –10, 15)

kji  Solution: area = ||uv||. uv =  782 = –36 + 30j + 24k  842

||–36i + 30j + 24k || = ( 36 2 () 30 2  () 24)2 = 6 77  52.6498

 Example: Find a vector orthogonal to u = (–4, 3, –4) and v = (4, 7, 1).

 kji Solution: uv =  434 = 31i – 12j – 40k 174

p. 155

8.3 Polar Form and De Moivre's Theorem. (Optional)

8.3 Polar Form and De Moivre's Theorem. (Optional)  Objective: Determine the polar form of a complex number and convert between the polar form and the standard form of a complex number.  Objective: Determine the exponential polar form of a complex number and convert between the exponential polar form and the standard form of a complex number using Euler’s formula.  Objective: Multiply and divide complex numbers in polar form and in exponential polar form.  Objective: Use DeMoivre’s Theorem to find powers of numbers in polar form and in exponential polar form.

Multiplication, division, and powers of complex number are tedious to calculate in standard form. However, when complex numbers are written in polar form, multiplication, division, and powers become easy to calculate and interpret.

The polar form of a nonzero complex number z = a + bi is given by

z = r (cos + i sin )  where a = r cos, b = r sin, r =  ba 22 = |z|, and tan = b/a.

The number r is the modulus of z and  is the argument of z.

There are infinitely many choices for the argument, because cosine and have a period of 2. Usually, we choose the principal argument, which satisfies  <   2.

The principal argument  = Arg(z) of a nonzero complex number z = a + bi is given by

tan = b/a and – <   

 Example: Graph and find the polar form (using the Arg(z) =  /2 principal argument) of

axis  a) –3 b) –4 – 3i c) 2 – 2 i Arg(z) = tan–1(b/a) +  Arg(z) = tan–1(b/a) nary

Solution: Imagi Real axis Arg(z) =  Arg(z) = 0

Arg(z) = tan–1(b/a) –  Arg(z) = tan–1(b/a)

a) Arg(z) = – /2

c)

b)

p. 157 8.3 Polar Form and De Moivre's Theorem. (Optional) a) –3 = 3(cos + i sin)

b) r = 2 )3()4(2 = 25 = 5

–1 3 Arg(z) =  + tan ( 4 )  3.3785 –4 – 3i  5(cos(3.3785) + i sin(3.3785))

c) r = 2  )2()2(2 = 4 = 2 Arg(z) = tan–1(  2 ) =  2  4 2 – 2 i = 2(cos( ) + i sin( ))

   Example: Find the standard form of 10 (cos( 6 ) + i sin( 6 ))  3 1   Solution: 10 (cos(  ) + i sin( )) = 10   i =  535i 6    2 2 

 Theorem. Euler’s Formula

 cos + i sin = ei

where e is Euler’s number, e  2.71828

Proof: From calculus, we know that the Maclaurin series (the Taylor series expansion around zero) of a function f is

1 1 1 1 1 1 f (x) = f (0) + f (0) x + f (0) x2 + f (0) x3 + f (4)(0) x4 + f (5)(0) x5 + … !0 !1 !2 !3 !4 !5

Therefore, the Maclaurin series for ex is

1 1 1 1 1 1 1 x 1 2 3 4 5 6 xxxxxxxe 7  ... !1 !2 !3 !4 !5 !6 !7

Substitute x = i. Note that i2 = –1, i3 = –i, i4 = 1, i5 = i, etc. So

1 1 1 1 1 1 1 i 1 ie 2 i 3 4 i 5 6 i  7  ... !1 !2 !3 !4 !5 !6 !7

Therefore, the Maclaurin series for cos and sin are

1 1 1  01)cos(  2 0  4 0  6 0  ... !2 !4 !6

1 1 1 1  0)sin(  0  3 0  5 0  7  ... !1 !3 !5 !7

p. 158 8.3 Polar Form and De Moivre's Theorem. (Optional) Comparing the series for ei, cos, sin, we see that ei = cos + i sin

 Example: Graph and find the polar form (using the principal argument) of a) –i b) 1 – 3 i c) 3 + 2i c)

Solution

a) –i = e–i /2 a) b) r = 2  )3(12 = 4 = 2 –1  = tan (– /1) = – /3 b) 1 – i = 2 e–i /3

c) r =  2322 = 13  3.606  = tan–1(2/3)  0.588 3 + 2i = 3.606 e0.588i d) Example: Graph and find the polar form (using the principal argument) of a) –i b) 1 – i c) 3 + 2i

 Example: Find the standard form of 3e5i/4

5 5  2 2  23 23 5i/4       Solution: 3e = 3 (cos   + i sin   ) = 3   i =  i  4   4   2 2  2 2

 Theorem 8.4 Product and Quotient of Two Complex Nubmers

er i1 er i2 = err i  21 )(  irir  )sin(cos)sin(cos = rr isin()[cos(   )]  1 2 21 11 221 2 2121 21

er i1 (cos  ir  )sin r 1 r1 i  21 )( 11 1 1 = e = [cos( 21 isin()   21 )] , z2  0 i2 2er r2 2 (cos 2  ir  2 )sin r2

Proof: The exponential formulas follow directly from the laws of exponents. The polar forms follow from the exponential polar forms and Euler’s formula.

 Example: Sketch and simplify.

3ei  i sin()[cos(3  )] i/3 i/2      a) 2e 3e = 2(cos 3 + i sin 3 )3(cos 2 + i sin 2 ) b)  i 4/3 3 3 4e 4  i sin()[cos(4  4  )]

p. 159 8.3 Polar Form and De Moivre's Theorem. (Optional) Solution:

a) 2ei/33ei/2 5 /6 =  /2 +  /3 i(/3+/2) = (2)(3)e 6 3

= 6ei5/6  /3 2  /2  /3     2(cos 3 + i sin 3 )3(cos 2 + i sin 2 )

5 5 = 6[cos( 6  ) + i sin( 6  )]

 /2 = 90,  /3 = 60, = 150

3ei 3 3 b) = ei  4/3(()) = ei  4/7 4e  i 4/3 4 4

3 3  7 /4 =  –(–3 /4) = ei   )4/72( = ei 4/ 4 4

 isin()[cos(3  )] 7 /4 3 3 3 4  isin()[cos(4  4  )]

3 ¾ – /4 = 2 – 7 /4 7  isin()[cos( 7  )] 4 4 4

3  i   4/sin()4/[cos( )] 4 –3 /4

4

 Theorem 8.5 DeMoivre’s Theorem

P (rei )n = r nein

[r (cos + i sin )]n = r n [cos(n ) + i sin(n )]

Proof: The exponential formula follows directly from the laws of exponents. The polar forms follow from the exponential polar forms and Euler’s formula.

p. 160 8.3 Polar Form and De Moivre's Theorem. (Optional)  Example: Sketch and simplify.

a) ei2k/3 for k = –1, 0, 1, 2, 3, 4

b) eik/2 for k = –2, –1, 0, 1, 2, 3, 4, 5

Solution

i(–1)/3 1 3 –i/3 a) e = – 2 – 2 i = e

ei(0)/3 = 1 = e0

i(1)/3 3 i/3 e = – + 2 i = e

ei(2)/3 = – – i = e–i/3

ei(3)/3 = 1 = e0

ei(4)/3 = – + i = ei/3

b) ei(–2)/2 = e–i = –1 = ei

ei(–1)/2 = –i = e–i/2

ei(0)/2 = 1 = e0

ei(1)/2 = i = ei/2

ei(2)/2 = –1 = ei

ei(3)/2 = –i = e–i/2

ei(4)/2 = 1 = e0

ei(5)/2 = i = ei/2

p. 161

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404 Chapter 8 Complex Vector Spaces

8.3 Polar Form and DeMoivre’s Theorem

Determine the polar form of a complex number, convert between the polar form and standard form of a complex number, and multiply and divide complex numbers in polar form. Use DeMoivre’s Theorem to find powers and roots of complex numbers in polar form.

POLAR FORM OF A COMPLEX NUMBER Imaginary axis At this point you can add, subtract, multiply, and divide complex numbers. However, there is still one basic procedure that is missing from the algebra of complex numbers. (a, b) To see this, consider the problem of finding the square root of a complex number such as i. When you use the four basic operations (addition, subtraction, multiplication, and division), there seems to be no reason to guess that r b 1 i Ίi . θ Ί2 Real a 0 axis That is, 1 i 2 i. ΂ Ί ΃ Standard Form: a + bi 2 θθ Polar Form: r(cos + i sin ) To work effectively with powers and roots of complex numbers, it is helpful to use a Figure 8.7 polar representation for complex numbers, as shown in Figure 8.7. Specifically, if a bi is a nonzero complex number, then let be the angle from the positive real axis to the radial line passing through the point ͑a, b͒ and let r be the modulus of a bi. This leads to the following. a r cos b r sin r Ίa2 b2 So,a bi ͑r cos ͒ ͑r sin ͒i, from which the polar form of a complex number REMARK is obtained. The polar form of z 0 is expressed as Definition of the Polar Form of a Complex Number z 0͑cos i sin ͒, where is any angle. The polar form of the nonzero complex number z a bi is given by z r͑cos i sin ͒ where a r cos , b r sin , r Ίa2 b2, and tan b͞a. The number r is the modulus of z and is the argument of z.

Because there are infinitely many choices for the argument, the polar form of a complex number is not unique. Normally, the values of that lie between and are used, although on occasion it is convenient to use other values. The value of that satisfies the inequality < Principal argument is called the principal argument and is denoted by Arg(z ). Two nonzero complex numbers in polar form are equal if and only if they have the same modulus and the same principal argument. 9781133110873_0803.qxp 3/10/12 6:54 AM Page 405

8.3 Polar Form and DeMoivre’s Theorem 405

Finding the Polar Form of a Complex Number

Find the polar form of each of the complex numbers. (Use the principal argument.) a.z 1 i b.z 2 3i c. z i SOLUTION a. Because a 1 and b 1, then r 2 12 ͑1͒2 2, which implies that r Ί2. From a r cos and b r sin , a 1 Ί2 b 1 Ί2 cos and sin . r Ί2 2 r Ί2 2 So, ͞4 and z Ί2 ΄cos ΂ ΃ i sin΂ ΃΅. 4 4 b. Because a 2 and b 3, then r2 22 32 13, which implies that r Ί13. So, a 2 b 3 cos and sin r Ί13 r Ί13 and it follows that Ϸ 0.98. So, the polar form is

z Ϸ Ί13 ΄cos͑0.98͒ i sin͑0.98͒΅.

c. Because a 0 and b 1, it follows that r 1 and ͞2, so z 1΂cos i sin ΃. 2 2 The polar forms derived in parts (a), (b), and (c) are depicted graphically in Figure 8.8.

Imaginary Imaginary axis axis

2 4 z = 2 + 3i 1 3 Real 2 θ axis −2 −1 2 − 1 1 z = 1 − i θ Real −2 axis 12 a.z Ί2 ΄cos΂ ΃ i sin΂ ΃΅ b. z Ϸ Ί13 ͓cos͑0.98͒ i sin͑0.98͔͒ 4 4

Imaginary axis

z = i 1

θ Real axis 1

c. z 1΂cos i sin ΃ 2 2 Figure 8.8 9781133110873_0803.qxp 3/10/12 6:54 AM Page 406

406 Chapter 8 Complex Vector Spaces

Converting from Polar to Standard Form

Express the complex number in standard form. z 8΄cos΂ ΃ i sin΂ ΃΅ 3 3 SOLUTION Because cos͑͞3͒ 1͞2 and sin͑͞3͒ Ί3͞2, obtain the standard form 1 Ί3 z 8΄cos΂ ΃ i sin΂ ΃΅ 8 i 4 4Ί3i. 3 3 ΄2 2 ΅

The polar form adapts nicely to multiplication and division of complex numbers. Suppose you have two complex numbers in polar form ͑ ͒ ͑ ͒ z1 r1 cos 1 i sin 1 and z2 r2 cos 2 i sin 2 .

Then the product of z1 and z2 is expressed as ͑ ͒͑ ͒ z1z 2 r1r2 cos 1 i sin 1 cos 2 i sin 2 ͓͑ ͒ ͑ ͔͒ r1r2 cos 1 cos 2 sin 1 sin 2 i cos 1 sin 2 sin 1 cos 2 . ͑ ͒ Using the trigonometric identities cos 1 2 cos 1 cos 2 sin 1 sin 2 and ͑ ͒ sin 1 2 sin 1 cos 2 cos 1 sin 2, you have ͓ ͑ ͒ ͑ ͔͒ z1z2 r1r2 cos 1 2 i sin 1 2 . This establishes the first part of the next theorem. The proof of the second part is left to you. (See Exercise 75.)

THEOREM 8.4 Product and Quotient of Two Complex Numbers Given two complex numbers in polar form ͑ ͒ ͑ ͒ z1 r1 cos 1 i sin 1 and z2 r2 cos 2 i sin 2 the product and quotient of the numbers are as follows. ͓ ͑ ͒ ͑ ͔͒ z1z2 r1r2 cos 1 2 i sin 1 2 Product z r 1 1 ͓ ͑ ͒ ͑ ͔͒ cos 1 2 i sin 1 2 , z2 0 Quotient z2 r2

LINEAR Elliptic curves are the foundation for elliptic curve ALGEBRA cryptography (ECC), a type of public key cryptography for APPLIED secure communications over the Internet. ECC has gained popularity due to its computational and bandwidth advantages over traditional public key algorithms. One specific variety of elliptic curve is formed using Eisenstein integers. Eisenstein integers are complex 1 Ί3 numbers of the form z a b, where i, 2 2 and and ba are integers. These numbers can be graphed as intersection points of a triangular lattice in the complex plane. Dividing the complex plane by the lattice of all Eisenstein integers results in an elliptic curve. thumb/Shutterstock.com 9781133110873_0803.qxp 3/10/12 6:54 AM Page 407

8.3 Polar Form and DeMoivre’s Theorem 407

Theorem 8.4 says that to multiply two complex numbers in polar form, multiply moduli and add arguments. To divide two complex numbers, divide moduli and subtract arguments. (See Figure 8.9.)

Imaginary Imaginary axis axis

z1z2 z2 z1 z2 θ θ z 1 + 2 r r r 1 2 z 2 1 z r r θ r1 1 θ r 2 1 2 2 r1 θ 2 2 1 θ − θ θ 1 2 1 Real Real axis axis

To multiply z1 and z 2 : To divide z12 and z : Multiply moduli and add arguments. Divide moduli and subtract arguments. Figure 8.9

Multiplying and Dividing in Polar Form

͞ Find z1z2 and z1 z2 for the complex numbers 1 z 5΂cos i sin ΃ and z ΂cos i sin ΃. 1 4 4 2 3 6 6 SOLUTION

Because z1 and z2 are in polar form, apply Theorem 8.4, as follows. multiply

1 5 5 5 z z ͑5͒΂ ΃ ΄cos΂ ΃ i sin΂ ΃΅ ΂cos i sin ΃ 1 2 3 4 6 4 6 3 12 12

add add divide z 5 1 ͞ ΄cos΂ ΃ i sin΂ ΃΅ 15΂cos i sin ΃ z2 1 3 4 6 4 6 12 12 subtract subtract

Use the standard forms of z1 and z2 to check the multiplication in Example 3. For instance, 5Ί2 5Ί2 Ί3 1 5Ί6 5Ί2 5Ί6 5Ί2 z z ΂ i΃΂ i΃ i i i2 1 2 2 2 6 6 12 12 12 12 5Ί6 5Ί2 5Ί6 5Ί2 i 12 12 5 Ί6 Ί2 Ί6 Ί2 ΂ i΃. REMARK 3 4 4 Try verifying the division in To verify that this answer is equivalent to the result in Example 3, use the formulas for Example 3 using the standard cos ͑u v͒ and sin ͑u v͒ to obtain forms of z1 and z2. 5 Ί6 Ί2 5 Ί6 Ί2 cos΂ ΃ cos΂ ΃ and sin΂ ΃ sin΂ ΃ . 12 6 4 4 12 6 4 4 9781133110873_0803.qxp 3/10/12 6:54 AM Page 408

408 Chapter 8 Complex Vector Spaces

DEMOIVRE’S THEOREM The final topic in this section involves procedures for finding powers and roots of complex numbers. Repeated use of multiplication in the polar form yields z r ͑cos i sin ͒ z2 r ͑cos i sin ͒r ͑cos i sin ͒ r 2͑cos 2 i sin 2͒ and z3 r ͑cos i sin ͒r 2 ͑cos 2 i sin 2͒ r3͑cos 3 i sin 3͒. Similarly, z4 r 4͑cos 4 i sin 4͒ and z 5 r 5͑cos 5 i sin 5͒. This pattern leads to the next important theorem, named after the French mathematician Abraham DeMoivre (1667Ð1754). You are asked to prove this theorem in Review Exercise 85.

THEOREM 8.5 DeMoivre’s Theorem If z r͑cos i sin ͒ and n is any positive integer, then zn r n͑cos n i sin n͒.

Raising a Complex Number to an Integer Power 12 Find ͑1 Ί3 i͒ and write the result in standard form. SOLUTION First convert to polar form. For 1 Ί3i, Ί3 r Ί͑1͒2 ͑Ί3͒2 2 and tan Ί3 1 which implies that 2͞3. So, 2 2 1 Ί3i 2΂cos i sin ΃. 3 3 By DeMoivre’s Theorem, 2 2 12 ͑1 Ί3i͒12 ΄2΂cos i sin ΃΅ 3 3 12͑2͒ 12͑2͒ 212΄cos i sin ΅ 3 3 4096͑cos 8 i sin 8͒ 4096͓1 i͑0͔͒ 4096. 9781133110873_0803.qxp 3/10/12 6:54 AM Page 409

8.3 Polar Form and DeMoivre’s Theorem 409

Recall that a consequence of the Fundamental Theorem of Algebra is that a polynomial of degree n has n zeros in the complex number system. So, a polynomial such as p͑x͒ x6 1 has six zeros, and in this case you can find the six zeros by factoring and using the Quadratic Formula. x6 1 ͑x3 1͒͑x3 1͒ ͑x 1͒͑x 2 x 1͒͑x 1͒͑x 2 x 1͒ Consequently, the zeros are 1 ± Ί3i 1 ± Ί3i x ±1, x , and x . 2 2 Each of these numbers is called a sixth root of 1. In general, the n th root of a complex number is defined as follows.

Definition of the nth Root of a Complex Number The complex number w a bi is an nth root of the complex number z when z wn ͑a bi͒n.

