Solutions Manual, Linear Algebra Theory and Applications

Exercises 1 Solutions Manual, Linear Algebra Theory And Applications

F.12 Exercises

1.6

1 1. Let z = 5 + i9. Find z− . 1 5 9 (5 + i9)− = i 106 − 106 2. Let z = 2 + i7 and let w = 3 i8. Find zw, z + w, z2, and w/z. − 62 + 5i, 5 i, 45 + 28i, and 50 37 i. − − − 53 − 53 3. Give the complete solution to x4 + 16 = 0. x4 + 16 = 0, Solution is: (1 i) √2, (1 + i) √2, (1 i) √2, (1 + i) √2. − − − − 4. Graph the complex cube roots of 8 in the complex plane. Do the same for the four fourth roots of 16. The cube roots are the solutions to z3 + 8 = 0, Solution is: i√3 + 1, 1 i√3, 2 − − The fourth roots are the solutions to z4 + 16 = 0, Solution is:

(1 i) √2, (1 + i) √2, (1 i) √2, (1 + i) − − − − √2. When you graph these, you will have three equally spaced points on the circle of radius 2 for the cube roots and you will have four equally spaced points on the circle of radius 2 for the fourth roots. Here are pictures which should result.

5. If z is a complex number, show there exists ω a complex number with ω = 1 and ωz = z . | | | | z If z = 0, let ω = 1. If z = 0, let ω = 6 z | | 6. De Moivre’s theorem says [r (cos t + i sin t)]n = rn (cos nt + i sin nt) for n a positive integer. Does this formula continue to hold for all integers, n, even negative integers? Explain. Yes, it holds for all integers. First of all, it clearly holds if n = 0. Suppose now that n is a negative integer. Then n > 0 and so − n 1 1 [r (cos t + i sin t)] = n = n [r (cos t + i sin t)]− r− (cos ( nt) + i sin ( nt)) − −

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rn rn (cos (nt) + i sin (nt)) = = (cos (nt) i sin (nt)) (cos (nt) i sin (nt)) (cos (nt) + i sin (nt)) − − = rn (cos (nt) + i sin (nt))

because (cos(nt) i sin (nt)) (cos (nt) + i sin (nt)) = 1. − 7. You already know formulas for cos(x + y) and sin (x + y) and these were used to prove De Moivre’s theorem. Now using De Moivre’s theorem, derive a formula for sin (5x) and one for cos(5x). sin (5x) = 5 cos4 x sin x 10 cos2 x sin3 x + sin5 x − cos (5x) = cos5 x 10 cos3 x sin2 x + 5 cos x sin4 x − 8. If z and w are two complex numbers and the polar form of z involves the angle θ while the polar form of w involves the angle φ, show that in the polar form for zw the angle involved is θ + φ. Also, show that in the polar form of a complex number, z, r = z . | | You have z = z (cos θ + i sin θ) and w = w (cos φ + i sin φ) . Then when you multiply these, you get| | | |

z w (cos θ + i sin θ) (cos φ + i sin φ) | | | | = z w (cos θ cos φ sin θ sin φ + i (cos θ sin φ + cos φ sin θ)) | | | | − = z w (cos (θ + φ) + i sin (θ + φ)) | | | | 9. Factor x3 + 8 as a product of linear factors. x3 + 8 = 0, Solution is: i√3 + 1, 1 i√3, 2 and so this polynomial equals − − (x + 2) x i√3 + 1 x 1 i√3 − − − 10. Write x3 + 27 in the form (x + 3) x2 + ax + b where x2 + ax + b cannot be factored any more using only real numbers. x3 + 27 = (x + 3) x2 3x + 9 − 11. Completely factor x4 + 16 as a product of linear factors. x4 + 16 = 0, Solution is: (1 i) √2, (1 + i) √2, (1 i) √2, (1 + i) √2. These are just the fourth roots of 16.− Then to− factor, this you− get− − x (1 i) √2 x (1 + i) √2 − − − − · x (1 i) √2 x (1 + i) √2 − − − − 12. Factor x4 + 16 as the product of two quadratic polynomials each of which cannot be factored further without using complex numbers. x4 + 16 = x2 2√2x + 4 x2 + 2√2x + 4 . You can use the information in the preceding problem.− Note that (x z)(x z) has real coeﬃcients. − − 13. If z, w are complex numbers prove zw = zw and then show by induction that z1 zm = m m ··· z1 zm. Also verify that k=1 zk = k=1 zk. In words this says the conjugate of a product··· equals the product of the conjugates and the conjugate of a sum equals the sum of the conjugates. P P (a + ib)(c + id) = ac bd + i (ad + bc) = (ac bd) i (ad + bc) − − −