DeMoivre’s Theorem is useful in determining roots of complex numbers. To see how this is done, let w be an n th root of z, where w s͑cos i sin ͒ and z r͑cos i sin ͒. Then, by DeMoivre’s Theorem wn sn͑cos n i sin n͒ and because wn z, it follows that sn ͑cos n i sin n͒ r͑cos i sin ͒. Now, because the right and left sides of this equation represent equal complex numbers, equate moduli to obtain sn r, which implies that s Ίn r, and equate principal arguments to conclude that and n must differ by a multiple of 2. Note that r is a positive real number and so s Ίn r is also a positive real number. Consequently, for some integer k, n 2k, which implies that 2k . n REMARK Finally, substituting this value of into the polar form of w produces the result stated Note that when k exceeds in the next theorem. n 1, the roots begin to repeat. For instance, when k n, the angle is THEOREM 8.6 The nth Roots of a Complex Number 2n 2 For any positive integer n, the complex number z r͑cos i sin ͒ has n n exactly n distinct roots. These n roots are given by which yields the same values 2k 2k for the sine and cosine as Ίn r ΄cos΂ ΃ i sin΂ ΃΅ k 0. n n where k 0, 1, 2, . . . , n 1. 9781133110873_0803.qxp 3/10/12 6:54 AM Page 410

410 Chapter 8 Complex Vector Spaces

Imaginary The formula for the n th roots of a complex Imaginary axis axis number has a nice geometric interpretation, as 1 3 1 3 − + i + i shown in Figure 8.10. Because the n th roots all 2 2 2 2 have the same modulus (length) Ίn r, they lie on a π 2 Ίn n circle of radius r with center at the origin. n π r 2 Furthermore, the n roots are equally spaced − n Real 1 1 Real axis around the circle, because successive n th roots axis have arguments that differ by 2͞n. You have already found the sixth roots of 1 by factoring and the Quadratic Formula. Try solving 1 3 1 3 the same problem using Theorem 8.6 to get the − − i − i 2 2 2 2 roots shown in Figure 8.11. When Theorem 8.6 is The nth Roots of a The Sixth Roots of Unity Complex Number applied to the real number 1, the n th roots have a special name—the nth roots of unity. Figure 8.11 Figure 8.10 Finding the nth Roots of a Complex Number

Determine the fourth roots of i. SOLUTION In polar form, i 1΂cos i sin ΃ 2 2 so r 1 and ͞2. Then, by applying Theorem 8.6, ͞2 2k ͞2 2k i1͞4 Ί4 1 ΄cos΂ ΃ i sin΂ ΃΅ 4 4 4 4 k k cos΂ ΃ i sin΂ ΃. 8 2 8 2 Setting k 0, 1, 2, and 3, z cos i sin 1 8 8 5 5 z cos i sin 2 8 8 9 9 z cos i sin 3 8 8 13 13 z cos i sin 4 8 8 REMARK as shown in Figure 8.12. In Figure 8.12, note that when Imaginary each of the four angles π π axis cos 5 + i sin 5 ͞8, 5͞8, 9͞8, and 13͞8 is 8 8 multiplied by 4, the result is of π π the form ͑͞2͒ 2k. cos + i sin 8 8 Real axis π π cos 9 + i sin 9 8 8 π π cos 13 + i sin 13 8 8 Figure 8.12 9781133110873_0803.qxp 3/10/12 6:54 AM Page 411

8.3 Exercises 411

8.3 Exercises

Converting to Polar Form In Exercises 1Ð4, express Multiplying and Dividing in Polar Form In Exercises the complex number in polar form. 27Ð34, perform the indicated operation and leave the 1.Imaginary 2. Imaginary result in polar form. axis axis 1 + 3i 27. 3 cos i sin 4 cos i sin Real 3 ΄ ΂ ΃΅΄ ΂ ΃΅ axis 3 3 6 6 1 2 3 2 28. ΄ ΂cos i sin ΃΅΄6΂cos i sin ΃΅ −1 4 2 2 4 4 1 29. ͓0.5͑cos i sin͔͓͒ 0.5͑cos͓͔ i sin͓͔͔͒ −2 Real 2 − 2i axis 1 2 2 −1 12 30. ΄3΂cos i sin ΃΅΄ ΂cos i sin ΃΅ 3 3 3 3 3 3.Imaginary 4. Imaginary axis axis 2͓cos͑2͞3͒ i sin͑2͞3͔͒ 3 3i 31. 3 4[cos͑5͞6͒ i sin͑5͞6͒] 2 cos͑5͞3͒ i sin͑5͞3͒ −6 1 2 Real 32. axis cos i sin −6 −5 −4 −3 −2 1 12͓cos͑͞3͒ i sin͑͞3͔͒ −2 Real 33. − axis ͓ ͑͞ ͒ ͑͞ ͔͒ −3 1 1 3 cos 6 i sin 6 9͓cos͑3͞4͒ i sin͑3͞4͔͒ 34. Graphing and Converting to Polar Form In Exercises 5͓cos͑͞4͒ i sin͑͞4͔͒ 5Ð16, represent the complex number graphically, and give the polar form of the number. (Use the principal Finding Powers of Complex Numbers In Exercises argument.) 35Ð44, use DeMoivre’s Theorem to find the indicated 5.2 2i 6. 2 2i powers of the complex number. Express the result in standard form. 7.2͑1 Ί3i ͒ 8. 5͑Ί3 i͒ 2 35.͑1 i͒4 36. ͑2 2i͒6 9.6i 10. 2i 37.͑1 i͒10 38. ͑Ί3 i͒7 11. 7 12. 4 39. ͑1 Ί3i͒3 13.3 Ί3i 14. 2Ί2 i 3 15.1 2i 16. 5 2i 40. ΄5΂cos i sin ΃΅ 9 9 Graphing and Converting to Standard Form In 5 5 4 41. ΄3΂cos i sin ΃΅ Exercises 17Ð26, represent the complex number 6 6 graphically, and give the standard form of the number. 5 5 10 42. ΂cos i sin ΃ 17. 2΂cos i sin ΃ 4 4 2 2 8 3 3 43. ΄2΂cos i sin ΃΅ 18. 5΂cos i sin ΃ 2 2 4 4 3 3 4 3 5 5 44. ΄5΂cos i sin ΃΅ 19. ΂cos i sin ΃ 2 2 2 3 3 3 7 7 Finding Square Roots of a Complex Number In 20. ΂cos i sin ΃ Exercises 45Ð52, find the square roots of the complex 4 4 4 number. 15 21.΂cos i sin ΃ 22. 8΂cos i sin ΃ 45.2i 46. 5i 4 4 4 6 6 47.3i 48. 6i 3 3 5 5 23.4΂cos i sin ΃ 24. 6΂cos i sin ΃ 49.2 2i 50. 2 2i 2 2 6 6 51.1 Ί3i 52. 1 Ί3i 25.7͑cos 0 i sin 0͒ 26. 9͑cos i sin ͒ 9781133110873_0803.qxp 3/10/12 6:54 AM Page 412

412 Chapter 8 Complex Vector Spaces

Finding and Graphing nth Roots In Exercises 53Ð64, 75. Proof When provided with two complex numbers ͑ ͒ ͑ ͒ (a) use Theorem 8.6 to find the indicated roots, (b) z 1 r1 cos 1 i sin 1 and z 2 r2 cos 2 i sin 2 , represent each of the roots graphically, and (c) express with z 2 0, prove that each of the roots in standard form. z r 1 1 ͓ ͑ ͒ ͑ ͔͒ cos 1 2 i sin 1 2 . z 2 r2 53. Square roots: 16΂cos i sin ΃ 3 3 76. Proof Show that the complex conjugate of ͑ ͒ 2 2 z r cos i sin is 54. Square roots: 9΂cos i sin ΃ ͓ ͑͒ ͔͑͒ 3 3 z r cos i sin . 4 4 77. Use the polar forms of z and z in Exercise 76 to find 55. Fourth roots: 16΂cos i sin ΃ each of the following. 3 3 (a) zz 5 5 56. Fifth roots: 32΂cos i sin ΃ (b) z͞z, z 0 6 6 78. Proof Show that the negative of z r͑cos i sin ͒ 57. Square roots: 25i is 58. Fourth roots: 625i z r͓cos͑ ͒ i sin͑ ͔͒. 59. Cube roots: 125 ͑1 Ί3i͒ 2 79. Writing Ί ͑ ͒ 60. Cube roots: 4 2 1 i 61. Cube roots: 8 (a) Let z r͑cos i sin ͒ 2΂cos i sin ΃. 6 6 62. Fourth roots: 81i Sketch z, iz, and z͞i in the complex plane. 63. Fourth roots: 1 (b) What is the geometric effect of multiplying a 64. Cube roots: 1000 complex number z by i? What is the geometric effect of dividing z by i? Finding and Graphing Solutions In Exercises 65Ð72, 80. Calculus Recall that the Maclaurin series for e x, find all the solutions of the equation and represent your sin , and cos xx are solutions graphically. x 2 x3 x 4 65.x 4 256i 0 66. x4 16i 0 ex 1 x . . . 2! 3! 4! 67.x 3 1 0 68. x3 27 0 x 3 x5 x 7 69.x 5 243 0 70. x 4 81 0 sin x x . . . 3! 5! 7! 71.x 3 64i 0 72. x 4 i 0 x 2 x 4 x6 cos x 1 . . . . 73. Electrical Engineering In an electric circuit, the 2! 4! 6! formula V I Z relates voltage drop V, current I, and (a) Substitute x i in the series for ex and show that impedance Z, where complex numbers can represent ei cos i sin . each of these quantities. Find the impedance when the (b) Show that any complex number z a bi can be voltage drop is i and the current is i 5 5 2 4 . expressed in polar form as z rei . (c) Prove that if z rei, then z rei. 74. Use the graph of the roots of a (d) Prove the formula ei 1. complex number. (a) Write each of the roots in trigonometric form. True or False? In Exercises 81 and 82, determine (b) Identify the complex number whose roots are whether each statement is true or false. If a statement is given. Use a graphing utility to verify your results. true, give a reason or cite an appropriate statement from the text. If a statement is false, provide an example that (i)Imaginary (ii) Imaginary axis axis shows the statement is not true in all cases or cite an appropriate statement from the text. 33 81. Although the square of the complex number bi is given 22 ° ° 2 2 30° 30° Real 45 45 Real by ͑bi͒ b , the absolute value of the complex axis axis Ί 2 2 2 1 45° 45° number z a bi is defined as Խa biԽ a b . −1 33 82. Geometrically, the n th roots of any complex number z are all equally spaced around the unit circle centered at the origin. 9781133110873_08_ANS.qxd 3/10/12 6:58 AM Page A2

Answer Key

13. Imaginary Section 8.3 axis ␲ ␲ 4 1. 8΄cos΂Ϫ ΃ ϩ i sin΂Ϫ ΅ 3. 6͑cos ␲ ϩ i sin ␲͒ z = 3 + 3i 4 4 2 5. Imaginary 1 axis Real axis −4 −1 12 4 −2 2 1 −4 Real axis ␲ ␲ 12 2Ί3΂cos ϩ i sin ΃ z = −2 − 2i 6 6 15. Imaginary axis 3␲ 3␲ 4 Ί8΄cos΂Ϫ ΃ ϩ i sin΂Ϫ ΃΅ 3 4 4 Imaginary 1 7. Real axis axis −3 −1 143

3 z = −1 − 2i 2 −4 1 Real Ί5 ͓cos͑Ϫ2.0344͒ ϩ i sin͑Ϫ2.0344͔͒ − − axis 3 2 123 17. Imaginary −2 axis −3 θ π = 2 z = −2 − 2 3i 1 r = 2 2␲ 2␲ Real 4΄cos΂Ϫ ΃ ϩ i sin΂Ϫ ΃΅ axis 3 3 −1 1 −1 9. Imaginary axis z = 6i 2i 4 19. Imaginary 2 axis Real axis −4 −2 24 2 3 −4 1 r = 2 Real axis ␲ ␲ −2 12 6΂cos ϩ i sin ΃ 2 2 5π − θ = 11. Imaginary 2 3 axis 3 3Ί3 Ϫ i 6 4 4 3 21. Imaginary axis Real axis π −3 36 θ = −3 3 4 z = 7 2 − 1 6 r = 3.75 Real − axis 7͑cos 0 ϩ i sin 0͒ 2 123 −2 −3

15Ί2 15Ί2 ϩ i 8 8 9781133110873_08_ANS.qxd 3/10/12 6:58 AM Page A3

Answer Key

23. Imaginary ␲ ␲ axis 51. 1 ϩ Ί3i ϭ 2΂cos ϩ i sin ΃ 3 3 Square roots: 3 ␲ ␲ Ί6 Ί2 2 r = 4 Ί2΂cos ϩ i sin ΃ ϭ ϩ i 1 Real 6 6 2 2 axis −3−2−1 123 7␲ 7␲ Ί6 Ί2 Ί2΂cos ϩ i sin ΃ ϭϪ Ϫ i −2 6 6 2 2 −3 π ␲ ␲ θ = 3 53. (a) 4΂cos ϩ i sin ΃ 2 6 6 Ϫ4i 7␲ 7␲ 4΂cos ϩ i sin ΃ 25. Imaginary 6 6 axis (b) Imaginary axis θ = 0 4 π 2 r = 7 θ = Real 2 1 6 axis r = 4 −4−2 246 Real −4 − axis − 2 6 −2 θ 7π 7 2 = 6 ␲ ␲ Ί ϩ i 27.12΂cos ϩ i sin ΃ 29. 0.25͑cos 0 ϩ i sin 0͒ (c) 2 3 2 2 2 Ϫ2Ί3 Ϫ 2i 1 ␲ ␲ ␲ ␲ 31. ΄cos΂Ϫ ΃ ϩ i sin΂Ϫ ΃΅ 55. (a) 2΂cos ϩ i sin ΃ 2 6 6 3 3 ␲ ␲ 5␲ 5␲ 33.4΂cos ϩ i sin ΃ 35.Ϫ4 37. Ϫ32i 2΂cos ϩ i sin ΃ 6 6 6 6 81 81Ί3 4␲ 4␲ 39.Ϫ8 41.Ϫ Ϫ i 43. 256 2΂cos ϩ i sin ΃ 2 2 3 3 ␲ ␲ 11␲ 11␲ 45. 2i ϭ 2΂cos ϩ i sin ΃ 2΂cos ϩ i sin ΃ 2 2 6 6 Square roots: (b) Imaginary ␲ ␲ axis Ί ϩ ϭ ϩ 5π 2΂cos i sin ΃ 1 i θ = θ π 4 4 2 6 1= 3 5␲ 5␲ Ί2΂cos ϩ i sin ΃ ϭϪ1 Ϫ i 1 4 4 r = 2 Real ␲ ␲ axis Ϫ ϭ 3 ϩ 3 −1 1 47. 3i 3΂cos i sin ΃ − 2 2 1 θ 11π 4 = 6 Square roots: π θ = 4 3␲ 3␲ Ί6 Ί6 3 3 Ί3΂cos ϩ i sin ΃ ϭϪ ϩ i 4 4 2 2 (c) 1 ϩ Ί3i ϪΊ ϩ 7␲ 7␲ Ί6 Ί6 3 i Ί3΂cos ϩ i sin ΃ ϭ Ϫ i 4 4 2 2 Ϫ1 Ϫ Ί3i Ί Ϫ 7␲ 7␲ 3 i 49. 2Ϫ2i ϭ 2Ί2΂cos ϩ i sin ΃ 4 4 Square roots: 7␲ 7␲ 81͞4΂cos ϩ i sin ΃ ϭϪ1.554 ϩ 0.644i 8 8 15␲ 15␲ 81͞4΂cos ϩ i sin ΃ ϭ 1.554 Ϫ 0.644i 8 8 9781133110873_08_ANS.qxd 3/10/12 6:58 AM Page A4

Answer Key

3␲ 3␲ 63. (a) cos 0 ϩ i sin 0 57. (a) 5΂cos ϩ i sin ΃ ␲ ␲ 4 4 cos ϩ i sin 2 2 7␲ 7␲ 5΂cos ϩ i sin ΃ cos ␲ ϩ i sin ␲ 4 4 ␲ ␲ 3 ϩ 3 (b) Imaginary cos i sin axis 2 2 (b) Imaginary θ = 3π 6 axis 1 4 4 π 1 θ = 2 r = 5 2 2 Real θ = 0 − − axis 1 6 2 246 r = 1 Real axis −4 θ 7π 1 2 = −6 4 θ = π π 3 −1 θ = 3 5Ί2 5Ί2 4 2 (c) Ϫ ϩ i 2 2 (c) 1 5Ί2 5Ί2 i Ϫ i 2 2 Ϫ1 Ϫ 4␲ 4␲ i 59. (a) 5΂cos ϩ i sin ΃ ␲ ␲ 9 9 65. 4΂cos ϩ i sin ΃ ␲ ␲ 8 8 10 ϩ 10 5΂cos i sin ΃ 5␲ 5␲ 9 9 4΂cos ϩ i sin ΃ 16␲ 16␲ 8 8 5΂cos ϩ i sin ΃ 9␲ 9␲ 9 9 4΂cos ϩ i sin ΃ 8 8 (b) Imaginary axis 13␲ 13␲ 4΂cos ϩ i sin ΃ θ = 4π 8 8 6 1 9 4 Imaginary axis 2 r = 5 θ 5π Real 2 = 8 − − axis π 6 2 246 θ = 2 1 8 −4 θ = 16π r = 4 θ 10π 3 9 Real 2 = axis 9 −2 2 (c) 0.868 ϩ 4.92i θ 9π −2 3 = 8 Ϫ4.70 Ϫ 1.71i π θ = 13 3.83 Ϫ 3.21i 4 8 ␲ ␲ 61. (a) 2͑cos 0 ϩ i sin 0͒ 67. cos ϩ i sin 2␲ 2␲ 3 3 2΂cos ϩ i sin ΃ cos ␲ ϩ i sin ␲ 3 3 5␲ 5␲ 4␲ 4␲ cos ϩ i sin 2΂cos ϩ i sin ΃ 3 3 3 3 Imaginary (b) Imaginary axis axis π θ 2π 1 θ = 2 = 3 1 3 θ = 0 θ = π 1 1 2 r = 1 r = 2 Real Real 1 axis axis −1 1 π −1 θ 5 −1 3 = 3 θ 4π 3 = 3 (c) 2 Ϫ1 ϩ Ί3i Ϫ1 Ϫ Ί3i 9781133110873_08_ANS.qxd 3/10/12 6:58 AM Page A5

Answer Key

␲ ␲ 69. 3΂cos ϩ i sin ΃ 5 5 3␲ 3␲ 3΂cos ϩ i sin ΃ 5 5 3͑cos ␲ ϩ i sin ␲͒ 7␲ 7␲ 3΂cos ϩ i sin ΃ 5 5 9␲ 9␲ 3΂cos ϩ i sin ΃ 5 5 Imaginary axis θ = 3π 2 5 θ = π 2 1 5 θ = π r = 3 3 Real axis −2 −1 12 − π 2 θ = 9 5 5 θ = 7π 4 5 ␲ ␲ 71. 4΂cos ϩ i sin ΃ 2 2 7␲ 7␲ 4΂cos ϩ i sin ΃ 6 6 11␲ 11␲ 4΂cos ϩ i sin ΃ 6 6 Imaginary axis θ π 1= 2

2 r = 4 Real axis −2 2 − 2 θ 11π 3= 6 θ 7π 2 = 6 3 Ϫ 1 73.2 2i 75. Proof 77. (a)r2 (b) cos͑2␪͒ ϩ i sin͑2␪͒ 79. (a) Imaginary axis iz = −1 + 3i z = 3 + i Real axis −3 −2 −1 123 − 2 z/i = 1 − 3i −3 ␲ ␲ (b) Counterclockwise rotation of ; clockwise rotation of 2 2 81. True

8.4 Complex Vector Spaces and Inner Products.

8.4 Complex Vector Spaces and Inner Products.  Objective: Represent a vector in Cn by a basis.  Objective: Find the row space and null space of a matrix.  Objective: Solve a system of linear equations.  Objective: Find the Euclidean inner product, Euclidean norm, and Euclidean distance in Cn

A complex vector space is a vector space in which the scalars are complex numbers. The complex version of Rn is the complex vector space Cn consisting of ordered n-tuples of complex numbers

v = (a1 + b1i, a2 + b2i, … an + bni)

We also represent vectors in Cn by their coordinate matrices relative to the standard basis e1 = (1, 0, 0, …, 0), e2 = (0, 1, 0, …, 0), …, en = (0, 0, 0, …, 1),

  11 iba    iba  v =  22         nn iba 

As in of Rn, the operations of vector addition and scalar multiplication in Cn are performed component by component.

The dimension of Cn is n. A basis of a subspace V of Cn is any linearly independent set of vectors that spans V. If V has dimension d, then any linearly independent set of d vectors in V is a basis of V. Also, any set of d vectors that spans V is a basis of V.

Example: Is R2 a subspace of the complex vector space C2 over C?

2 2 2 Solution: R is a subset of C because R ={(a1 + b1i, a2 + b2i) where b1 = b2 = 0}. Moreover,

p. 163 8.4 Complex Vector Spaces and Inner Products. 3 Example: Show that S1 = {v1, v2, v3} = {(i, 2, 0), (2, i, 0), (1 + i, 0, 1 – i)} is a basis for C .

3 Solution: Because the dimension of C is 3, the set S1 will be a basis if it is linearly independent.

So we must check that

 Example: Write u = (2 – i, 8 – 2i, 5 – i) as a linear combination of the vectors in the set S1 = {v1, v2, v3} = {(i, 2, 0), (2, i, 0), (1 + i, 0, 1 – i)}, i.e. the same set as in the previous example.

Solution:

Example: Find the dimension of and a basis for S2 = span{(–2 – 2i, –3i, –9 – 6i), (–1 – i, – 3, –3), (–3 – 2i, –1 + 3i, –6 – 5i)}.