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(a ib)(c id) = ac bd i (ad + bc) which is the same thing. Thus it holds for a product− − of two complex− − numbers. Now suppose you have that it is true for the product of n complex numbers. Then

z z = z z z 1 ··· n+1 1 ··· n n+1 and now, by induction this equals

z z z 1 ··· n n+1 As to sums, this is even easier.

n n n (xj + iyj ) = xj + i yj j=1 j=1 j=1 X X X n n n n = x i y = x iy = (x + iy ). j − j j − j j j j=1 j=1 j=1 j=1 X X X X n n 1 14. Suppose p (x) = anx + an 1x − + + a1x + a0 where all the ak are real numbers. Suppose also that p (z) = 0− for some···z C. Show it follows that p (z) = 0 also. ∈ You just use the above problem. If p (z) = 0, then you have

n n 1 p (z) = 0 = anz + an 1z − + + a1z + a0 − ··· n n 1 = anz + an 1z − + + a1z + a0 − ··· n n 1 = an z + an 1 z − + + a1 z + a0 − ··· n n 1 = anz + an 1z − + + a1z + a0 − ··· = p (z)

15. I claim that 1 = 1. Here is why. − 1 = i2 = √ 1√ 1 = ( 1)2 = √1 = 1. − − − − q This is clearly a remarkable result but is there something wrong with it? If so, what is wrong? Something is wrong. There is no single √ 1. − 16. De Moivre’s theorem is really a grand thing. I plan to use it now for rational exponents, not just integers.

1 = 1(1/4) = (cos 2π + i sin 2π)1/4 = cos (π/2) + i sin (π/2) = i.

Therefore, squaring both sides it follows 1 = 1 as in the previous problem. What does this tell you about De Moivre’s theorem?− Is there a profound diﬀerence between raising numbers to integer powers and raising numbers to non integer powers? It doesn’t work. This is because there are four fourth roots of 1. 17. Show that C cannot be considered an ordered ﬁeld. Hint: Consider i2 = 1. − It is clear that 1 > 0 because 1 = 12. (In general a2 > 0. This is clear if a > 0. If a < 0, then adding a to both sides, yields that 0 < a. Also recall that a = ( 1) a and − − − − that ( 1)2 = 1. Therefore, ( a)2 = ( 1)2 a2 = a2 > 0. ) Now it follows that if C can be ordered,− then 1 > 0, but− this is a− problem because it implies that 0 > 1 = 12 > 0. −

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18. Say a + ib < x + iy if a < x or if a = x, then b < y. This is called the lexicographic order. Show that any two different complex numbers can be compared with this order. What goes wrong in terms of the other requirements for an ordered field. From the definition of this funny order, 0 < i and so if this were an order, you would need to have 0 < i2 = 1. Now add 1 to both sides and obtain 0 > 1 = 12 > 0, a contradiction. −

1 19. With the order of Problem 18, consider for n N the complex number 1 n . Show that with the lexicographic order just described,∈ each of 1 im is an upper− bound to all these numbers. Therefore, this is a set which is “bounded− above” but has no least upper bound with respect to the lexicographic order on C. This follows from the deﬁnition. 1 im > 1 1/n for each m. Therefore, if you 1 − − consider the numbers 1 n you have a nonempty set which has an upper bound but no least upper bound. −