Solution: S2 = the column space of

p. 164 8.4 Complex Vector Spaces and Inner Products. Example: Find the rank, nullity, and bases for the row space and null space of

Solution:

p. 165 8.4 Complex Vector Spaces and Inner Products.  Example: Solve

Solution:

 Let u and v be vectors in Cn. The Euclidean inner product of u and v is given by

u•v = u1v1* + u2v2* + … + unvn* = 푢1푣̅̅1̅ + 푢2푣̅̅2̅ + ⋯ + 푢푛푣̅̅푛̅

Theorem 8.7 Properties of the Euclidean inner product Let u, v, and w be vectors in Cn and let k be a complex scalar. Then

1) u•v = (v•u)* = v∙u̅̅̅̅ Different from Rn! 2) (u + v)•w = u•w + v•w 3) (k u)•v = k (u•v) 4) u•(k v) = k*(u•v) = 푘̅(u∙v) Different from Rn! 5) u•u  0 (This also says that u•u is real.) 6) u•u = 0 if and only if u = 0

p. 166 8.4 Complex Vector Spaces and Inner Products.

 The Euclidean norm of u in Cn is ||u|| =  uu

 The Euclidean distance between u and v in Cn is d(u, v) = ||u – v||

To find a dot product u•v on the TI-89, use MatrixVector opsdotP(u,v)

To find a dot product u•v in Mathematica, use u.vconj which displays as u.v*

To find a dot product u•v in PocketCAS, use dot(v,u) (PocketCAS takes the complex conjugate of the first vector, rather than the second.)

To find a norm ||v|| on the TI-89, use MatrixNormsnorm(v)

To find a norm ||v|| in Mathematica, use Norm[v]

To find a norm ||v|| in PocketCAS, use l2norm(v) or norm(v)

 Example: Let u = (i, 1, 0), v = (–i, 1, 0), and w = (–2 + 3i, –5 + i, –4 – 6i).

Find ||u||

Find ||w||

Find u•v

Find v•w

Find d(u, w)

p. 167

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8.4 Complex Vector Spaces and Inner Products 413

8.4 Complex Vector Spaces and Inner Products

Recognize and perform vector operations in complex vector spaces C n, represent a vector in C n by a basis, and find the Euclidean inner product, Euclidean norm, and Euclidean distance in C n. Recognize complex inner product spaces.

COMPLEX VECTOR SPACES All the vector spaces studied so far in the text have been real vector spaces because the scalars have been real numbers. A complex vector space is one in which the scalars are

complex numbers. So, if v1, v2, . . . , vm are vectors in a complex vector space, then a linear combination is of the form c1v1 c2v2 cmvm n where the scalars c1, c2, . . . , cm are complex numbers. The complex version of R is the complex vector space C n consisting of ordered n -tuples of complex numbers. So, a vector in C n has the form ͑ ͒ v a1 b1i, a 2 b2i, . . . , an bni . It is also convenient to represent vectors in C n by column matrices of the form a1 b1i a b i v 2 . 2 . ΄ . ΅ an bni As with Rn, the operations of addition and scalar multiplication in C n are performed component by component.

Vector Operations in Cn

Let v ͑1 2i, 3 i͒ and u ͑2 i, 4͒ be vectors in the complex vector space C 2. Determine each vector. a. v u b. ͑2 i͒v c. 3v ͑5 i͒u

SOLUTION a. In column matrix form, the sum v u is 1 2i 2 i 1 3i v u ΄ ΅ ΄ ΅ ΄ ΅. 3 i 4 7 i b. Because ͑2 i͒͑1 2i͒ 5i and ͑2 i͒͑3 i͒ 7 i, ͑2 i͒v ͑2 i͒͑1 2i, 3 i͒ ͑5i, 7 i͒. c. 3 v ͑5 i͒u 3͑1 2i, 3 i͒ ͑5 i͒͑2 i, 4͒ ͑3 6i, 9 3i͒ ͑9 7i, 20 4i͒ ͑12 i, 11 i͒ 9781133110873_0804.qxp 3/10/12 6:56 AM Page 414

414 Chapter 8 Complex Vector Spaces

Many of the properties of Rn are shared by Cn. For instance, the scalar multiplicative identity is the scalar 1 and the additive identity in Cn is 0 ͑0, 0, 0, . . . , 0͒. The standard basis for Cn is simply ͑ ͒ e1 1, 0, 0, . . . , 0 e ͑0, 1, 0, . . . , 0͒ 2 . . . ͑ ͒ en 0, 0, 0, . . . , 1 which is the standard basis for Rn. Because this basis contains n vectors, it follows that the dimension of Cn is n. Other bases exist; in fact, any linearly independent set of n vectors in Cn can be used, as demonstrated in Example 2.

Verifying a Basis

ͭ ͮ ͭ͑ ͒ ͑ ͒ ͑ ͒ͮ 3 Show that S v1, v2, v3 i, 0, 0 , i, i, 0 , 0, 0, i is a basis for C . SOLUTION 3 ͭ ͮ Because the dimension of C is 3, the set v1, v2, v3 will be a basis if it is linearly independent. To check for linear independence, set a linear combination of the vectors in S equal to 0, as follows. ͑ ͒ c1v1 c2v2 c3v3 0, 0, 0 ͑ ͒ ͑ ͒ ͑ ͒ ͑ ͒ c1i, 0, 0 c2i, c2i, 0 0, 0, c3i 0, 0, 0 ͑͑ ͒ ͒ ͑ ͒ c1 c2 i, c2i, c3i 0, 0, 0 This implies that ͑ ͒ c1 c2 i 0 c 2i 0 c 3i 0. ͭ ͮ So, c1 c2 c3 0, and v1, v2, v3 is linearly independent.

Representing a Vector in Cn by a Basis

Use the basis S in Example 2 to represent the vector v ͑2, i, 2 i͒. SOLUTION By writing v c1v1 c2v2 c3v3 ͑͑ ͒ ͒ c1 c2 i, c2i, c3i ͑2, i, 2 i͒ you can obtain ͑ ͒ c1 c2 i 2 c2i i c3i 2 i which implies that REMARK i i Try verifying that this linear 2 2 c2 1, c1 1 2i, and c3 1 2i. combination yields ͑2, i, 2 i͒. i i ͑ ͒ ͑ ͒ So, v 1 2i v1 v2 1 2i v3. 9781133110873_0804.qxp 3/10/12 6:56 AM Page 415

8.4 Complex Vector Spaces and Inner Products 415

Other than Cn, there are several additional examples of complex vector spaces. For instance, the set of m n complex matrices with matrix addition and scalar multiplication forms a complex vector space. Example 4 describes a complex vector space in which the vectors are functions.

The Space of Complex-Valued Functions

Consider the set S of complex-valued functions of the form ͑ ͒ ͑ ͒ ͑ ͒ f x f1 x if 2 x

where f1 and f2 are real-valued functions of a real variable. The set of complex numbers forms the scalars for S, and vector addition is defined by ͑ ͒ ͑ ͒ ͓ ͑ ͒ ͑ ͔͒ ͓ ͒ ͑ ͔͒ f x g x f1 x if 2 x g1(x i g2 x ͓ ͑ ͒ ͑ ͔͒ ͓ ͑ ͒ ͑ ͔͒ f1 x g1 x i f 2 x g2 x . It can be shown that S, scalar multiplication, and vector addition form a complex vector space. For instance, to show that S is closed under scalar multiplication, let c a bi be a complex number. Then ͑ ͒ ͑ ͓͒ ͑ ͒ ͑ ͔͒ c f x a bi f1 x if 2 x ͓ ͑ ͒ ͑ ͔͒ ͓ ͑ ͒ ͑ ͔͒ af1 x bf 2 x i bf1 x af 2 x is in S.

The definition of the Euclidean inner product in Cn is similar to the standard dot n REMARK product in R , except that here the second factor in each term is a complex conjugate. Note that if u and v happen to be “real,” then this definition Definition of the Euclidean Inner Product in Cn agrees with the standard inner (or dot) product in Rn. Let u and v be vectors in Cn. The Euclidean inner product of u and v is given by u v u1v1 u2v2 unvn.

Finding the Euclidean Inner Product in C3

Determine the Euclidean inner product of the vectors u ͑2 i, 0, 4 5i͒ and v ͑1 i, 2 i, 0͒. SOLUTION u v u1v1 u2v2 u 3v3 ͑2 i͒͑1 i͒ 0͑2 i͒ ͑4 5i͒͑0͒ 3 i

Several properties of the Euclidean inner product Cn are stated in the following theorem.

THEOREM 8.7 Properties of the Euclidean Inner Product Let u, v, and w be vectors in C n and let k be a complex number. Then the following properties are true. 1.u v v u 2. ͑u v͒ w u w v w 3.͑ku͒ v k͑u v͒ 4. u ͑kv͒ k͑u v͒ 5.u u 0 6. u u 0 if and only if u 0. 9781133110873_0804.qxp 3/10/12 6:56 AM Page 416

416 Chapter 8 Complex Vector Spaces

PROOF The proof of the first property is shown below, and the proofs of the remaining properties have been left to you (see Exercises 59Ð63). Let ͑ ͒ ͑ ͒ u u1, u 2, . . . , un and v v1, v2, . . . , vn . Then . . . v u v1u1 v2u2 vnun . . . v1u1 v2u2 vnun . . . v1u1 v2u 2 vnun u1v1 u2v2 unvn u v.

The Euclidean inner product in Cn is used to define the Euclidean norm (or length) of a vector in Cn and the Euclidean distance between two vectors in Cn .

Definitions of the Euclidean Norm and Distance in Cn The Euclidean norm (or length) of u in Cn is denoted by ʈuʈ and is ʈuʈ ͑u u͒1͞2. The Euclidean distance between and vu is d͑u, v͒ ʈu vʈ .

The Euclidean norm and distance may be expressed in terms of components, as follows (see Exercise 51). ʈ ʈ ͑ 2 2 . . . 2͒1͞2 u Խu1Խ Խu2Խ ԽunԽ ͑ ͒ ͑ 2 2 . . . 2͒1͞2 d u, v Խu1 v1Խ Խu2 v2Խ Խun vnԽ

Finding the Euclidean Norm and Distance in Cn

Let u ͑2 i, 0, 4 5i͒ and v ͑1 i, 2 i, 0͒. a. Find the norms of u and v. b. Find the distance between u and v. SOLUTION ʈ ʈ ͑ 2 2 2͒1͞2 a. u Խu1Խ Խu2Խ Խu3Խ ͓͑22 12͒ ͑02 02͒ ͑42 ͑5͒2͔͒1͞2 ͑5 0 41͒1͞2 Ί46

ʈ ʈ ͑ 2 2 2͒1͞2 v Խv1Խ Խv2Խ Խv3Խ ͓͑12 12͒ ͑22 12͒ ͑02 02͔͒1͞2 ͑2 5 0͒1͞2 Ί7

b. d͑u, v͒ ʈu vʈ ʈ͑1, 2 i, 4 5i͒ʈ ͓͑12 02͒ ͑͑2͒2 ͑1͒2͒ ͑42 ͑5͒2͔͒1͞2 ͑1 5 41͒1͞2 Ί47 9781133110873_0804.qxp 3/10/12 6:56 AM Page 417

8.4 Complex Vector Spaces and Inner Products 417

COMPLEX INNER PRODUCT SPACES The Euclidean inner product is the most commonly used inner product in Cn . On occasion, however, it is useful to consider other inner products. To generalize the notion of an inner product, use the properties listed in Theorem 8.7.

Definition of a Complex Inner Product Let and vu be vectors in a complex vector space. A function that associates u and v with the complex number ͗u, v͘ is called a complex inner product when it satisfies the following properties. 1. ͗u, v͘ ͗v, u͘ 2. ͗u v, w͘ ͗u, w͘ ͗v, w͘ 3. ͗ku, v͘ k͗u, v͘ 4. ͗u, u͘ 0 and ͗u, u͘ 0 if and only if u 0.

A complex vector space with a complex inner product is called a complex inner product space or unitary space.

A Complex Inner Product Space

͑ ͒ ͑ ͒ 2 Let u u1, u2 and v v1, v2 be vectors in the complex space C . Show that the function defined by ͗ ͘ u, v u1v1 2u2v2 is a complex inner product. SOLUTION Verify the four properties of a complex inner product, as follows. ͗ ͘ ͗ ͘ 1. v, u v1u1 2v2u2 u1v1 2u2v2 u, v ͗ ͘ ͑ ͒ ͑ ͒ 2. u v, w u1 v1 w1 2 u2 v2 w2 ͑ ͒ ͑ ͒ u1w1 2u2w2 v1w1 2v2w2 ͗u, w͘ ͗v, w͘ ͗ ͘ ͑ ͒ ͑ ͒ ͑ ͒ ͗ ͘ 3. ku, v ku1 v1 2 ku2 v2 k u1v1 2u2v2 k u, v ͗ ͘ 2 2 4. u, u u1u1 2u2u2 Խu1Խ 2Խu2Խ 0 ͗ ͘ Moreover,u, u 0 if and only if u1 u2 0. Because all the properties hold,͗u, v͘ is a complex inner product.

LINEAR Complex vector spaces and inner products have an important ALGEBRA application called the Fourier transform, which decomposes APPLIED a function into a sum of orthogonal basis functions. The given function is projected onto the standard basis functions for varying frequencies to get the Fourier amplitudes for each frequency. Like Fourier coefficients and the Fourier approximation, this transform is named after the French mathematician Jean-Baptiste Joseph Fourier (1768–1830). The Fourier transform is integral to the study of signal processing. To understand the basic premise of this transform, imagine striking two piano keys simultaneously. Your ear receives only one signal, the mixed sound of the two notes, and yet your brain is able to separate the notes. The Fourier transform gives a mathematical way to take a signal and separate out its frequency components.

Eliks/Shutterstock.com 9781133110873_0804.qxp 3/10/12 6:56 AM Page 418

418 Chapter 8 Complex Vector Spaces

8.4 Exercises

Vector Operations In Exercises 1Ð8, perform the Finding the Euclidean Norm In Exercises 27Ð34, indicated operation using determine the Euclidean norm of v. 4i, 6ͨ. 27.v ͑i, i͒ 28. v ͑1, 0͒ͧ ؍ ؉ i, 3 ؉ i͒, and w 2ͧ ؍ i, 3 ؊ iͨ, vͧ ؍ u 1.3u 2. 4iw 29.v 3͑6 i, 2 i͒ 30. v ͑2 3i, 2 3i͒ 3.͑1 2i͒w 4. iv 3w 31.v ͑1, 2 i, i͒ 32. v ͑0, 0, 0͒ 5.u ͑2 i͒v 6. ͑6 3i͒v ͑2 2i͒w 33. v ͑1 2i, i, 3i, 1 i͒ 7.u iv 2iw 8. 2iv ͑3 i͒w u 34. v ͑2, 1 i, 2 i, 4i͒

Linear Dependence or Independence In Exercises Finding the Euclidean Distance In Exercises 35Ð40, 9Ð12, determine whether the set of vectors is linearly determine the Euclidean distance between u and v. independent or linearly dependent. 35. u ͑1, 0͒, v ͑i, i͒ ͭ͑ ͒ ͑ ͒ͮ 9. 1, i , i, 1 36. u ͑2 i, 4, i͒, v ͑2 i, 4, i͒ ͭ͑ ͒ ͑ ͒ ͑ ͒ͮ 10. 1 i, 1 i, 1 , i, 0, 1 , 2, 1 i, 0 37. u ͑i, 2i, 3i͒, v ͑0, 1, 0͒ ͭ͑ ͒ ͑ ͒ ͑ ͒ͮ 11. 1, i, 1 i , 0, i, i , 0, 0, 1 38. u ͑Ί2, 2i, i͒, v ͑i, i, i͒ ͭ͑ ͒ ͑ ͒ ͑ ͒ͮ 12. 1 i, 1 i, 0 , 1 i, 0, 0 , 0, 1, 1 39. u ͑1, 0͒, v ͑0, 1͒ ͑ ͒ ͑ ͒ Verifying a Basis In Exercises 13Ð16, determine 40. u 1, 2, 1, 2i , v i, 2i, i, 2 whether S is a basis for Cn. Complex Inner Products In Exercises 41Ð44, determine ͭ͑ ͒ ͑ ͒ͮ 13. S 1, i , i, 1 whether the function is a complex inner product, where ͨ ͧ ؍ ͨ ͧ ؍ ͮ͒ ͑ ͒ ͑ͭ 14. S 1, i , i, 1 u u1, u2 and v v1, v2 . ͭ͑ ͒ ͑ ͒ ͑ ͒ͮ ͗ ͘ 15. S i, 0, 0 , 0, i, i , 0, 0, 1 41. u, v u1 u2v2 ͭ͑ ͒ ͑ ͒ ͑ ͒ͮ ͗ ͘ ͑ ͒ ͑ ͒ 16. S 1 i, 0, 1 , 2, i, 1 i , 1 i, 1, 1 42. u, v u1 v1 2 u2 v2 43. ͗u, v͘ 4u v 6u v Representing a Vector by a Basis In Exercises 17Ð20, 1 1 2 2 ͗ ͘ express v as a linear combination of each of the following 44. u, v u1v1 u2v2 basis vectors. Finding Complex Inner Products In Exercises 45Ð48, ؉ ؍ a) {ͧi, 0, 0ͨ, ͧi, i, 0ͨ, ͧi, i, iͨ} ͗ ͘) use the inner product u, v u1v1 2u2v2 to find .b) {ͧ1, 0, 0ͨ, ͧ1, 1, 0ͨ, ͧ0, 0, 1 ؉ iͨͮ ͗u, v͘) 17.v ͑1, 2, 0͒ 18. v ͑1 i, 1 i, 3͒ 45. u ͑2i, i͒ and v ͑i, 4i͒ 19.v ͑i, 2 i, 1͒ 20. v ͑i, i, i͒ 46. u ͑3 i, i͒ and v ͑2 i, 2i͒ 47. u ͑2 i, 2 i͒ and v ͑3 i, 3 2i͒ Finding Euclidean Inner Products In Exercises 21 and ͑ ͒ ͑ ͒ 22, determine the Euclidean inner product u v. 48. u 4 2i, 3 and v 2 3i, 2 ͑ ͒ ͑ ͒ 21.u i, 2i, 1 i 22. u 4 i, i, 0 Finding Complex Inner Products In Exercises 49 and v ͑3i, 0, 1 2i͒ v ͑1 3i, 2, 1 i͒ 50, use the inner product u v u v u v u v ؍ u, v͗͘ Properties of Euclidean Inner Products In Exercises 11 11 12 12 21 21 22 22 ؉ i, 0ͨ, where 1ͧ ؍ 2i, 2 ؉ iͨ, wͧ ؍ ؊ i, 3i͒, v 1ͧ ؍ 23Ð26, let u ؊i. Evaluate the expressions in parts (a) and (b) u u v v؍ and k [12 11 ] ؍ and v [12 11 ] ؍ u to verify that they are equal. u21 u22 v21 v22 23. (a) u v 24. (a) ͑u v͒ w to find ͗u, v͘. (b)v u (b) u w v w 0 i 1 1 2i 49. u ΄ ΅ v ΄ ΅ 25. (a) ͑ku͒ v 26. (a) u ͑kv͒ 1 2i 0 i ͑ ͒ ͑ ͒ 1 2i i 2i (b)k u v (b) k u v 50.u ΄ ΅ v ΄ ΅ 1 i 0 3i 1 9781133110873_0804.qxp 3/10/12 6:56 AM Page 419

8.4 Exercises 419

͑ ͒ 51. Let u a1 b1i, a2 b2i, . . . , an bni . Finding an and a Preimage In Exercises 67Ð70, m → n (a) Use the definitions of Euclidean norm and Euclidean the linear transformation T : C C is shown by ؍ ͨ ͧ inner product to show that T v Av. Find the image of v and the preimage of w. ʈ ʈ ͑ 2 2 2͒1͞2 1 0 1 i 0 u Խu1Խ Խu2Խ . . . ԽunԽ . 67. A ΄ ΅, v ΄ ΅, w ΄ ΅ i i 1 i 0 (b) Use the results of part (a) to show that i d͑u, v͒ ͑Խu v Խ2 Խu v Խ2 0 i 1 1 1 1 2 2 68. A ΄ ΅, v 0 , w ΄ ΅ . . . Խu v Խ2͒1͞2. i 0 0 ΄ ΅ 1 n n 1 i 1 0 2 2 i 52. The complex Euclidean inner 69. A i 0 , v ΄ ΅, w 2i ΄ ΅ 3 2i ΄ ΅ product of and vu is sometimes called the i i 3i complex dot product. Compare the properties of the complex dot product in Cn and those of the dot 0 1 1 2 1 i product in Rn. 70. A ΄ i i 1΅, v ΄5΅, w ΄1 i΅ 0 i 0 0 i (a) Which properties are the same? Which properties are different? 71. Find the kernel of the linear transformation in Exercise 68. (b) Explain the reasons for the differences. 72. Find the kernel of the linear transformation in Exercise 69.