F.13 Exercises

1.11

1. Give the complete solution to the system of equations, 3x y + 4z = 6, y + 8z = 0, and 2x + y = 4. − − − x = 2 4t, y = 8t, z = t. − − 2. Give the complete solution to the system of equations, x+3y +3z = 3, 3x+2y +z = 9, and 4x + z = 9. − − x = y = 2, z = 1 − 3. Consider the system 5x + 2y z = 0 and 5x 2y z = 0. Both equations equal zero and so 5x + 2y− z = 5−x 2y z which− − is equivalent− to y = 0. Thus x and z can equal− anything.− But when− −x =− 1, z = 4, and y = 0 are plugged in to the equations, it doesn’t work. Why? − These are invalid row operations. 4. Give the complete solution to the system of equations, x+2y+6z = 5, 3x+2y+6z = 7 , 4x + 5y + 15z = 7. − − No solution. 5. Give the complete solution to the system of equations

x + 2y + 3z = 5, 3x + 2y + z = 7, 4x + 5y + z = 7, x + 3z = 5. − − x = 2, y = 0, z = 1. 6. Give the complete solution of the system of equations,

x + 2y + 3z = 5, 3x + 2y + 2z = 7 4x + 5y + 5z = 7, x = 5 − − No solution.

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7. Give the complete solution of the system of equations x + y + 3z = 2, 3x y + 5z = 6 − 4x + 9y + z = 8, x + 5y + 7z = 2 − − x = 2 2t, y = t, z = t. − − 8. Determine a such that there are infinitely many solutions and then find them. Next determine a such that there are no solutions. Finally determine which values of a correspond to a unique solution. The system of equations is 3za2 3a + x + y + 1 = 0 3x a −y + z a2 + 4 5 = 0 za− 2 −a 4x + 9y + 9− = 0 − − If a = 1, there are infinitely many solutions of the form x = 2 2t, y = t, z = t. If a = 1, t then there are no solutions. If a is anything else, there− is a unique− solution. − 2a 1 1 x = , y = , z = a + 1 −a + 1 a + 1 9. Find the solutions to the following system of equations for x, y, z, w. y + z = 2, z + w = 0, y 4z 5w = 2, 2y + z w = 4 − − − x = t, y = s + 2, z = s, w = s − 10. Find all solutions to the following equations. x + y + z = 2, z + w = 0, 2x + 2y + z w = 4, x + y 4z 5z = 2 − − − x = t + s + 2, y = t, z = s, w = s, where s, t are each in F. − − F.14 Exercises

1.14 1. Verify all the properties 1.11-1.18. 2. Compute 5 (1, 2 + 3i, 3, 2) + 6 (2 i, 1, 2, 7) . − − − 3. Draw a picture of the points in R2 which are determined by the following ordered pairs. (a) (1, 2) (b) ( 2, 2) − − (c) ( 2, 3) − (d) (2, 5) − 4. Does it make sense to write (1, 2) + (2, 3, 1)? Explain. 5. Draw a picture of the points in R3 which are determined by the following ordered triples. If you have trouble drawing this, describe it in words. (a) (1, 2, 0) (b) ( 2, 2, 1) − − (c) ( 2, 3, 2) − −

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1.17

1. Show that (a b) = 1 a + b 2 a b 2 . · 4 | | − | − | h i 2. Prove from the axioms of the inner product the parallelogram identity, a + b 2 + | | a b 2 = 2 a 2 + 2 b 2 . | − | | | | | n n 3. For a, b R , define a b k=1 βkakbk where βk > 0 for each k. Show this satisfies the axioms∈ of the inner· product.≡ What does the Cauchy Schwarz inequality say in this case. P The product satisfies all axioms for the inner product so the Cauchy Schwarz inequality holds.

4. In Problem 3 above, suppose you only know βk 0. Does the Cauchy Schwarz inequality still hold? If so, prove it. ≥ Yes, it does. You don’t need the part which says that the only way a a = 0 is for a = 0 in the argument for the Cauchy Schwarz inequality. · 5. Let f, g be continuous functions and deﬁne

1 f g f (t) g (t)dt · ≡ Z0 show this satisﬁes the axioms of a inner product if you think of continuous functions in the place of a vector in Fn. What does the Cauchy Schwarz inequality say in this case? The only part which is not obvious for the axioms is the one which says that if

1 f 2 = 0 | | Z0 then f = 0. However, this is obvious from continuity considerations. 6. Show that if f is a real valued continuous function,

2 b b f (t) dt (b a) f (t)2 dt. ≤ − Za ! Za

1/2 1/2 b b b f (t) dt 12dt f (t) 2 dt a ≤ a ! a | | ! Z Z Z 1/2 b = (b a)1/2 f (t) 2 dt − | | Za ! which yields the desired inequality when you square both sides.

Saylor URL: http://www.saylor.org/courses/ma212/ The Saylor Foundation Exercises 7 F.16 Exercises

2.2

1. In 2.1 - 2.8 describe A and 0. − 2. Let A be an n n matrix. Show A equals the sum of a symmetric and a skew symmetric matrix. × A+AT A AT A = 2 + −2 3. Show every skew symmetric matrix has all zeros down the main diagonal. The main diagonal consists of every entry of the matrix which is of the form aii. It runs from the upper left down to the lower right. You know that A = A . Let j = i to conclude that A = A and so A = 0. ij − ji ii − ii ii 4. Using only the properties 2.1 - 2.8 show A is unique. − Suppose that B also works. Then

A = A + (A + B) = ( A + A) + B = B − − − 5. Using only the properties 2.1 - 2.8 show 0 is unique.

If 0′ is another additive identity, then 0′ = 0 + 0′ = 0 6. Using only the properties 2.1 - 2.8 show 0A = 0. Here the 0 on the left is the scalar 0 and the 0 on the right is the zero for m n matrices. × 0A = (0 + 0) A = 0A + 0A. Now add the additive inverse of 0A to both sides. 7. Using only the properties 2.1 - 2.8 and previous problems show ( 1) A = A. − − 0 = 0A = (1 + ( 1)) A = A + ( 1) A. Hence, ( 1) A is the unique additive inverse of A. Thus A = (− 1) A. − − − − 8. Prove 2.17. (AB)T (AB) = A B = BT AT = BT AT . Hence the formula holds. ij ≡ ji k jk ki k ik kj ij 9. Prove that I A = APwhere A is anPm n matrix. m × (I A) I A = A and so I A = A. m ij ≡ k ik kj ij m n m 10. Let A andP be a real m n matrix and let x R and y R . Show (Ax, y)Rm = T × ∈ k ∈ x,A y where ( , ) k denotes the dot product in R . Rn · · R (Ax, y) = i (Ax)i yi = i k Aikxkyi x T y T ,A =P k xk i APkiPyi = k i xkAikyi, the same as above. Hence the two are equal. P P P P 11. Use the result of Problem 10 to verify directly that (AB)T = BT AT without making any reference to subscripts. (AB)T x, y (x, (AB) y) = AT x,By = BT AT x, y . Since this holds for every ≡ x y y , , you have for all (AB)T x BT AT x, y − Let y = (AB)T x BT AT x. Then since x is arbitrary, the result follows. −

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12. Let x = ( 1, 1, 1) and y = (0, 1, 2) . Find xT y and xyT if possible. − − 1 0 1 2 xT y = −1 0 1 2 = 0 −1 −2  −1   0− 1− 2    0   xyT = 1 1 1 1 = 1 − −  2    13. Give an example of matrices, A, B, C such that B = C, A = 0, and yet AB = AC. 6 6 1 1 1 1 0 0 = 1 1 1− 1 0 0 − 1 1 1 1 0 0 = 1 1 −1 1 0 0 − 1 1 1 1 3 1 1 2 − 14. Let A = 2 1 , B = − − , and C = 1 2 0 . Find  − −  2 1 2  −  1 2 − 3 1 0 − − if possible.   

(a) AB (b) BA (c) AC (d) CA (e) CB (f) BC

15. Consider the following digraph.

1 2

4 3

Write the matrix associated with this digraph and ﬁnd the number of ways to go from 3 to 4 in three steps. 0 1 1 0 1 0 0 1 The matrix for the digraph is  1 1 0 2   0 1 0 1     

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3 0 1 1 0 1 5 2 4 1 0 0 1 3 2 0 5 =  1 1 0 2   4 5 1 8   0 1 0 1   1 3 1 3      Thus it appears that there are 8 ways to do this.