Finding an Image In Exercises 73 and 74, find the 53. Let v ͑i, 0, 0͒ and v ͑i, i, 0͒. If v ͑z , z , z ͒ i, iͨ for the indicated composition, whereͧ ؍ image of v 3 2 1 3 2 1 and the set ͭv , v , v ͮ is not a basis for C3, what does 1 2 3 T and T are the matrices below. this imply about z , z , and z ? 1 2 ؊ 3 2 1 i i ؍ i 0 ؍ ͒ ͑ ͒ ͑ 54. Let v1 i, i, i and v2 1, 0, 1 . Determine a vector T1 and T2 [i 0] [ i ؊i ] 3 ͮ ͭ v3 such that v1, v2, v3 is a basis for C .

73.T2 T1 74. T1 T2 Properties of Complex Inner Products In Exercises 55Ð58, verify the statement using the properties of a 75. Determine which of the sets below are subspaces of the complex inner product. vector space of 2 2 complex matrices. 55. ͗u, kv w͘ k͗u, v͘ ͗u, w͘ (a) The set of 2 2 symmetric matrices. 56. ͗u, 0͘ 0 (b) The set of 2 2 matrices A satisfying ͑A͒T A. 57. ͗u, v͘ ͗u, v͘ ͗v, 2u͘ (c) The set of 2 2 matrices in which all entries are 58. ͗u, kv͘ k͗u, v͘ real. (d) The set of 2 2 diagonal matrices. Proof In Exercises 59Ð63, prove the property, where n 76. Determine which of the sets below are subspaces of , v, and wu are vectors in and kC is a complex number. the vector space of complex-valued functions (see 59. ͑u v͒ w u w v w Example 4). 60. ͑ku͒ v k͑u v͒ (a) The set of all functions f satisfying f͑i͒ 0. 61. u ͑kv͒ k͑u v͒ (b) The set of all functions f satisfying f͑0͒ 1. 62. u u 0 (c) The set of all functions f satisfying f͑i͒ f͑i͒. 63. u u 0 if and only if u 0. True or False? In Exercises 77 and 78, determine 64. Writing Let ͗u, v͘ be a complex inner product and let whether each statement is true or false. If a statement is k be a complex number. How are ͗u, v͘ and ͗u, kv͘ true, give a reason or cite an appropriate statement from related? the text. If a statement is false, provide an example that shows the statement is not true in all cases or cite an Finding a Linear Transformation In Exercises 65 and appropriate statement from the text. m → n 66, determine the linear transformation T : C C that 77. Using the Euclidean inner product of and vu in C n, has the given characteristics. u v u v u v . . . u v . ͑ ͒ ͑ ͒ 1 1 2 2 n n 65. T 1, 0 2 i, 1 78. The Euclidean norm of u in Cn denoted by ʈuʈ is T͑0, 1͒ ͑0, i͒ ͑u u͒2. 66. T͑i, 0͒ ͑2 i, 1͒ T͑0, i͒ ͑0, i͒ Answer Key

Section 8.4 1. ͑3i, 9Ϫ 3i͒ 3. ͑Ϫ8 ϩ 4i, 6 ϩ 12i͒ 5. ͑Ϫ5 ϩ i, Ϫ4͒ 7. ͑Ϫ9 ϩ 3i, 2 ϩ 14i͒ 9. Linearly dependent 11. Linearly independent 13. S is not a basis for C 2. 15. S is a basis for C 3. 17. (a) ͑1, 2, 0͒ ϭ i͑i, 0, 0͒ Ϫ 2i͑i, i, 0͒ ϩ 0͑i, i, i͒ (b) ͑1, 2, 0͒ ϭϪ͑1, 0, 0͒ ϩ 2͑1, 1, 0͒ ϩ 0͑0, 0, 1 ϩ i͒ 19. (a) ͑ Ϫi, 2 ϩ i, Ϫ1͒ ϭ ͑Ϫ2 ϩ 2i͒͑i, 0, 0͒ ϩ ͑ 1 Ϫ 3i͒͑i, i, 0͒ ϩ i͑i, i, i͒ (b) ͑ Ϫi, 2 ϩ i, Ϫ1͒ ϭ ͑Ϫ2 Ϫ 2i͒͑1, 0, 0͒ ϩ ͑ 2 ϩ i͒͑1, 1, 0͒ ϩ ͑Ϫ1 ϩ 1 ͒͑ ϩ ͒ 2 2i 0, 0, 1 i 21. Ϫ4 Ϫ 3i 23. (a) 1 ϩ 4i (b) 1 ϩ 4i The expressions are equal. 25. (a) 4 Ϫ i (b) 4 Ϫ i The expressions are equal. 27. Ί2 29. 3Ί42 31. Ί7 33. Ί17 35. Ί3 37. Ί15 39. Ί2 41. Not a complex inner product 43. A complex inner product 45. Ϫ6 47. 23 Ϫ 3i 49. Ϫ4 ϩ i 51. (a) and (b) Proofs ϭ 53. z3 0, z1 and z2 can be any complex numbers. 55. ͗u, kv ϩ w͘ ϭ ͗u, kv͘ ϩ ͗u, w͘ ϭ k͗u, v͘ ϩ ͗u, w͘ 57. ͗u, v͘ ϩ ͗u, v͘ ϭ ͗v, u͘ ϩ ͗v, u͘ ϭ ͗v, u ϩ u͘ ϭ ͗v, 2u͘ ϩ u1 ϭ 2 i 0 u1 59Ð63. Proofs 65. T΄ ΅ ΄ Ϫ ΅΄ ΅ u2 1 i u2 2 Ϫ i 1 ϩ i 0 2 67. ΄ ΅, ΄ ΅ 69. 1 ϩ 2i , ΄ ΅ 2i 0 ΄ ΅ 1 Ϫ1 ϩ 5i 71. ker͑T͒ ϭ ͭ͑0, t, Ϫti͒, where t ʦ Rͮ 0 73. ΄ ΅ 75. (a), (b), and (d) are subspaces. 0 77. False. ϭ ϩ ϩ . . . ϩ u и v u1v1 u2v2 unvn Chapter 6: Linear Transformations. 6.1 Introduction to Linear Transformations.

Chapter 6: Linear Transformations.

6.1 Introduction to Linear Transformations.  Objective: Find the image and preimage of a function  Objective: Show that a function is a linear transformation.  Recognize and describe some common linear transformations.

 A function T: V  W maps a vector space V to a W. V is the domain and W is the codomain of S*T . If v  V and w  W such that T (v) = w then w is the image of v under T. The set of all images of vectors in V is called the range of T. The set of all vectors in V such that T (v) = w is the preimage of w.

preimage  domain v  V T T T T

image  range  codomain w  range  W

 Let V and W be vector spaces. Then the function (or map) T : V  W is a linear transformation when the following two properties are true for all v  V and w  W and for any scalar c. And T()* 1) T(u + v) = T(u) + T(v) 2) T(cu) = c T(u)

A linear transformation is “operation preserving” because applying an operation in V (before T ) gives the same result as applying the corresponding operation in W (after T ).

scalar scalar addition addition multiplication multiplication in V in W in V in W

T(u + v) = T(u) + T(v) T(cu) = c T(u)

2 2  Example: Show that T : R  R given by T(v1, v2) = (v1 – v2, v1 + v2) is a linear transformation, find the image of (13, –4), and find the preimage of (2, 8). [Note: technically, we should write

T((v1, v2)) because v = (v1, v2).]

p. 169 Chapter 6: Linear Transformations. 6.1 Introduction to Linear Transformations.

Solution: T(u + v) = T(u1 + v1, u2 + v2) = ((u1 + v1) – (u2 + v2), (u1 + v1) + (u2 + v2)) = (u1 – u2, u1 + u2) + (v1 – v2, v1 + v2) = T(u) + T(v)

T(cu) = T(cv1, cv2) (cv1 – cv2, cv1 + cv2) (c(v1 – v2), c(v1 + v2)) c (v1 – v2, v1 + v2) =c T(u)

Image of (13, –4): T(13, –4) = (13 + (–4), 13 – (–4)) = (9, 17)

Preimage of (2, 8): T(v1, v2) = (v1 – v2, v1 + v2) = (2, 8) v1 – v2 = 2 2 v1 = 10 v1 + v2 = 8 –2 v2 = –6 (v1, v2) = (5, 3)

Example: Show that f : R  R given by f (x) = x2 is not a linear transformation.

S* Solution: f (x + y) = (x + y)2 = x2 + 2xy + y2. Usually, this is not equal to f (x) + f (y) = x2 + y2. In particular, a specific counterexample is f (1 + 1) = 4, but f (1) + f (1) = 2.

Example: Show that f : R  R given by f (x) = x + 1 is not a linear transformation. S* Solution: f (x + y) = x + y + 1, which is never equal to f (x) + f (y) = x + 1 + y + 1. [However, f (x) = x + 1 is a linear function, because its graph is a straight line.]

Two simple linear transformations are the zero transformation T : V  W, T(v) = 0 for all v  V and the identity transformation T : V  V, T(v) = v for all v  V.

 Theorem 6.1 Properties of Linear Transformations

Let T: V  W be a linear transformation, and u and v be vectors in V. Then

1) T(0) = 0 2) T(–v) = –T(v) 3) T(u – v) = T(u) – T(v) 4) If v = c1v1 + c2v2 + …+ cnvn then T(v) = T(c1v1 + c2v2 + …+ cnvn) = c1T(v1) + c2T(v2) + …+ cnT(vn)

p. 170 Chapter 6: Linear Transformations. 6.1 Introduction to Linear Transformations. Proof:

1) T(0) =

2) T(–v) =

3) T(u – v) =

4) Prove T(c1v1 + c2v2 + …+ cnvn) = c1T(v1) + c2T(v2) + …+ cnT(vn) by induction:

 T(c1v1) = c1T(v1) by the second property of linearity

 Suppose T(c1v1 + c2v2 + …+ cnvn) = c1T(v1) + c2T(v2) + …+ cnT(vn).

Then T(c1v1 + c2v2 + …+ cnvn + cn+1vn+1)

= T(c1v1 + c2v2 + …+ cnvn) + T(cn+1vn+1) by the first property of linearity

= c1T(v1) + c2T(v2) + …+ cnT(vn) + T(cn+1vn+1) by supposition

= c1T(v1) + c2T(v2) + …+ cnT(vn) + cn+1T(vn+1) by the second property of linearity

Property 4 of Theorem 6.1 tells us that a linear transformation is completely determined by its action on a basis for V. In other words, if {v1, v2, …, vn} is a basis for V and if T(v1), T(v2), …, T(vn) are known, then T(v) is determined for any v  V.

Example: Let T: R3  R3 be a linear transformation such that T(1, 0, 0) = (3, 1, 4) T(0, 1, 0) = (1, 5, 9) T(0, 0, 1) = (2, 6, 5) Find T(2, 7, 1).

Solution: T(2, 7, 1) = 2 T(1, 0, 0) + 7 T(0, 1, 0) + 1 T(0, 0, 1)

= 2(3, 1, 4) + 7(1, 5, 9) + 1(2, 6, 5)

= (17, 46, 51)

p. 171 Chapter 6: Linear Transformations. 6.1 Introduction to Linear Transformations. Example: Let T: R2  R3 the function such that  7 5  v  T(v) = Av =   1  4  3   v2   6  7 Show that T is a linear function. Find T(1, 0) and T(0, 1).

Solution: For any u and v in R2, T(u + v) = A(u + v) = Au + Av = T(u) + T(v).

Also, for any v in R2 and any scalar c, T(cv) = A(cv) = c(Av) = cT(v).

 7  1   T(1, 0) =   =  4 = (7, –4, –6) 0    6 and  5  0   T(0, 1) =   =  3 = (5, –3, –7) 1    7

Notice that T(1, 0) and T(0, 1) are just the two columns of A.

Theorem 6.2: The Linear Transformation Given by a Matrix

Let A be an mn matrix. The function T: Rn  Rm the defined by T(v) = Av is a linear transformation. Vectors in Rn are represented by n1 (column) matrices and vectors in Rm are represented by m1 (column) matrices.

a11 a12  a1n  v1   a11v1  a12v2  a1nvn  a a  a  v   a v  a v  a v  T(v) =  21 22 2n   2  =  21 1 22 2 2n n                    a a  a v a v  a v  a v  m1 m2 mn  mn  n  n1  m1 1 m2 2 mn n  m1

Example: Rotation. We show that the linear transformation T: R2  R2 represented by the matrix

cos()  sin() 2 A =   rotates every vector in R counterclockwise by an angle  about the sin() cos()  origin.

p. 172 Chapter 6: Linear Transformations. 6.1 Introduction to Linear Transformations.

r cos() cos()  sin() r cos()  T   =     r sin() sin() cos()  r sin() cos()cos()  sin()sin()  = r   sin()cos()  cos()sin() r cos( ) =   r sin(  )

T(1.5v) T (u + v) u + v

T(v) u 1.5v  v T(u) Observe that T is a linear transformation. It  preserves vector addition and scalar multiplication.

Example: projection. The linear transformation T: R3  R3 represented by the matrix 1 0 0 A =   projects a vector v = (x, y, z) to T(v) = (x, y, 0). In other words, T maps every 0 1 0 0 0 0 vector in R3 to its orthogonal projection in the x-y plane.

T Example: Transpose. We show that the linear transformation T: Mm,n Mn,m given by T(A) = A is a linear transformation.

If A and B are mn matrices, then T(A + B) = (A + B)T = AT + BT = T(A) + T(B).

If A is an mn matrix and c is a scalar, then T(cA = (cA)T = c(AT) = c T(A).

p. 173 Chapter 6: Linear Transformations. 6.1 Introduction to Linear Transformations. Example: The Differential Operator. Let C[a, b], also written as C1[a, b], be the set of all functions whose derivatives are continuous on the interval [a, b]. We show that the differential operator Dx defines a linear transformation from C[a, b] into C[a, b].

d Using operator notation, we write Dx( f ) = [ f ], where f is in C[a, b]. dx From calculus, we know that for any functions f and g in C[a, b],

Dx( f + g) = [ f + g] = [ f ] + [ g] = Dx( f ) + Dx( g)  C[a, b] and for any function f in C[a, b] and for any scalar c,

Dx(c f ) = [c f ] = c [ f ] = c Dx( f )  C[a, b]

Example: The Definite Integral Operator. Let P be the vector space of all polynomial functions. b We show that the definite integral operator T: P  R defined by T(p) =  p(x)dx a is a linear transformation.

From calculus, we know that for any polynomials p and q in P, b b b b T(p + q) = [ p  q](x)dx = [ p(x)  q(x)]dx =  p(x)dx +  q(x)dx = T(p) + T(q) a a a a and for any polynomial p in P and for any scalar c, b b b T(cp) = [cp](x)dx =  cp(x)dx = c p(x)dx = cT(p). a a a

p. 174 Chapter 6: Linear Transformations. 6.2 The Kernel and Range of a Linear Transformation.

6.2 The Kernel and Range of a Linear Transformation.  Objective: Find the kernel of a linear transformation.  Objective: Find a basis for the range, the rank, and the nullity of a linear transformation.  Objective: Determine whether a linear transformation is one-to-one or onto.  Objective: Prove basic results about one-to-one and/or onto linear transformations.  Objective: Determine whether vector spaces are isomorphic.

S* Let T: V  W be a linear transformation. Then the set of all vectors v in V such that T(v) = 0 is the kernel of T and is denoted by ker(T).

Remember that for any linear transformation, T(0) = 0, so ker(T) always contains 0.

 Example Find the kernel of the projection T: R3  R3 defined by T(x, y, z) = (x, y, 0).

Solution: The solution to T(x, y, z) = (0, 0, 0) is x = 0, y = 0. So ker(T) = {(0, 0, z): z  R}

 Example Find the kernel of the zero transformation T: V  W defined T(v) = 0.

Solution: T(v) = 0 is true for all v  V so ker(T) = V

 Example Find the kernel of the identity transformation T: V  V defined T(v) = v.

Solution: T(v) = 0 means v = 0 so ker(T) = {0}.

 Example Find the kernel of the linear transformation T: R3  R2 defined by

T(x1, x2, x3) = (5x1 – 4x2 – 6x3, 4x2 –8x3).

Solution: T(x1, x2, x3) = (0, 0) yields the system

xxx  0645  0645 14 321   rref   5 001        xx 32  084   0840    0210 

14 which has the solution x3 = t, x2 = 2t, x1 = 5 t so ker(T) = {( , 2t, t): t  R}

Theorem 6.3: The Kernel of T: V  W is a Subspace of V

The kernel of a linear transformation T: V  W is a subspace of the domain V.

Proof: 0  ker(T) so ker(T) is not empty.

If u, v  ker(T), then T(u) = 0 and T(v) = 0

so T(u + v) = T(u) + T(v) = 0 so u + v  ker(T)

p. 175 Chapter 6: Linear Transformations. 6.2 The Kernel and Range of a Linear Transformation. If c is a scalar and v  ker(T), then T(v) = 0

so T(cv) = cT(v) = 0 so cv  ker(T)

 Example Find a basis for the kernel of the linear transformation T: R5  R4 defined by T(x) = Ax, where

Solution: ker(T) = the nullspace of A (i.e. the solution space of Ax = 0).

S*

S* Let T: V  W be a linear transformation. Then the set of all vectors w in W that are images of vectors in V is the range of T, and is denoted by range(T). That is, range(T) = {T(v): v  V}

kernel  domain T T

0  range  codomain

p. 176 Chapter 6: Linear Transformations. 6.2 The Kernel and Range of a Linear Transformation.  Theorem 6.4: The Range of T: V  W is a Subspace of W

The range of a linear transformation T: V  W is a subspace of the codomain W.

Proof: 0  range(T) because T(0) = 0, so range(T) is not empty.

If T(u) and T(v) are vectors in range(T), then u and v are vectors in V. V is closed under addition, so u + v  V, so T(u + v)  range(T). T(u + v) = T(u) + T(v), so T(u) and T(v)  range(T) implies that T(u) + T(v)  range(T).

Let c be a scalar and T(v)  range(T) Then v  V. Since V is closed under scalar multiplication, cv  V, so T(cv)  range(T). T(cv) = cT(v), so Let c be a scalar and T(v)  range(T) implies that cT(v)  range(T).

so T(cv) = cT(v) = 0 so cv  ker(T)

 Example Find a basis for the range of the linear transformation T: R5  R4 defined by T(x) = Ax, where

Solution: range(T) = the column space of A.

S*

 Let T: V  W be a linear transformation. The dimension of the kernel of T is called the nullity S* of T and is denoted by nullity(T), the dimension of the range of T is called the rank of T and is denoted by rank(T).

p. 177 Chapter 6: Linear Transformations. 6.2 The Kernel and Range of a Linear Transformation.  Theorem 6.5: The Sum of Rank and Nullity

S*Let T: V  W be a linear transformation from the n-dimensional vector space V into a vector space W. Then rank(T ) + nullity(T ) = n i.e. dim(range(T )) + dim(kernel(T )) = dim(domain(T ))

 Example Find the rank and nullity of the linear transformation T: R8  R5 defined by

T(x) = Ax, where

Solution: S*

 A function T: V  W is one-to-one (injective) if and only if every vector in the range of T has a single (unique) preimage. An equivalent definition is that T is one-to one if and only if T(u) = T(v) implies that u = v. S*  A function T: V  W is onto (surjective) if and only if the range of T is W (the codomain of T ). And equivalent definition is that that T is one-to one if and only if every element in W has a preimage in V.

 Example: T: R2  R3 defined by T(x, y) = (x, y, x) is one-to-one but not onto (points with z  x have no preimage).

 Example: The projection T: R3  R2 defined by T(x, y, z) = (x, y) is onto but not one-to-one.

p. 178 Chapter 6: Linear Transformations. 6.2 The Kernel and Range of a Linear Transformation.  Example: The projection T: R2  R2 defined by T(x, y) = (x – 2y, 3x – 6y) = (x – 2y) (1, 3) is neither one-to-one (because there are many different (x, y) pairs that give the same value of x – 2y ) nor onto (because points with y  3x are not in the range of T).

 Example: The transformation T: Rn  R n defined by T(v) = Av, where A is an invertible matrix, is one-to-one and onto. T is one-to-one because T(u) = T(v) implies Au = Av. Since A is invertible, we can pre-multiply by A–1 to obtain A–1Au = A–1Av or u = v. T is onto because the rank of an invertible nn matrix (namely, A) is n. So we also have rank(T ) = n. By Theorem 6.7 (below), T is onto because its rank equals the dimension of its codomain.

 Theorem 6.6: One-to-One Linear Transformations

Let T: V  W be a linear transformation. Then T is one-to-one if and only if ker(T ) = {0}.

 Proof If T is one-to-one, then T(v) = 0 has only one solution, namely, v = 0. Thus, ker(T ) = {0}. S* Conversely, suppose ker(T ) = {0} and T(u) = T(v). Because T is a linear transformation,

T(u – v) =

so u – v = because

Therefore,

p. 179 Chapter 6: Linear Transformations. 6.2 The Kernel and Range of a Linear Transformation.  Theorem 6.7: Onto Linear Transformations

Let T: V  W be a linear transformation, where W is finite dimensional. Then T is onto if and only rank(T ) = dim(W).

 Proof Let T: V  W be a linear transformation.

If T is onto, then W is equal to the range of T, so rank(T ) = dim(range(T)) = dim(W). S* Let dim(W) = n. If rank(T ) = n, then there are n linearly independent vectors T(v1), T(v2), …,

T(vn) in the range of T. Since the range of T is in W, the vectors T(v1), T(v2), …, T(vn) are linearly independent W. By Thm. 4.12 (n linearly independent vectors in an n-dimensional

space form a basis), the vectors T(v1), T(v2), …, T(vn) form a basis for W. So any vector w  W can be written as a linear combination

w = c1T(v1) + c2T(v2) + … + cnT(vn) = T(c1v1 + c2v2 + … + cnvn), which is in the range of T. Therefore, T is onto.

 Theorem 6.8: One-to-One and Onto Linear Transformations

Let T: V  W be a linear transformation, where vector spaces V and W both have dimension n. Then T is one-to-one if and only if it is onto.

 Proof If T is one-to-one, then by Thm 6.6, ker(T ) = {0}, and dim(ker(T )) = 0. By Thm 6.5,

S* dim(range(T)) =

Therefore, by Thm 6.7, T is onto.

Conversely, if T is onto, then dim(range(T)) = dim(W) = n

so by Thm 6.5, dim(ker(T)) =

Therefore, ker(T ) = {0}, and by Thm 6.6, T is one-to-one.

p. 180 Chapter 6: Linear Transformations. 6.2 The Kernel and Range of a Linear Transformation.  Example: Consider the linear transformations Let T: V  W represented by T(x) = Ax. Find the rank and nullity of T, and determine whether T is one-to-one or onto. (Notice that all of the matrices are in row-echelon form.)

1 3  5 1 3 1 3  5     1 3  5   a) A = 0 1 4 b) A = 0 1 c) A =   d) A = 0 1 4     0 1 4    0 0 1  0 0 0 0 0 

dim(codomain) dim(domain) dim(range) dim(kernel) One-to-one? Onto? m n r = rank(T)  = nullity(T)

a)

b)

c)

d)

A function that is both one-to-one (injective) and onto (surjective) is called bijective. A bijective function is invertible.

 A linear transformation T: V  W that is one-to-one and onto is called an isomorphism. If V and S*W are vector spaces such that there exists an isomorphism from V to W, then V and W are said to be isomorphic to each other.

p. 181 Chapter 6: Linear Transformations. 6.2 The Kernel and Range of a Linear Transformation.  Theorem 6.9: Isomorphic Spaces and Dimension

Two finite-dimensional vector spaces V and W are isomorphic if and only if they are of the same dimension.

 Proof Assume V is isomorphic to W, and V has dimension n. Then there exists a linear

transformation T: V  W that is one-to-one and onto. Because T is one-to-one,

dim(ker(T)) = , so dim(range(T)) =

Moreover, because T is onto, dim(W) =

so V and W are of the same dimension.

Conversely, assume V and W both have dimension n. Let {v1, v2, … vn} be a basis for V, and

{w1, w2, … wn} be a basis for W. Then an arbitrary vector v  V can be uniquely written as S*

v = c1v1 + c2v2 + … + cnvn. Define the linear transformation T: V  W by

T(v) = c1w1 + c2w2 + … + cnwn.

Then T is one-to-one: If T(v) = 0,

so ker(T ) = {0}, and by Thm 6.6, T is one-to-one.

Also, T is onto: Since

the basis {w1, w2, … wn} is contained in the range of T, which is contained in W.

Thus, rank(T ) = n and Thm 6.7, T is onto.

p. 182 Chapter 6: Linear Transformations. 6.3 Matrices for Linear Transformations.

6.3 Matrices for Linear Transformations.  Objective: Find the standard matrix for a linear transformation.  Objective: Find the standard matrix for the composition of linear transformations and find the inverse on an invertible linear transformation.  Objective: Find the matrix for a linear transformation relative to a nonstandard basis.  Objective: Prove results in linear algebra using linearity.

Recall Theorem 6.1, Part 4: If v = c1v1 + c2v2 + …+ cnvn then T(v) = T(c1v1 + c2v2 + …+ cnvn) = c1T(v1) + c2T(v2) + …+ cnT(vn)

This says that a linear transformation T: V  W is completely determined by its action on a basis of V. In other words, if {v1, v2, …, vn} is a basis of V and T(v1), T(v2), …, T(vn) are given, then T(v) is determined for any v in V. This is the key to representing T by a matrix.

Recall that the standard basis for Rn in column vector notation is 1 0 0       0 1 0 S = {e1, e2, …, en} =  , ,,             0 0 1

 Theorem 6.10: Standard Matrix for a Linear Transformation

n m n Let T: R  R be a linear transformation such that for the standard basis vectors ei of R ,

a11   a12  a1n        a21 a22 a2n T(e1) =   , T(e2) =   , …, T(en) =   .                am1  am2  amn  S*

Then the mn matrix whose n columns correspond to T(ei)

 11 12  aaa 1n    aaa  A =  21 22 2n        mm 31  aaa mn  is such that T(v) = Av for every v in Rn. A is called the standard matrix for T.

 Proof n Choose any v = v1e1 + v2e2 + …+ vnen in R .

Then T(v) = T(v1e1 + v2e2 + …+ vnen) = v1 T(e1) + v2 T(e2) + …+ vn T(en)

p. 183 Chapter 6: Linear Transformations. 6.3 Matrices for Linear Transformations.

v1  v  On the other hand, in column vector notation, v =  2       vn 

 11 12  aaa1n  v1   a11v1  a12v2  a1nvn    aaa v   a v  a v  a v  Av =  21 22 2n   2  =  21 1 22 2 2n n                 mm31  aaamn  vn  am1v1  am2v2  amn vn 

a11   a12  a1n        a21 a22 a2n = v1   + v2   + … + vn   = v1 T(e1) + v2 T(e2) + …+ vn T(en).                am1  am2  amn 

 Example Find the standard matrix for the linear transformation T(x, y) = (2x – 3y, x – y, y – 4x).

Solution: . T: R2  R3.

 Example Find the standard matrix A for the linear transformation T that is the reflection in the line y = x in R2, use A to find the image of the vector v = (3, 4) and sketch the graph of v and its image T(v).

S Solution:

A = [T(e1), T(e2)]

=

T(v) = T(3, 4) =

p. 184 Chapter 6: Linear Transformations. 6.3 Matrices for Linear Transformations.

n m m p S* The composition of T1: R  R with T2: R  R is defined by T(v) = T2(T1(v)) where v is a n vector in R . This composition is denoted by T2 ◦ T1.

 Theorem 6.11: Compostion of Linear Transformations

S* n m m p Let T1: R  R with T2: R  R be linear transformations with standard matrices A1 and A2, n m respectively. The composition T: R  R , defined by T(v) = T2(T1(v)), is a linear transformation. Moreover, the standard matrix A for T is given by the matrix product A = A2A1.

 Proof To show that T is a linear transformation, let u and v be vectors in R n and let c be a scalar.

Then T(u + v) = T2(T1(u + v)) =

T2(T1(u) + T1(v)) = T2(T1(u)) + T2(T1(v)) = T(u) + T(v)

Also, T(cv) = T2(T1(cv)) =

T2(cT1(v)) = cT2(T1(v)) = cT(v)

To show that A2A1 is the standard matrix for T,

T(v) = T2(T1(v)) = T2(A1v)

= A2(A1v)

= (A2A1)v using the associative property of matrix multiplication

Composition is not commutative because matrix multiplication is not commutative. We can generalize to compositions of three or more linear transformations. For example, T = T3 ◦ T2 ◦ T1 is defined by T(v) = T3(T2(T1(v))) and is represented by A3A2A1.

p. 185 Chapter 6: Linear Transformations. 6.3 Matrices for Linear Transformations.

3 2 2 3  Example Given the linear transformations T1: R  R with T2: R  R defined by  x   x  1 1 1  1 0 2 1 x   x  T x  = x  and T 1 = 0 2 1 1  2     2  1       1 1 0 x2  x2  x3  x3  1 0 Find the standard matrix for T = T ◦ T . S 2 1  Solution:

n n A linear transformation T1: R  R is invertible if and only if there exists a linear n n n transformation T2: R  R such that for every v in R ,

T2(T1(v)) = v and T1(T2(v)) = v

–1 T2 is the inverse of T1. When the inverse exists, it is unique, and we denote it by T1 .

Theorem 6.12 Existence of an Inverse Transformation

Let T: Rn  Rn be a linear transformation with a standard matrix A. Then the following conditions are equivalent.

1) T is invertible. 2) T is an isomorphism. 3) A is invertible.

If T is invertible with standard matrix A, then the standard matrix for T–1 is A–1.

 Example Given the linear transformations T: R3  R3 defined by

T(x1, x2, x3) = (x1, 2x1 + x2, 3x1 + x2 + x3). Show that T is invertible, and find its inverse.

 Solution: S

p. 186 Chapter 6: Linear Transformations. 6.3 Matrices for Linear Transformations.

Let V be an n-dimensional vector space with basis B = {v1, v2, …, vn}and W be an m- dimensional vector space with basis B = {w1, w2, …, wn}. Let T: V  W is a linear transformation such that

a11   a12   a1n        a21 a22 a2n [T(v1)]B =   , [T(v2)]B =   , …, [T(vn)]B =   ,                am1  am2  amn 

i.e. T(v1) = a11w1 + a21w2 + … + am1wm T(v2) = a12w1 + a22w2 + … + am2wm ⁞ ⁞ ⁞ ⁞ T(vn) = a1nw1 + a2nw2 + … + amnwm

Then the mn matrix whose n columns correspond to [T(vi)]B

 11 12  aaa 1n    aaa  A =  21 22 2n        mm 31  aaa mn 

is such that [T(v)]B = A[v]B for every v in V. A is called the matrix of T relative to the bases B and B.

 Example Let Dx: P3  P2 be the differential operator that maps a polynomial p of degree 3 or 2 3 less to its derivative p. Find the matrix for Dx using the bases B = {1, x, x , x } and B  = {1, x, x2}.

 Solution: The derivatives of the basis vectors in B are 0 [D (1)] = [0] = [0(1) + 0x + 0x2] =   x B B B 0 0

S0 [Dx(x)]B =

p. 187 Chapter 6: Linear Transformations. 6.3 Matrices for Linear Transformations.

p. 188 Chapter 6: Linear Transformations. 6.3 Matrices for Linear Transformations.

 Example Let B = B  = {1, cos(x), sin(x), cos(2x), sin(2x)} and let Dx: span(B)  span(B) be the

differential operator that maps a function f of to its derivative f . Find the matrix for Dx using the bases B and B.

 Solution: The derivatives of the basis vectors in B are 0   0 [Dx(1)]B = [0]B = [0(1) + 0cos(x) + 0sin(x) + 0cos(2x) + 0sin(2x)]B = 0 ;   0 0

S0

p. 189

Chapter 6: Linear Transformations. 6.4 Transition Matrices and Similarity.

6.4 Transition Matrices and Similarity.  Objective: Find and use a matrix for a linear transformation relative to a basis.  Objective: Use properties of similar matrices and diagonal matrices.

In 6.3 we saw that the matrix for a linear transformation T: V  W depends on the choice of basis B1 for V and B2 for W. Oftentimes, we can find bases that give a very simple matrix for a given linear transformation. In this section, we will consider linear transformations T: V  V and we will use the same basis for the domain as for the codomain. We will find matrices relative to different bases. (In Chapter 7, we will learn how to find the bases that give us simple matrices.)

The matrix for T: V  V relative to the basis B = {v1, v2, …, vn} is A, where [T(x)]B = A[x]B

The matrix for T: V  V relative to the basis B = {v1, v2, …, vn} is A, where [T(x)]B = A[x]B

  The transition matrix from B to B is P =  vv   v ][][][  , so P[x] = [x] BB  1 B 2 B Bn  B B  

–1 –1 A The transition matrix from B to B is P , so P [x]B = [x]B [x]B [T(x)]B

So we have [T(x)] = A[x] B B –1 and we also have P P –1 [T(x)]B = P BB A PBB [x]B Therefore, A [x]B [T(x)]B

–1 S0 A  = P AP

 826   Example: Let T: R3  R3 which has the standard matrix A =   . Find the matrix A 20 6 25   60  13 for T relative to the basis B = {(1, 3, –2), (2, 7, –3), (3, 10, –4)}. S0  Solution: A is the matrix relative to the standard basis S.

PBS =

p. 191 Chapter 6: Linear Transformations. 6.4 Transition Matrices and Similarity.

 4 0 0 (4) 2 0 0    Diagonal matrices: Observe that A2 =   = 0 (1) 2 0  0 1 0    2   0 0 2  0 0 (2)  (4) k 0 0  (4) k 1 0 0  k+1  k   k 1  and A =  0 (1) 0  =  0 (1) 0  .  k   k 1   0 0 (2)   0 0 (2)  Also, AT = A.

d1 0 0 0   0 d 0 0  If D =  2  is a diagonal matrix and none of the diagonal elements are zero,  0 0  0     0 0 0 d n 

1/ d1 0 0 0   0 1/ d 0 0  then D–1 =  2   0 0  0     0 0 0 1/ d n 

Two square matrices A and B are similar if and only there is an invertible matrix P such that B = P–1AP. For example, the matrix representations of a linear transformation relative to two different bases are similar.

Theorem 6:13: Properties of Similar Matrices

Let A, B, and C be square matrices. Then

1) A is similar to A. 2) If A is similar to B, then B is similar to A. 3) If A is similar to B and B is similar to C, then A is similar to C.

p. 192 Chapter 6: Linear Transformations. 6.5 Applications of Linear Transformations.

6.5 Applications of Linear Transformations.  Objective: Identify linear transformations defined by reflections, expansions, contractions, or shears in R2.  Objective: Use a linear transformation to rotate a figure in R3.

 Reflections in R2.

Reflection in y-axis Reflection in x-axis Horizonal reflection Vertical reflection T(x, y) = (–x, y) T(x, y) = (x, –y)

The matrix A for Tv = Av is A = [ Te1 | Te2 ]

p. 193 Chapter 6: Linear Transformations. 6.5 Applications of Linear Transformations. Reflection in the line y = x T(x, y) = (y, x) T(x, y) = (y, x)

The matrix A for Tv = Av is A = [ Te1 | Te2 ]

 Expansions and Contractions in R2.

Horizonal expansion Vertical expansion T(x, y) = (kx, y) for k > 1 T(x, y) = (x, ky) for k > 1

p. 194 Chapter 6: Linear Transformations. 6.5 Applications of Linear Transformations. Horizonal contraction Vertical contraction T(x, y) = (kx, y) for 0 < k < 1 T(x, y) = (x, ky) for 0 < k < 1

The matrix A for Tv = Av is A = [ Te1 | Te2 ]

p. 195 Chapter 6: Linear Transformations. 6.5 Applications of Linear Transformations.  Shears in R2.

Horizonal shear Vertical shear T(x, y) = (–x, y) T(x, y) = (x, –y)

The matrix A for Tv = Av is A = [ Te1 | Te2 ]

p. 196 Chapter 6: Linear Transformations. 6.5 Applications of Linear Transformations.  Rotation in R2.

T(x, y) = (x cos( ) – y sin( ), x sin( ) + y cos( ))

p. 197 Chapter 6: Linear Transformations. 6.5 Applications of Linear Transformations.

 Rotation in R3.

The matrix A for Tv = Av is A = [ Te1 | Te2 | Te3 ]

Rotation about x-axis by /6 rad = 30 counterclockwise (right-hand rule).

p. 198 Chapter 6: Linear Transformations. 6.5 Applications of Linear Transformations.

Rotation about y-axis by /6 rad = 30 Rotation about y-axis by /6 rad = 30

The matrix A for Tv = Av is A = [ Te1 | Te2 | Te3 ]

p. 199

Chapter 7: Eigenvalues and Eigenvectors. 7.1 Eigenvalues and Eigenvectors.

Chapter 7: Eigenvalues and Eigenvectors.

7.1 Eigenvalues and Eigenvectors.  Objective: Prove properties of eigenvalues and eigenvectors.  Objective: Verify eigenvalues and corresponding eigenvectors.  Objective: Find eigenvalues and corresponding eigenspaces.  Objective: Use the characteristic equation to find eigenvalues and eigenvectors.  Objective: Use software to find eigenvalues and eigenvectors.

 10 1 0 Recall the example A =   from Section 6.5. Then Ae1 =   =   = e2  01 0 1 0 1 and Ae2 =   =   = e1 1 0

 5.1   5.1  Let x1 =   . Then Ax1 =   = = 1x1  5.1   5.1 

 5.1   5.1   5.1  Let x2 =   . Then Ax2 =   =   = –1x2  5.1   5.1   5.1 

 Given square matrix A, if we have a nonzero vector x and a scalar  (“lambda”) such that S* Ax = x “Eigen” is German for “particular.” then x is an eigenvector of A with the corresponding eigenvalue .

Example: has eigenvectors x1 = , x2 = with eigenvalues 1= 1, 2 = –1.

p. 201 Chapter 7: Eigenvalues and Eigenvectors. 7.1 Eigenvalues and Eigenvectors.

1 0.65 0.05  Example Show that   is an eigenvector of   , and find the corresponding  2 0.30 0.60 eigenvalue.

S*

Solution:

 Theorem 7.1: Eigenvectors of  Form a Subspace

If A is an nn matrix with an eigenvalue , then the set of all eigenvectors of , together with the zero vector

{x: x is an eigenvector of }  {0} = {x: Ax = x}

is a subspace of Rn. S* Proof: {x: Ax = x} is not empty because it contains the zero vector. {x: Ax = x} is closed under scalar multiplication because for any vector x  {x: Ax = x} and for any scalar c, A(cx) = c(Ax) = c(x) = (cx)

{x: Ax = x} is closed under addition because for any vectors x1, x2  {x: Ax = x}, A(x1 + x2) = Ax1 + Ax2 = x1 + x2 = (x1 + x2)

To find an eigenvalue of a matrix A, we start with the equation Ax = x. Ax = x 0 = (I – A)x Ax = Ix I – A is singular 0 = Ix – Ax det(I – A) = 0

p. 202 Chapter 7: Eigenvalues and Eigenvectors. 7.1 Eigenvalues and Eigenvectors.  Theorem 7.2: Eigenvalues and Eigenvectors of a Matrix

Let A be an nn matrix.

1) An eigenvalue of A is a scalar  such det(I – A) = 0.

2) The eigenvectors of A corresponding to  are the nonzero solutions of (I – A)x = 0 i.e. the nullspace of I – A.

Proof: S* A has an eigenvector x  0 with eigenvalue  if and only if

if and only if A – I is singular (so its determinant is zero) and x is in the nullspace of A – I.

det(I – A) = 0 is called the characteristic equation of A. The characteristic polynomial det(I – A) is an nth-degree polynomial in .

The Fundamental Theorem of Algebra states that an nth-degree polynomial has exactly n roots (i.e. an nth-degree polynomial equation has exactly n solutions). In other words, we will always have

det(I – A) = ( – 1) ( – 2) … ( – n)

where the n constants 1, 2, …, n are the eigenvalues of A. Note that some i may be complex or some i may be repeated, for example 1 = 2.

 21   Example Find the eigenvalues and corresponding eigenvectors of A =   .  13 

Solution by hand: 0 = det(

p. 203 Chapter 7: Eigenvalues and Eigenvectors. 7.1 Eigenvalues and Eigenvectors.

so 1 =2 =

Eigenvector for 1 =

Eigenvector for 2 =

0 1  Example Find the eigenvalues and corresponding eigenvectors of A =   . 1 0 

Solution: 0 = det(

so 1 = 2 =

p. 204 Chapter 7: Eigenvalues and Eigenvectors. 7.1 Eigenvalues and Eigenvectors.

Eigenvector for 1 =

Eigenvector for 2 =

 1.6 0.8  0.8  Example Find the eigenvalues and corresponding eigenvectors of A =   .  0.4 1.2 0.8   0.4 0.8 1.2 

Solution: 0 = det(

so 1 = 2 = 3 =

Eigenvector for 1 =

p. 205 Chapter 7: Eigenvalues and Eigenvectors. 7.1 Eigenvalues and Eigenvectors.

Using rref we find which has a solution x =

E’vector for 2 = 3 =

p. 206 Chapter 7: Eigenvalues and Eigenvectors. 7.1 Eigenvalues and Eigenvectors.

Using rref we find which has solution

Note that this eigenvalue, which has multiplicity two, has two linearly independent eigenvectors.

a b  Example Find the eigenvalues and corresponding eigenvectors of A =   where b  0. 0 a

Solution: 0 = det(

so 1 = 2 =

Eigenvector for 1 =

Eigenvector for 2 =

Note that this eigenvalue has multiplicity two but has only one linearly independent eigenvector.

1 4 0 1  Using software to find eigenvalues and eigenvectors. Let A =   and B =   . 3 2 1 0 

TI-89: Use MatrixeigVl(A) and MatrixeigVc(A)

p. 207 Chapter 7: Eigenvalues and Eigenvectors. 7.1 Eigenvalues and Eigenvectors. The eigenvectors will be the columns of the matrix returned by eigVc, corresponding in order to the eigenvalues from eigVl. The eigenvectors are normalized (unit vectors).

However, eigVl and eigV will not work with complex eigenvalues and eigenvectors. In this case you must find the characteristic polynomial using

Matrixdet(xMatrixidentity(2)-b)

Then use the Algebra menu to solve: ComplexcSolve(x^2+1=0,x) or factor : ComplexcFactor(x^2+1)

Use rref to find the eigenvectors: Matrixrref(Matrixidentity(2)-b)

Mathematica: On the Basic Math Assistant Palette, under   Basic Commands |   | More   use Eigenvalues[a] and Eigenvectors[a]. For example, the eigenvalue 5 corresponds to the eigenvector (1, 1).

Mathematica will handle real and complex eigenvalues and eigenvectors. However, if you want the characteristic polynomial, you can use Det[x*IdentityMatrix[2]-a] Use the Factor[ ], Simplify[ ], and Solve[lhs ==rhs,var] funtions as appropriate. To find the eigenvectors the long way, you can use MatrixForm[RowReduce[ ]]

 Theorem 7.3: Eigenvalues of Triangular Matrices

If A is an nn triangular matrix, then its eigenvalues are the entries on the main diagonal.

S*

Proof: If A is an nn triangular matrix, then I – A is a triangular matrix with  – a11,  – a22, …,  – ann on the diagonal. Since I – A is triangular, its determinant is the product of the entries on the main diagonal ( – a11)( – a22)…( – ann) by Theorem 3.2. Therefore, the eigenvalues of A are the entries on the main diagonal: a11, a22, …, ann.

p. 208 Chapter 7: Eigenvalues and Eigenvectors. 7.2 Diagonalization.

7.2 Diagonalization.  Objective: Prove properties of eigenvalues and eigenvectors.  Objective: Find the eigenvalues of similar matrices, determine whether a matrix is diagonalizable, and find a matrix P such that P–1AP is diagonal.  Objective: Find, for a linear transformation T: V  V a basis B for V such that the matrix for T relative to B is diagonal.

An nn matrix A is diagonalizable when there exists an invertible matrix P such that D = P–1AP is a diagonal matrix.

Recall from Section 6.4 that two square matrices A and B are similar if and only there is an invertible matrix P such that B = P–1AP. So we can also say that A is diagonalizable when it it similar to a diagonal matrix D.

 Theorem 7.4: Similar Matrices Have the Same Eigenvalues

If A and B are similar nn matrices, then they have the same eigenvalues.

S*Proof: Because A and B are similar, there exists an invertible matrix P such that B = P–1AP, so

1 |I – B| = | P–1IP – P–1AP | = |P–1(I – A)P| = |P–1| |(I – A)| |P| = |(I – A)| |P| = |(I – A)| P ||

Since A and B have the same characteristic polynomial, they must have the same eigenvalues.

 Theorem 7.5: Condition for Diagonalization

An nn matrix A is diagonalizable if and only if it has n linearly independent eigenvectors.

Proof:

First, assume A is diagonalizable. Then there exists an invertible matrix P such that D = P–1AP is a diagonal matrix. Then PD = AP. Let the column vectors of P be p1, p2, …, pn and the main diagonal entries of D be 1, 2, …, n.

1  00    0 2  0 Now PD = [ p1 | p2 | … | pn ]   = [ 1p1 | 2p2 | … | npn ]       000 n 

and AP = A[ p1 | p2 | … | pn ] = [ Ap1 | Ap2 | … | Apn ]

p. 209 Chapter 7: Eigenvalues and Eigenvectors. 7.2 Diagonalization.

Since PD = AP, we have Api = ipi for each of the n column vectors of P. Since P is invertible, the n column vectors are also linearly independent. So A has n linearly independent eigenvectors.

Conversely, assume A has n linearly independent eigenvectors p1, p2, …, pn with corresponding eigenvalues 1, 2, …, n. Let P be the matrix whose columns are the n eigenvectors: P = [ p1 | p2 | … | pn ]. Then

AP = A[ p1 | p2 | … | pn ] = [ Ap1 | Ap2 | … | Apn ] = [ 1p1 | 2p2 | … | npn ]

1  00   0 2  0 = [ p1 | p2 | … | pn ]        000n 

Thus, AP = PD, where D = is a diagonal matrix. Since the are linearly

independent, P is invertible, so D = P–1AP and A is diagonalizable.

3 2 1  Example Diagonalize A =   by finding an invertible matrix P and a diagonal matrix D 0 0 2 0 2 0 such that D = P–1AP. Use the characteristic polynomial to find the eigenvalues and the reduced row-echechelon function of your software to find the eigenvectors.

Solution: 0 = det(

=

so 1 = 2 = 3 =

p. 210 Chapter 7: Eigenvalues and Eigenvectors. 7.2 Diagonalization.

Eigenvector for 1 =

Eigenvector for 2 =

Eigenvector for 3 =

p. 211 Chapter 7: Eigenvalues and Eigenvectors. 7.2 Diagonalization.

 1 2  2  Example: Diagonalize A =   by finding an invertible matrix P and a diagonal  2 5  2  6 6  3 matrix D such that D = P–1AP. Use eigVl and EigVc on the TI-89, or Eigenvalues[ ] and Eigenvectors[ ] in Mathematica, or eigenvalues and eigenvectors in PocketCAS.

 Solution:

TI-89: MatrixeigVl(A) MatrixeigVc(A).P  P^-1*A*P

S*  Mathematica: Eigenvalues[a]

MatrixForm[p=Transpose[Eigenvectors[a]]] (Note that Mathematica gives the transpose of P.)

MatrixForm[Inverse[p].a.p]

PocketCAS eigenvalues(a) [3, 3, –3]  3 0 1 p:=eigenvectors(a)    0 3 1  3 3 3 3 0 0  p^-1*a*p   0 3 0  0 0  3

p. 212 Chapter 7: Eigenvalues and Eigenvectors. 7.2 Diagonalization.

1 0 0 Example: Try to diagonalize A =   Use eigVl and EigVc on the TI-89, or 1 2 1 1 0 2 Eigenvalues[ ] and Eigenvectors[ ] in Mathematica, or eigenvalues and eigenvectors in PocketCAS.

 Solution:

TI-89: MatrixeigVl(A) MatrixeigVc(A).P  P^(-1)*A*P Error: Singular Matrix

Mathematica: Eigenvalues[a]

MatrixForm[p=Transpose[Eigenvectors[a]]] (Note that Mathematica gives the transpose of P.)

MatrixForm[Inverse[p].a.p]

PocketCAS eigenvalues(a)  [2, 2, 1] p:=eigenvectors(a)  Not diagonalizable at eigenvalue 2

p. 213

Chapter 7: Eigenvalues and Eigenvectors. 7.3 Symmetric Matrices and Orthogonal Diagonalization.

7.3 Symmetric Matrices and Orthogonal Diagonalization.  Objective: Recognize, and apply properties of, symmetric matrices.  Objective: Recognize, and apply properties of, orthogonal matrices.  Objective: Find an orthogonal matrix P that orthogonally diagonalizes a symmetric matrix A.

A square matrix A is symmetric when A = AT.

 Theorem 7.7: Properties of Symmetric Matrices

If A is a symmetric nn matrix, then S*1) A is diagonalizable. 2) All eigenvalues of A are real. 3) If  is an eigenvalue of A with multiplicity k, then  has k linearly independent eigenvectors. That is, the eigenspace of  has dimension k.

The proof will be deferred until Ch. 8.5

 Theorem 7.9: Property of Symmetric Matrices

Let A be an nn symmetric matrix. if 1 and 2 are distinct eigenvalues of A, then their corresponding eignevectors x1 and x2 are orthogonal.

Proof:

Let 1 and 2 are distinct eigenvalues of A with corresponding eignevectors x1 and x2, so T Ax1 = 1x1 and Ax2 = 2x2. Recall that we can write x1•x2 = x1 x2. Then

1 (x1•x2) = (1x1)•x2 S* = (Ax1)•x2

= 2(x1•x2)

Therefore, (1 – 2)(x1•x2) = 0. Since 1 – 2  0, we must have x1•x2 = 0 so x1 and x2 are orthogonal.

p. 215 Chapter 7: Eigenvalues and Eigenvectors. 7.3 Symmetric Matrices and Orthogonal Diagonalization.

0 1 1  Example: Find the eigenvalues and eigenvectors of   . Are the eigenvalues real? What 1 0 1 1 1 0 are the dimensions of the eigenspaces? Find the angles between the eigenvectors.

Solution:

1 0 0 0 1 1 0 = det(   –   ) 0 1 0 1 0 1 0 0 1 1 1 0

(using the calculator to take the determinant and factor it.)

1 = 2, 2 = –1, 3 = –1. All eigenvalues are real.

Solve (1I – )x1 = 0 using rref.

1  Write x =   .  = t is a free variable. We solve for  = t and  = t. 1  2  3 1 1    3   is the t 1 Greek letter The solution is   = t   so x = . xi t 1 1 t 1 The eigenspace of  = 2 (multiplicity = 1) is one-dimesional.

Solve (2I – )x2 = 0 and

Solve (3I – )x3 = 0 using rref. (Note 2 = 3.)

2 = s and 3 = t are free variables. We solve for 1 = –s – t.

 s  t 1 1 1 1 The solution is   = s   + t   so x =   and x =   .  s   1   0  2  1  3  0   t   0   1   0   1 

p. 216 Chapter 7: Eigenvalues and Eigenvectors. 7.3 Symmetric Matrices and Orthogonal Diagonalization. The eigenspace of  = –1 (multiplicity = 2) is two-dimensional.

x1•x2 = (1, 1, 1)•(–1, 1, 0) = 0 so x1 is orthogonal to x2.

x1•x3 = (1, 1, 1)•(–1, 0, 1) = 0 so x1 is orthogonal to x3.

(1,1,0)  (1,0,1) 1 1  x2•x3 = = = so the angle between x2 and x3 is arccos( ) = . (1,1,0) (1,0,1) 2 2 2 3

Alternative method:

Use eigVl and eigVc on the calculator to obtain

1 = 3 = –1 and 2 = 2

with eigenvectors  .816497 .57735  .007355 p =   , p =   , p =   1  .408248  2 .57735 3  .703401  .408248  .57735  .710756 

–14 We can calculate ||p1|| = ||p2|| = ||p3|| = 1, p1•p2 = 10 , p1•p3 = 0.009008,

–15 and p2•p3 = –6.810 ,

so the eigenvectors are almost orthonormal.

A square matrix P is called orthogonal when it is invertible and P–1 = PT.

 Theorem 7.8: Property of an Orthogonal Matrix

An nn matrix P is orthogonal if and only if its column vectors form an orthonormal set.

Proof:

Write P = [ p1 | p2 | … | pn ] in terms of its column vectors. Then S* T T T p1  p1 p1 p1 p 2  p1 p n     T T T  p p p p p  p p PT =  2  and PTP =  2 1 2 2 2 n  .              p T T T  n  p n p1 p n p 2  p n p n  T 2 Thus, P P = I if and only if ||pi|| = pi•pi = 1 for each i and pi•pj = 0 whenever i  j. Therefore, PT = P –1 if and only if its column vectors form an orthonormal set.

p. 217 Chapter 7: Eigenvalues and Eigenvectors. 7.3 Symmetric Matrices and Orthogonal Diagonalization. Because of this theorem, it would have been better if orthogonal matrices had been called “orthonormal matrices.” but we are stuck with the name “orthogonal.”

 Theorem 7.10: Fundamental Property of an Symmetric Matrices

Let A be a real nn matrix. Then A is orthogonally diagonalizable and has real eigenvalues if and only if A is symmetric.

Proof: Suppose A is orthogonally diagonalizable and real, so there exists a real, orthogonal matrix P such that D = P –1AP is diagonal and real. Then = PDP –1 = PDPT so AT = (PDPT)T = (PT)TDTPT = PDPT S* = A

Conversely, Theorems 7.7 and 7.9 together tell us that all symmetric matrices are orthogonally diagonalizable with real eigenvalues.

An nn symmentric matrix has n eigenvalues, counting multiplicity. Because the dimension of each eigenspace equals the multiplicity of the corresponding eigenvalue, we have n linearly independent eigenvectors, so the matrix is diagonalizable (Theorem 7.5).

Eigenvectors with different eigenvalues are orthogonal. Within each eigenspace, we can use the Gram-Schmidt process to find orthonormal eigenvectors. The set of all of the orthonormal eigenvectors from all of the eigenspaces make up the columns of an orthogonal matrix P, such that

D = P –1AP

is diagonal.

0 1 1  Example (continued): Diagonalize   using… 1 0 1 1 1 0 1) the characteristic equation and rref 2) eigVl and eigVc on the TI-89 3) Eigenvalues[ ] and Eigenvectors[ ] in Mathematica 4) eigenvalues and eigenvectors in PocketCAS

p. 218 Chapter 7: Eigenvalues and Eigenvectors. 7.3 Symmetric Matrices and Orthogonal Diagonalization.  Solution:

1) From the first part of the example, 1 = 2, 2 = –1, 3 = –1. 0 1 1 Solve ( I –   )x = 0 using rref. 1 1 0 1 1 1 1 0

3 = t is a free variable. We solve for 1 = t and 1 = t. t 1 The solution is   = t   so x = . t 1 1 t 1

Solve (2,3I – )x2,3 = 0

2 = s and 3 = t are free variables. We solve for 1 = –s – t.  s  t 1 1 1 1 The solution is   = s   + t   so x =   and x =   .  s   1   0  2  1  3  0   t   0   1   0   1  1 x 1 Normalize p = 1 =   . 1 1 x1 3 1 1 x 1 Gram-Schmidt: p = 2 =   2  1  x 2 2  0  1/ 2 1  1  u 1 u = x – (x •p )p = – =   ; p = 2 =   3 3 3 2 2   1/ 2 2 1  2  u 2 6  1   2  2 0 0  1/ 3 1/ 2 1/ 6   so D =   and P = 1/ 3 1/ 2 1/ 6 0 1 0        0 0 1 1/ 3 0 2 / 6 

Check that PTP = I and multiply P –1AP = PTAP to check that it gives you D.

p. 219 Chapter 7: Eigenvalues and Eigenvectors. 7.3 Symmetric Matrices and Orthogonal Diagonalization. 2) TI-89: Use eigVl and eigVc to obtain

1 0 0   .816497 .57735  .077355 D =   and P =    0 2 0   .408248 .57735  .703401  0 0 1  .408248 .57735 .710756 

 1. 1.E 14 .009008  Check PTP =   so P is almost orthogonal. 1.E 14 1.  6.8E 15 .009008  6.8E 15 1. 

By looking at the ratios between components, it appears that

p1 = unitV([-2,1,1]) =  6 /3 6 / 6 6 / 6

p2 = unitV([1,1,1]) =  3 /3 3 /3 3 /3

p3 = unitV([0,-1,1]) = 0  2 / 2 2 / 2

 6 / 3 3 / 3 0    Try a new P =  6 / 6 3 / 3  2 / 2    6 / 6 3 / 3 2 / 2 

.816497 .57735 0  p gives    .408248 .57735  .707107  .408248 .57735 .707107 

Check that PTP = I and multiply P –1AP = PTAP to check that it gives you D.

2 0 0  3) Mathematica: Eigenvalues[a] yields {2,-1,-1} so D =   0 1 0  0 0 1

Eigenvalues[a] gives {{1,1,1},{-1,0,1},{-1,1,0}} so

1 1 1 x =   , x =   , x =   1 1 2  0  3  1  1  1   0 

Warning: MatrixForm[Eigenvalues[a]] gives which has the eigenvectors in the rows (instead of in the columns)!

p. 220 Chapter 7: Eigenvalues and Eigenvectors. 7.3 Symmetric Matrices and Orthogonal Diagonalization. Orthogonalize[Eigenvalues[a]] gives the orthonormal basis

1/ 3 1/ 2 1/ 6   so P = 1/ 3 0 2 / 3    1/ 3 1/ 2 1/ 6 Warning: MatrixForm[Orthogonalize[Eigenvalues[a]]] gives PT, not P!

Check that PTP = I and multiply P –1AP = PTAP to check that it gives you D.

4) PocketCAS: Computing M TM shows in the off-diagonal elements that the vectors are orthogonal, and the diagonal elements give the norm squared of each vector.

2 0 0  so D =   0 1 0  0 0 1 gramschmidt( ) will construct an eigenvectors( ) orthonormal basis from the rows of a matrix, gives orthogonal, so use gramschmidt(transpose(p)) to but unnormalized obtain PT. eigenvectors.

1/ 3 1/ 2 1/ 6   so P = 1/ 3 0 2 / 6    1/ 3 1/ 2 1/ 6 Check that PTP = I and multiply P –1AP = PTAP to check that it gives you D.

p. 221 Chapter 7: Eigenvalues and Eigenvectors. 7.3 Symmetric Matrices and Orthogonal Diagonalization.

 2 2 4 Example: Diagonalize   .  2  2 4  4 4 4

p. 222 Chapter 7: Eigenvalues and Eigenvectors. 8.5 Unitary and Hermitian Matrices.

8.5 Unitary and Hermitian Matrices.  Objective: Find the conjugate transpose (Hermitian conjugate) of a complex matrix A.  Objective: Determine if a matrix A is unitary.  Objective: Find the eigenvalues and eigenvectors of a Hermitian matrix, and diagonalize a Hemitian matrix.

In order to prove the theorems in Section 7.4 about real symmetric matrices, we must consider complex “Hermitian” matices.

The conjugate transpose (or Hermitian conjugate) of a complex matrix A, denoted by A† or AH or A*, is given by A† = (퐴̅ )푇 = 퐴̅̅̅푇̅. That is, Take the complex conjugate of each of entry of A, and transpose the matrix.

Notice that if A is real (all entries are real), then A† = AT.

A is Hermitian if and only if A† = A. Notice that if A is real, then Hermitian is the same as symmetric.

To take the Hermitian conjugate A† on the TI-89, use aMatrixT

To take the Hermitian conjugate A† in Mathematica, use act

To take the Hermitian conjugate A† in PocketCAS, use trn(a)

A is unitary if and only if A† = A –1. Notice that if A is real, then unitary is the same as orthogonal.

 Example: Classify the following matrices as Hermitian, unitary, both, or neither.

 0 i  †  0 i  † † 1) A =   : A =   . A = A so A is Hermitian. AA = I so A is also unitary.  i 0  i 0

2) B = : B† = . B†  B so B is not Hermitian.

BB† = I so B is unitary.

1   234 i † 1   234 i † 3) C =   : C =   . C = C so C is Hermitian. 7   i 523  7   i 523 

CC† =  I so C is not unitary.

p. 223 Chapter 7: Eigenvalues and Eigenvectors. 8.5 Unitary and Hermitian Matrices.

1 0 † 1 0  † † 1 0  4) D =   : D =   . D  D so D is not Hermitian. D D =    I so D is not 0 i  0  i 0 1 unitary.

Theorem 8.8: Properties of the Hermitian Conjugate (Complex Transpose)

If A, B, and C complex matrices with dimensions mn, mn, and np respectively, and k is a complex number, then the following properties are true. S* 1) (A†)† = A 2) (A + B)† = A† + B† 3) (kA)† = 푘̅A† 4) (AC)† = C†A†

Theorem 8.9: Unitary Matrices

An nn complex matrix U is unitary if and only if its row vectors form an orthonormal set in Cn using the Euclidean inner product.

Proof: Recall that the Euclidean inner product of u and v in Cn is given by u•v = 푢1푣̅̅1̅ + 푢2푣̅̅2̅ + ⋯ + 푢푛푣̅̅푛̅ Note that if we write u and v as row vectors, then u•v = u v† ||u|| = u  u and u is orthogonal to v if and only if u•v = 0.

u1  u  Write A =  2  in terms of its row vectors. Then      u n 

u1  u1 u1  u 2  u1  u n    T † † † † u 2  u1 u 2  u 2  u 2  u n A = [ u1 | u 2 | … | un ] and AA =   .         u n  u1 u n  u 2  u n  u n  † 2 Thus, AA = I if and only if ||ui|| = ui•ui = 1 for each i and ui•uj = 0 whenever i  j. Therefore, A† = A –1 if and only if its row vectors form an orthonormal set.

Theorem 8.9.1: A is unitary if and only if A† is unitary.

p. 224 Chapter 7: Eigenvalues and Eigenvectors. 8.5 Unitary and Hermitian Matrices. Corollary 8.9.2: Unitary Matrices

An nn complex matrix A is unitary if and only if its column vectors form an orthonormal set in Cn using the Euclidean inner product.

Theorem 8.10: The Eigenvalues of a Hermitian Matrix

If H is a Hermitian matrix, then its eigenvalues are real.

Proof: Let  be an eigenvalue of H and let v  0 be its corresponding eigenvalue, so Hv = v. Then (v†Hv)† = v†H †(v†)† = v†Hv Now v†Hv = v†(Hv) = v†(v) = (v†v) = ||v||2 So (v†Hv)† = ̅̅‖̅̅퐯̅̅‖̅2̅ = ̅||v||2 because ||v|| is real. Since ||v||  0 and ||v||2 = ̅||v||2,  must be real.

p. 225

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420 Chapter 8 Complex Vector Spaces

8.5 Unitary and Hermitian Matrices

Find the conjugate transpose A* of a complex matrix A. Determine if a matrix A is unitary. Find the eigenvalues and eigenvectors of a Hermitian matrix, and diagonalize a Hermitian matrix.

CONJUGATE TRANSPOSE OF A MATRIX Problems involving diagonalization of complex matrices and the associated eigenvalue problems require the concepts of unitary and Hermitian matrices. These matrices roughly correspond to orthogonal and symmetric real matrices. In order to define unitary and Hermitian matrices, the concept of the conjugate transpose of a complex matrix must first be introduced.

Definition of the Conjugate Transpose of a Complex Matrix The conjugate transpose of a complex matrix A, denoted by A *, is given by A* A T where the entries of A are the complex conjugates of the corresponding entries of A.

Note that if A is a matrix with real entries, then A * AT . To find the conjugate transpose of a matrix, first calculate the complex conjugate of each entry and then take the transpose of the matrix, as shown in the following example.

Finding the Conjugate Transpose of a Complex Matrix Determine A * for the matrix 3 7i 0 A ΄ ΅. 2i 4 i SOLUTION 3 7i 0 3 7i 0 A ΄ ΅ ΄ 2i 4 i΅ 2i 4 i 3 7i 2i A* AT ΄ ΅ 0 4 i

Several properties of the conjugate transpose of a matrix are listed in the following theorem. The proofs of these properties are straightforward and are left for you to supply in Exercises 47Ð50.

THEOREM 8.8 Properties of the Conjugate Transpose If A and B are complex matrices and k is a complex number, then the following properties are true. 1.͑A*͒* A 2. ͑A B͒* A* B* 3.͑kA͒* kA* 4. ͑AB͒* B*A* 9781133110873_0805.qxp 3/10/12 6:55 AM Page 421

8.5 Unitary and Hermitian Matrices 421

UNITARY MATRICES Recall that a real matrix A is orthogonal if and only if A1 AT. In the complex system, matrices having the property that A1 A * are more useful, and such matrices are called unitary.

Definition of Unitary Matrix A complex matrix A is unitary when A1 A*.

A Unitary Matrix

Show that the matrix A is unitary. 1 1 i 1 i A ΄ ΅ 2 1 i 1 i SOLUTION Begin by finding the product AA *. 1 1 i 1 i 1 1 i 1 i AA* ΄ ΅ ΄ ΅ 2 1 i 1 i 2 1 i 1 i 1 4 0 ΄ ΅ 4 0 4 1 0 ΄ ΅ 0 1 Because 1 0 AA* ΄ ΅ I 0 1 2 it follows that A* A1. So,A is a unitary matrix.

Recall from Section 7.3 that a real matrix is orthogonal if and only if its row (or column) vectors form an orthonormal set. For complex matrices, this property characterizes matrices that are unitary. Note that a set of vectors ͭ ͮ v1, v2, . . . , vm in Cn (a complex Euclidean space) is called orthonormal when the statements below are true. ʈ ʈ 1. vi 1, i 1, 2, . . . , m 2. vi vj 0, i j The proof of the next theorem is similar to the proof of Theorem 7.8 presented in Section 7.3.

THEOREM 8.9 Unitary Matrices An n n complex matrix A is unitary if and only if its row (or column) vectors form an orthonormal set in Cn. 9781133110873_0805.qxp 3/10/12 6:55 AM Page 422

422 Chapter 8 Complex Vector Spaces

The Row Vectors of a Unitary Matrix

Show that the complex matrix A is unitary by showing that its set of row vectors forms an orthonormal set in C 3. 1 1 i 1 2 2 2 i i 1 A ΄ Ί3 Ί3 Ί3΅ 5i 3 i 4 3i 2Ί15 2Ί15 2Ί15 SOLUTION

Let r1, r2, and r3 be defined as follows. 1 1 i 1 i i 1 r ΂ , , ΃, r ΂ , , ΃, 1 2 2 2 2 Ί3 Ί3 Ί3 5i 3 i 4 3i r ΂ , , ΃ 3 2Ί15 2Ί15 2Ί15

Begin by showing that r1, r2, and r3 are unit vectors. 1 1 1 i 1 i 1 1 1͞2 1 2 1 1͞2 ʈr ʈ ΄΂ ΃΂ ΃ ΂ ΃΂ ΃ ΂ ΃΂ ΃΅ ΄ ΅ 1 1 2 2 2 2 2 2 4 4 4 i i i i 1 1 1͞2 ʈr ʈ ΄΂ ΃΂ ΃ ΂ ΃΂ ΃ ΂ ΃΂ ΃΅ 2 Ί3 Ί3 Ί3 Ί3 Ί3 Ί3 1 1 1 1͞2 ΄ ΅ 3 3 3 1 5i 5i 3 i 3 i 4 3i 4 3i 1͞2 ʈr ʈ ΄΂ ΃΂ ΃ ΂ ΃΂ ΃ ΂ ΃΂ ΃΅ 3 2Ί15 2Ί15 2Ί15 2Ί15 2Ί15 2Ί15 25 10 25 1͞2 ΄ ΅ 60 60 60 1 Then show that all pairs of distinct vectors are orthogonal. 1 i 1 i i 1 1 r r ΂ ΃΂ ΃ ΂ ΃΂ ΃ ΂ ΃΂ ΃ 1 2 2 Ί3 2 Ί3 2 Ί3 i i 1 1 2Ί3 2Ί3 2Ί3 2Ί3 0 1 5i 1 i 3 i 1 4 3i r r ΂ ΃΂ ΃ ΂ ΃΂ ΃ ΂ ΃΂ ΃ 1 3 2 2Ί15 2 2Ί15 2 2Ί15 5i 4 2i 4 3i 4Ί15 4Ί15 4Ί15 0 i 5i i 3 i 1 4 3i r r ΂ ΃΂ ΃ ΂ ΃΂ ΃ ΂ ΃΂ ΃ REMARK 2 3 Ί3 2Ί15 Ί3 2Ί15 Ί3 2Ί15 5 1 3i 4 3i Try showing that the column vectors of A also form an 6Ί5 6Ί5 6Ί5 orthonormal set in C 3. 0 ͭ ͮ So,r1, r2, r3 is an orthonormal set. 9781133110873_0805.qxp 3/10/12 6:55 AM Page 423

8.5 Unitary and Hermitian Matrices 423

HERMITIAN MATRICES A real matrix is symmetric when it is equal to its own transpose. In the complex system, the more useful type of matrix is one that is equal to its own conjugate transpose. Such a matrix is called Hermitian after the French mathematician Charles Hermite (1822Ð1901).

Definition of a Hermitian Matrix A square matrix A is Hermitian when A A*.

As with symmetric matrices, you can recognize Hermitian matrices by inspection. To see this, consider the 2 2 matrix A. a1 a2i b1 b2i A ΄ ΅ c1 c2i d1 d2i The conjugate transpose of A has the form A* AT a a i c c i 1 2 1 2 ΄ ΅ b1 b2i d1 d2i a1 a2i c1 c2i ΄ ΅. b1 b2i d1 d2i If A is Hermitian, then A A*. So,A must be of the form a1 b1 b2i A ΄ ΅. b1 b2i d1 Similar results can be obtained for Hermitian matrices of order n n. In other words, a square matrix A is Hermitian if and only if the following two conditions are met. 1. The entries on the main diagonal of A are real.

2. The entry aij in the i th row and the j th column is the complex conjugate of the entry aji in the j th row and the i th column.

Hermitian Matrices

Which matrices are Hermitian? 1 3 i 0 3 2i a.΄ ΅ b. ΄ ΅ 3 i i 3 2i 4 3 2 i 3i 1 2 3 c.΄2 i 0 1 i΅ d. ΄ 2 0 1΅ 3i 1 i 0 3 1 4 SOLUTION a. This matrix is not Hermitian because it has an imaginary entry on its main diagonal. b. This matrix is symmetric but not Hermitian because the entry in the first row and second column is not the complex conjugate of the entry in the second row and first column. c. This matrix is Hermitian. d. This matrix is Hermitian because all real symmetric matrices are Hermitian. 9781133110873_0805.qxp 3/10/12 6:55 AM Page 424

424 Chapter 8 Complex Vector Spaces

REMARK One of the most important characteristics of Hermitian matrices is that their Note that this theorem implies eigenvalues are real. This is formally stated in the next theorem. that the eigenvalues of a real symmetric matrix are real, as stated in Theorem 7.7. THEOREM 8.10 The Eigenvalues of a Hermitian Matrix If A is a Hermitian matrix, then its eigenvalues are real numbers.

PROOF Let be an eigenvalue of A and let a1 b1i a b i v 2 2 ΄ . ΅ . an bni be its corresponding eigenvector. If both sides of the equation Av v are multiplied by the row vector v*, then ͑ ͒ ͑ ͒ ͑ 2 2 2 2 . . . 2 2͒ v*Av v* v v*v a1 b1 a2 b2 an bn . Furthermore, because ͑v*Av͒* v*A*͑v*͒* v*Av it follows that v *Av is a Hermitian 1 1 matrix. This implies that v *Av is a real number, so is real.

To find the eigenvalues of complex matrices, follow the same procedure as for real matrices.

Finding the Eigenvalues of a Hermitian Matrix

Find the eigenvalues of the matrix A. 3 2 i 3i A ΄2 i 0 1 i΅ 3i 1 i 0 SOLUTION The characteristic polynomial of A is 3 2 i 3i Խ I AԽ 2 i 1 i Խ Խ 3i 1 i ͑ 3͒͑2 2͒ ͑2 i͓͒͑2 i͒ ͑3i 3͔͒ 3i ͓͑1 3i͒ 3i͔ ͑3 32 2 6͒ ͑5 9 3i͒ ͑3i 9 9͒ 3 32 16 12 ͑ 1͒͑ 6͒͑ 2͒. So, the characteristic equation is ͑ 1͒͑ 6͒͑ 2͒ 0, and the eigenvalues of A are 1, 6, and 2. 9781133110873_0805.qxp 3/10/12 6:55 AM Page 425

8.5 Unitary and Hermitian Matrices 425

To find the eigenvectors of a complex matrix, use a procedure similar to that used for a real matrix. For instance, in Example 5, to find the eigenvector corresponding to the eigenvalue 1, substitute the value for into the equation 3 2 i 3i v1 0 2 i 1 i v 0 ΄ ΅΄ 2΅ ΄ ΅ 3i 1 i v3 0 to obtain 4 2 i 3i v1 0 2 i 1 1 i v 0 . ΄ ΅΄ 2΅ ΄ ΅ 3i 1 i 1 v3 0 Solve this equation using Gauss-Jordan elimination, or a graphing utility or software program, to obtain the eigenvector corresponding to 1 1, which is shown below. TECHNOLOGY 1 Some graphing utilities and v 1 2i software programs have built-in 1 ΄ ΅ programs for finding the 1 eigenvalues and corresponding eigenvectors of complex Eigenvectors for 2 6 and 3 2 can be found in a similar manner. They are matrices. 1 21i 1 3i ΄ 6 9i΅ and΄2 i΅, respectively. 13 5

LINEAR Quantum mechanics had its start in the early 20th century ALGEBRA as scientists began to study subatomic particles and light. APPLIED Collecting data on energy levels of atoms, and the rates of transition between levels, they found that atoms could be induced to more excited states by the absorption of light. German physicist Werner Heisenburg (1901–1976) laid a mathematical foundation for quantum mechanics using matrices. Studying the dispersion of light, he used vectors to represent energy levels of states and Hermitian matrices to represent “observables” such as momentum, position, and acceleration. He noticed that a measurement yields precisely one real value and leaves the system in precisely one of a set of mutually exclusive (orthogonal) states. So, the eigenvalues are the possible values that can result from a measurement of an observable, and the eigenvectors are the corresponding states of the system following the measurement. Let matrix A be a diagonal Hermitian matrix that represents an observable. Then consider a physical system whose state is represented by the column vector u. To measure the value of the observable A in the system of state u, you can find the product

a11 0 0 u1 u*Au ͓u u u ͔ 0 a 0 u 1 2 3 ΄ 22 ΅΄ 2΅ 0 0 a33 u3 ͑ ͒ ͑ ͒ ͑ ͒ u1u1 a11 u2u2 a22 u3u3 a33. Because A is Hermitian and its values along the diagonal are real, u*Au is a real number. It represents the average of the values given by measuring the observable A on a system in the state u a large number of times. Jezper/Shutterstock.com 9781133110873_0805.qxp 3/10/12 6:55 AM Page 426

426 Chapter 8 Complex Vector Spaces

Just as real symmetric matrices are orthogonally diagonalizable, Hermitian matrices are unitarily diagonalizable. A square matrix A is unitarily diagonalizable when there exists a unitary matrix P such that P1AP is a diagonal matrix. Because P is unitary,P1 P *, so an equivalent statement is that A is unitarily diagonalizable when there exists a unitary matrix P such that P *AP is a diagonal matrix. The next theorem states that Hermitian matrices are unitarily diagonalizable.

THEOREM 8.11 Hermitian Matrices and Diagonalization If A is an n n Hermitian matrix, then 1. eigenvectors corresponding to distinct eigenvalues are orthogonal. 2. A is unitarily diagonalizable.

PROOF

To prove part 1, let v1 and v2 be two eigenvectors corresponding to the distinct (and real) eigenvalues 1 and 2. Because Av1 1v1 and Av2 2v2, you have the ͑ ͒ equations shown below for the matrix product Av1 *v2. ͑ ͒ Av1 **v2 v1 A**v2 v1 Av2 v1 *2v2 2v1 *v2 ͑ ͒ ͑ ͒ Av1 ****v2 1v1 v2 v1 1v2 1v1 v2 So, 2v1*v2 1v1*v2 0 ͑ ͒ 2 1 v1*v2 0 v1*v2 0 because 1 2

and this shows that v1 and v2 are orthogonal. Part 2 of Theorem 8.11 is often called the Spectral Theorem, and its proof is left to you.

The Eigenvectors of a Hermitian Matrix

The eigenvectors of the Hermitian matrix shown in Example 5 are mutually orthogonal because the eigenvalues are distinct. Verify this by calculating the Euclidean inner

products v1 v2, v1 v3, and v2 v3. For example, v1 v2 ͑1͒͑1 21i͒ ͑1 2i͒͑6 9i͒ ͑1͒͑13͒ ͑1͒͑1 21i͒ ͑1 2i͒͑6 9i͒ 13 1 21i 6 9i 12i 18 13 0.

The other two inner products v1 v3 and v2 v3 can be shown to equal zero in a similar manner.

The three eigenvectors in Example 6 are mutually orthogonal because they correspond to distinct eigenvalues of the Hermitian matrix A . Two or more eigenvectors corresponding to the same eigenvalue may not be orthogonal. Once any set of linearly independent eigenvectors is obtained for an eigenvalue, however, the Gram-Schmidt orthonormalization process can be used to find an orthogonal set. 9781133110873_0805.qxp 3/10/12 6:55 AM Page 427

8.5 Unitary and Hermitian Matrices 427

Diagonalization of a Hermitian Matrix

Find a unitary matrix P such that P *AP is a diagonal matrix where 3 2 i 3i A ΄2 i 0 1 i΅. 3i 1 i 0 SOLUTION The eigenvectors of A are shown after Example 5. Form the matrix P by normalizing these three eigenvectors and using the results to create the columns of P. ʈ ʈ ʈ͑ ͒ʈ Ί Ί v1 1, 1 2i, 1 1 5 1 7 ʈ ʈ ʈ͑ ͒ʈ Ί Ί v2 1 21i, 6 9i, 13 442 117 169 728 ʈ ʈ ʈ͑ ͒ʈ Ί Ί v3 1 3i, 2 i, 5 10 5 25 40 So, 1 1 21i 1 3i Ί7 Ί728 Ί40 1 2i 6 9i 2 i P . ΄ Ί7 Ί728 Ί40 ΅ 1 13 5 Ί7 Ί728 Ί40 Try computing the product P *AP for the matrices A and P in Example 7 to see that 1 0 0 P*AP ΄ 0 6 0΅ 0 0 2 where 1, 6, and 2 are the eigenvalues of A. You have seen that Hermitian matrices are unitarily diagonalizable. It turns out that there is a larger class of matrices, called normal matrices, that are also unitarily diagonalizable. A square complex matrix A is normal when it commutes with its conjugate transpose:AA* A*A. The main theorem of normal matrices states that a complex matrix A is normal if and only if it is unitarily diagonalizable. You are asked to explore normal matrices further in Exercise 56. The properties of complex matrices described in this section are comparable to the properties of real matrices discussed in Chapter 7. The summary below indicates the correspondence between unitary and Hermitian complex matrices when compared with orthogonal and symmetric real matrices.

Comparison of Symmetric and Hermitian Matrices A is a symmetric matrix A is a Hermitian matrix (real) (complex) 1. Eigenvalues of A are real. 1. Eigenvalues of A are real. 2. Eigenvectors corresponding 2. Eigenvectors corresponding to distinct eigenvalues are to distinct eigenvalues are orthogonal. orthogonal. 3. There exists an orthogonal 3. There exists a unitary matrix P matrix P such that such that PTAP P*AP is diagonal. is diagonal. 9781133110873_0805.qxp 3/10/12 6:55 AM Page 428

428 Chapter 8 Complex Vector Spaces

8.5 Exercises

Finding the Conjugate Transpose In Exercises 1Ð4, i i i determine the conjugate transpose of the matrix. Ί2 Ί3 Ί6 i i 1 2i 2 i i i i 1.΄ ΅ 2. ΄ ΅ 16. A 2 3i 1 1 ΄ Ί2 Ί3 Ί6΅ i i Ί 0 0 5 i 2i 2 i Ί3 Ί6 3.΄ 5 i 6 4΅ 4. ΄5 3i΅ Ί2i 4 3 0 6 i Row Vectors of a Unitary Matrix In Exercises 17Ð20, (a) verify that A is unitary by showing that its rows are Finding the Conjugate Transpose In Exercises 5 and orthonormal, and (b) determine the inverse of A. 6, use a software program or graphing utility to find the 4 3 1 i 1 i conjugate transpose of the matrix. i 5 5 2 2 17.A 18. A 1 i 0 1 i ΄ 3 4 ΅ ΄ 1 1 ΅ 2 i 1 0 2i i 5. 5 5 Ί2 Ί2 ΄1 i i 2 4i΅ i i 1 Ί3 i 1 Ί3i 2 1 0 19. A ΄ ΅ 2Ί2 Ί3 i 1 Ί3i 2 i 1 1 2i 0 2 i 2i 1 i 0 1 0 6. ΄ i 2 i i 1΅ 1 i 1 i 20. A 0 1 2i 4 0 2i ΄ Ί6 Ί3 ΅ 2 1 Non-Unitary Matrices In Exercises 7Ð10, explain why 0 Ί6 Ί3 the matrix is not unitary. i 0 1 i Identifying Hermitian Matrices In Exercises 21Ð26, 7.A ΄ ΅ 8. A ΄ ΅ 0 0 i 1 determine whether the matrix is Hermitian. 0 i i 0 1 i i 21.΄ ΅ 22. ΄ ΅ 0 i i 9. A ΄ Ί2 Ί2΅ 0 0 0 1 0 0 2 i 1 0 i 1 23.2 i i 0 24. 2 i i 0 1 1 1 i ΄ ΅ ΄ ΅ 2 2 2 1 0 1 0 1 0 i 1 i 10. A 1 2 i 3 i Ί3 Ί3 Ί3 25. ΄ ΅ ΄ ΅ 2 i 2 3 i 1 1 1 i 2 2 2 1 Ί2 i 5 26. Ί2 i 2 3 i Identifying Unitary Matrices In Exercises 11Ð16, ΄ ΅ 5 3 i 6 determine whether A is unitary by calculating AA *. 1 i 1 i 1 i 1 i Finding Eigenvalues of a Hermitian Matrix In 11.A ΄ ΅ 12. A ΄ ΅ 1 i 1 i 1 i 1 i Exercises 27Ð32, determine the eigenvalues of the matrix A. i i 0 i 3 i 27.A ΄ ΅ 28. A ΄ ΅ i 0 Ί2 Ί2 i 0 i 3 13.A ΄ ΅ 14. A 0 i ΄ i i ΅ 3 1 i 0 2 i 29.A ΄ ΅ 30. A ΄ ΅ Ί2 Ί2 1 i 2 2 i 4 4 3 1 4 1 i 5 5 15. A 31. A 0 i 3i ΄ 3 4 ΅ ΄ ΅ i i 0 0 2 i 5 5 9781133110873_0805.qxp 3/10/12 6:55 AM Page 429

8.5 Exercises 429

i i Unitary Matrices In Exercises 45 and 46, use the result of 2 Ί2 Ί2 Exercise 44 to determine , b, and ca such that A is unitary. i 6 3i 32. A 2 0 1 1 a 1 a Ί2 45.A ΄ ΅ 46. A Ί45 ΄ ΅ Ί b c Ί ΄ ΅ i 2 2 0 2 b c Ί2 Proof In Exercises 47Ð50, prove the formula, where A .Finding Eigenvectors of a Hermitian Matrix In and are n ؋ nB complex matrices Exercises 33Ð36, determine the eigenvectors of the 47.͑A*͒* A 48. ͑A B͒* A* B* matrix in the indicated exercise. 49.͑kA͒* kA* 50. ͑AB͒* B*A* 33. Exercise 27 34. Exercise 30 51. Proof Let A be a matrix such that A* A O. Prove 35. Exercise 31 36. Exercise 28 that iA is Hermitian. Diagonalization of a Hermitian Matrix In Exercises 52. Show that det͑A͒ det͑A͒, where A is a 2 2 matrix. 37Ð41, find a unitary matrix P that diagonalizes the Determinants In Exercises 53 and 54, assume that the matrix A. result of Exercise 52 is true for matrices of any size. 0 i 0 2 i 37.A ΄ ΅ 38. A ΄ ΅ 53. Show that det͑A*͒ det͑A͒. i 0 2 i 4 54. Prove that if A is unitary, then Խdet͑A͒Խ 1. i i 2 55. (a) Prove that every Hermitian matrix A can be written as Ί2 Ί2 the sum A B iC, where B is a real symmetric i 39. A 2 0 matrix and C is real and skew-symmetric. Ί2 ΄ ΅ (b) Use part (a) to write the matrix i 0 2 Ί2 2 1 i A ΄ ΅ 1 i 3 4 2 2i 40. A ΄ ΅ 2 2i 6 as the sum A B iC, where B is a real symmetric matrix and C is real and skew-symmetric. 1 0 0 (c) Prove that every n n complex matrix A can be 41. A 0 1 1 i ΄ ΅ written as A B iC, where and CB are 0 1 i 0 Hermitian. (d) Use part (c) to write the complex matrix 42. Consider the following matrix. i 2 2 3 i 4 i A ΄ ΅ 2 i 1 2i A ΄3 i 1 1 i΅ 4 i 1 i 3 as the sum A B iC, where and CB are Hermitian. (a) Is A unitary? Explain. 56. (a) Prove that every Hermitian matrix is normal. (b) Is A Hermitian? Explain. (b) Prove that every unitary matrix is normal. (c) Are the row vectors of A orthonormal? Explain. (c) Find a 2 2 matrix that is Hermitian, but not unitary. (d) The eigenvalues of A are distinct. Is it possible to determine the inner products of the pairs of (d) Find a 2 2 matrix that is unitary, but not eigenvectors by inspection? If so, state the Hermitian. value(s). If not, explain why not. (e) Find a 2 2 matrix that is normal, but neither (e) Is A unitarily diagonalizable?Explain. Hermitian nor unitary. (f) Find the eigenvalues and corresponding eigenvectors of your matrix in part (e). 43. Show that A I is unitary by computing AA*. n (g) Show that the complex matrix 44. Let z be a complex number with modulus 1. Show that the matrix A is unitary. ΄ i 1΅ 0 i 1 z z A ΄ ΅ Ί2 iz iz is not diagonalizable. Is this matrix normal? 9781133110873_08_ANS.qxd 3/10/12 6:58 AM Page A6

Answer Key

Section 8.5 Ϫi 2 1 1 1 1. ΄ ΅ 37. P ϭ ΄ ΅ i Ϫ3i Ί2 Ϫi i 0 5 ϩ i Ί2i Ί2 0 Ί2 1 3. 5 Ϫ i 6 4 39. P ϭ Ϫi Ί2 i ΄ ΅ 2 ΄ ΅ ϪΊ2i 4 3 i Ί2 Ϫi 1 Ϫ i 2 Ϫ i 1 ϩ i Ϫi Ί6 0 0 1 0 1 Ϫi 2 Ϫ i 41. P ϭ 0 Ϫ1 ϩ i 2 5. Ί6 ΄ ΅ ΄ 1 0 2 Ϫ1΅ 0 2 1 ϩ i Ϫ Ϫ i 2i 4i 0 1 0 0 . . . 7. A is not unitary because it is singular. 0 1 0 . . . 43. A ϭ 9. A is not unitary because it is not a square matrix. ΄ 0 0 1 . . .΅ . . . 4 4i . . . 11. AA* ϭ ΄ ΅  I . . . Ϫ4i 4 2 1 0 0 . . . So, A is not unitary. 0 1 0 . . . A* ϭ * ϭ ϭ 1 0 ΄ 0 0 1 . . .΅ 13. AA I ΄ ΅ . 2 . . . 0 1 . . . So, A is unitary. 1 0 0 . . . 1 0 15. AA* ϭ ΄ ΅ ϭ I 0 1 0 . . . 0 1 2 AA* ϭ ϭ I ΄ 0 0 1 . . .΅ n . So, A is unitary...... ϭ ͑Ϫ4 3 ͒ ϭ ͑3 4 ͒ ϭ Ϫ1 17. (a) r1 5, 5i , r2 5, 5i Therefore A* A and In is unitary. ʈr ʈ ϭ 1, ʈr ʈ ϭ 1, r и r ϭ 0 1 Ϫ1 Ϫ1 1 2 1 2 45. A ϭ ΄ ΅ 47Ð53. Proofs Ί2 Ϫi i Ϫ4 3 5 5 (b) AϪ1 ϭ ΄ 3 4 ΅ A ϩ A A Ϫ A Ϫ i Ϫ i 55. (a) A ϭ ϩ i 5 5 2 2i ϭ 1 ͑Ί Ϫ ϩ Ί ͒ 2 1 0 1 19. (a) r1 3 i, 1 3i , (b) ΄ ΅ ϩ i΄ ΅ 2Ί2 1 3 Ϫ1 0 ϭ 1 ͑Ί ϩ Ϫ Ί ͒ A ϩ A* A Ϫ A* r2 3 i, 1 3i (c) A ϭ ϩ i 2Ί2 2 2i ʈ ʈ ϭ ʈ ʈ ϭ и ϭ r1 1, r2 1, r1 r2 0 i 1 0 2 Ϫ 1 1 Ί3 ϩ i Ί3 Ϫ i 2 2 (b) AϪ1 ϭ ΄ ΅ (d) ϩ i 2Ί2 1 Ϫ Ί3i 1 ϩ Ί3i ΄ i ΅ ΄1 ΅ 2 ϩ 1 Ϫ2 21. A is Hermitian because A ϭ A*. 2 2

23. A is not Hermitian because the entry a22, on the main diagonal, is not a real number. So, A  A*. 25. A is not Hermitian because the matrix is not square. ␭ ϭ ␭ ϭ ␭ ϭ 27. 1 1 29. 1 1 31. 1 1 ␭ ϭϪ ␭ ϭ ␭ ϭ 2 1 2 4 2 i ␭ ϭ ϩ 3 2 i ϭ ͑ Ϫ ͒ ϭ ͑ ͒ 33. v1 1, i 35. v1 1, 0, 0 ϭ ͑ ͒ ϭ ͑ ϩ Ϫ ͒ v2 1, i v2 2 2i, 1, 0 ϭ ͑ ϩ ͒ v3 6 4i, 3i, 2

Chapter 7: Eigenvalues and Eigenvectors. 7.4 Applications of Eigenvalues and Eigenvectors.

7.4 Applications of Eigenvalues and Eigenvectors.  Objective: Model population growth using an age transion matrix and age distribution vector, and find a stable age distribution vector.  Objective: Use a matrix equation to solve a system of first-order linear differential equations.

To model population growth of a population with a lifespan of L years, we partition the lifespan into n equal-size classes

1st age class 2nd age class … ith age class … nth age class  L   L 2L    )1(Li iL    )1(Ln   ,0   ,   ,   , L  n   n n  …  n n  …  n 

 x1    x2 th The age distribution vector is x =   where xi is the number of individuals in the i age class.      xn 

th Let pi be the probability that after L/n years, a member of the i age class will survive to become th a member of the (i + 1) age class. (Note that 0  pi  1 for i = 0, 1, …, n – 1, and pn = 0.) Let bi th be the average number of offspring produced by a member of the i age class. (Note that 0  pi for i = 0, 1, …, n.)

 321  1 bbbbb nn   p  0000   1 

The age transition matrix or Leslie matrix is A =  p2  0000          000  pn1 0 

If xi is the age distribution vector at a specific time, then the age distribution vector L/n years later is xi+1 = Axi

 Example: 80% of a population of mice survives the first year. Of that 80%, 25% survives the second year. The maximum lifespan is three years. The number of offspring for each member of the population is 3 in the first year, 6 in the second, and 3 in the third year.

The population now consists of 120 members in each age class. How many members will there be in each age class in one year? In two years?

p. 227 Chapter 7: Eigenvalues and Eigenvectors. 7.4 Applications of Eigenvalues and Eigenvectors.

Solution: x1 = and A =

0  age  1

After one year, x2 = = for 1  age  2 2  age  3

0  age  1

After two years, x3 = = for 1  age  2 2  age  3

 Example: A certain type of lizard has a maximum lifespan of two years. Only 8% of lizards survive from their first year to their second year. The average number of offspring for each member of the population is 1.5 in the first year, and 2 in the second year. Find a stable age distribution vector for this population.

 Solution: A = .

Using the TI-89, we find eigenvalues

with correspronding eigenvectors

A negative eigenvalue (and an eigenvector with some positive and some negative entries) does not make sense. x and Ax = x must both have all non-negative entries, because the entries represent the number of individuals in each age class. We therefore use the eigenvector for  = The number of individuals in each class

should be a whole number, so try x =

Now check: Ax =

The ratio of the age classes is stable at (first year) : (second year) =

p. 228 Chapter 7: Eigenvalues and Eigenvectors. 7.4 Applications of Eigenvalues and Eigenvectors. A system of first-order linear differential equations has the form

y1  a11y1  a12 y2  a1n yn y1 (0)  C1 y  a y  a y  a y y (0)  C 2 21 1 22 2 2n n with initial conditions 2 2  

yn  an1 y1  an2 y2  ann yn yn (0)  Cn dy Where each y is a function of t and y  i . i i dt

 y1   y1   y   y  This is a linear system because the y′ =  2  is a linear transformation of y =  2  .            yn   yn  y′(t) = Ay(t), where A is a matrix of constants.

These are first-order differential equations because they contain first (and not higher) derivatives.

 Example: Solve the system

y1  1 y1  0y2  0y3 y1 (0)  C1

y2  0y1  2 y2  0y3 with initial conditions y2 (0)  C2

y3  0y1  0y2  3 y3 y3 (0)  C3 Solution:

From calculus, we know that the general solution to y′(t) = ky(t) is y(t) = Cekt. So the solution to the system (with a diagonal matrix) is

1t y1 (t)  C1e

2t y2 (t)  C2e

3t y3 (t)  C3e

 Example: Solve the system

y1  2y1  y2  y3 y1 (0)  4

y2  y1  y2 with initial conditions y2 (0)  1

y3  y1  y3 y3 (0)  2

p. 229 Chapter 7: Eigenvalues and Eigenvectors. 7.4 Applications of Eigenvalues and Eigenvectors. Solution:

2 1 1 4     y′(t) = Ay(t), where A = 1 1 0 and y(0) = 1 1 0 1 2 Try a substitution y(t) = Pw(t), where P is some invertible constant matrix. Then y′(t) = Pw′(t) and P–1y′(t) = w′(t) so

y′(t) = A y(t) P–1y′(t) = P–1A y(t) w′(t) = P–1A(Pw(t)) w′(t) = (P–1AP)w(t)

1t 1 0 0   B1e  –1 –1     t  If is P AP diagonal, i.e. P AP = D = 0  0 , then w(t) = B e 2 for unknown  2   2   0 0    3t   3  B3e  constants B1, B2, B3. Using the TI-89 to find the eigenvalues and eigenvectors of A, we find

0.816497 0.57735 0  3 0 0     P = 0.408248  0.57735  0.707107 and D = 0 0 0 0.408248  0.57735 0.707107  0 0 1

3t 3t B1e  B1e      So w(t) =  B2  , y(t) = Pw(t) = P  B2  , and  t   t   B3e   B3e 

4 B1  B1  4  4.49073  y(0) = 1 = P B  , so B  = P–1 1 =  0.57735     2   2      2 B3  B3  2 0.707107  4.49073et   0.3333  3.667e3t   1  11 e3t       3 3  Then y(t) = Pw(t) = P = 3t = 1 1 11 3t  0.57735   0.3333  0.5 1.833e   3  2  6 e  0.707107et   0.3333  0.5 1.833e3t   1  1  11 e3t       3 2 6  Other software may yield other P and D matrices, but w(t) will always be the same.

p. 230 Chapter 7: Eigenvalues and Eigenvectors. 7.4 Applications of Eigenvalues and Eigenvectors.  Example: Solve the system

y1   3y2  5y3 y1 (0)  2

y2  4y1  4y2 10y3 with initial conditions y2 (0)  3

y3  4y3 y3 (0)  0

Solution: y′(t) = Ay(t), where A = and y(0) =

Try a substitution y(t) = Pw(t) y′(t) = A y(t) Pw′(t) = A Pw(t) w′(t) = (P–1AP) w(t)

1 0 0  We want diagonal P–1AP =  0  0  , so w(t) = for unknowns B , B , B .  2  1 2 3  0 0 3  Using the TI-89 to find the eigenvalues and eigenvectors of A, we find

P = and D =

So w(t) = , y(t) =

B1  B1  y(0) = = P B  , so B  =  2   2  B3  B3 

Then y(t) = Pw(t) = P

Use Expand on the TI to obtain y(t) =

p. 231 Chapter 7: Eigenvalues and Eigenvectors. 7.4 Applications of Eigenvalues and Eigenvectors. An example from continuum mechanics of a symmetric matrix is the stress tensor

 11  12  13  T =     .  21 22 23  31  32  33

The figure uses the notation that T: R3  R3 (ei) is the linear transformation T  Tei

ij is the force per unit area on the plan perpendicular to ei, in the ej direction. ii is is a pressure or normal stress. ij is a shearing stress for i  j. T is symmetric because of conservation of angular momentum.

1 4 3    Example: Let T = 4 1 0 . 3 0 1

 0.707107  0.707107 0   5  5 0   5 2 5 2  Use the TI-89 to diagonalize T: P =  0.565685 0.565685  0.6 = 4 4  3    5 2 5 2 5   0.424264 0.424264 0.8   3 3 4     5 2 5 2 5  11 0 0   and D =  0 6 0  0 0 1  0.707107   0.707107  0        The mutually orthogonal vectors p1 =  0.565685  , p1 =  0.565685  , and p3 =  0.6  0.424264  0.424264   0.8  (pi) are called the principal directions, where the stress vector T is parallel to pi and where there are no shear stresses dij for i  j. The three stresses 1= 11, 2=6, and 3= 1, are called principal stresses.

p. 232 Chapter 7: Eigenvalues and Eigenvectors. 7.4 Applications of Eigenvalues and Eigenvectors. Example (multivariable calculus): The Hessian matrix H of a function of several variables is symmetric.

If f (x, y): R2  R is a (not necessarily linear) function of two variables, then   2 f  2 f   2  x xy H =   has orthogonal eigenvectors. Near a critical point, the level curves of   2 f  2 f   2  yx y  f (x, y) = constant are ellipses if the eigenvalues of H have the same sign, a straight lines if one eigenvalue is zero, or if the eigenvalues have opposite signs. See the text for more discussion.

f (x, y) = 13x2 –8 xy + 7y2 – 45

Eigenvalues = 5, 15

 2 1 Eigenvectors =   ,    1  2

f (x, y) = 2푥2 + 4푥푦 + 2푦2 + 4√2푥 + 4√2푦 + 45

Eigenvalues = 0, 4

1 1 Eigenvectors =   ,   1  1 

f (x, y) = 17x2 + 32xy – 7y2 – 75

Eigenvalues = 25, –15

2 1 Eigenvectors =   ,   1  2 

p. 233