1 16. Show that if A− exists for an n n matrix, then it is unique. That is, if BA = I and 1 × AB = I, then B = A− . 1 1 From the given equations, multiply on the right by A− . Then B = A− .

1 1 1 17. Show (AB)− = B− A− . 1 1 1 ABB− A− = AIA− = I 1 1 1 B− A− AB = B− IB = I 1 Then by the deﬁnition of the inverse and its uniqueness, it follows that (AB)− exists and 1 1 1 (AB)− = B− A−

T T 1 1 T 18. Show that if A is an invertible n n matrix, then so is A and A − = A− . × T 1 T 1 T A A− = A− A = I

1 T T 1 T A− A = AA− = I Then from the deﬁnition of the inverse and its uniqueness, T 1 1 T it follows that A − exists and equals A− . 19. Show that if A is an n n invertible matrix and x is a n 1 matrix such that Ax = b × 1 × for b an n 1 matrix, then x = A− b. × 1 Multiply both sides on the left by A− . 20. Give an example of a matrix A such that A2 = I and yet A = I and A = I. 2 6 6 − 0 1 1 0 = 1 0 0 1 21. Give an example of matrices, A, B such that neither A nor B equals zero and yet AB = 0. 1 1 1 1 0 0 = 1 1 1− 1 0 0 −

x1 x2 + 2x3 x1 2−x + x x 22. Write  3 1  in the form A  2  where A is an appropriate matrix. 3x3 x3  3x + 3x + x   x   4 2 1   4      1 1 2 0 x1 x1 x2 + 2x3 1− 0 2 0 x x− + 2x    2  =  1 3  0 0 3 0 x3 3x3  1 3 0 3   x   x + 3x + 3x     4   1 2 4        23. Give another example other than the one given in this section of two square matrices, A and B such that AB = BA. 6 Almost anything works. 1 2 1 2 5 2 = 3 4 2 0 11 6

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1 2 1 2 7 10 = 2 0 3 4 2 4 24. Suppose A and B are square matrices of the same size. Which of the following are correct?

(a) (A B)2 = A2 2AB + B2 Note this. − − (b) (AB)2 = A2B2 Not this. (c) (A + B)2 = A2 + 2AB + B2 Not this. (d) (A + B)2 = A2 + AB + BA + B2 This is all right. (e) A2B2 = A (AB) B This is all right. (f) (A + B)3 = A3 + 3A2B + 3AB2 + B3 Not this. (g) (A + B)(A B) = A2 B2 Not this. − − (h) None of the above. They are all wrong. (i) All of the above. They are all right.

1 1 25. Let A = − − . Find all 2 2 matrices, B such that AB = 0. 3 3 × 1 1 x y x z w y = −3− 3 z w 3−x +− 3z 3−w +− 3y z w , z, w arbitrary. −z− w 1 26. Prove that if A− exists and Ax = 0 then x = 0. 1 Multiply on the left by A− . 27. Let 1 2 3 A = 2 1 4 .  1 0 2 

1 1   Find A− if possible. If A− does not exist, determine why. 1 1 2 3 − 2 4 5 2 1 4 = −0 1 −2  1 0 2   1 2− 3  −     28. Let 1 0 3 A = 2 3 4 .  1 0 2 

1 1   Find A− if possible. If A− does not exist, determine why. 1 1 0 3 − 2 0 3 − 1 2 2 3 4 = 0 3 3  1 0 2   1 0 −1  −     29. Let 1 2 3 A = 2 1 4 .  4 5 10   

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1 1 Find A− if possible. If A− does not exist, determine why. 5 1 2 3 1 0 3 2 2 1 4 , row echelon form: 0 1 3 A has no inverse.  4 5 10   0 0 0      30. Let 1 2 0 2 1 1 2 0 A =  2 1 3 2  −  1 2 1 2    1 1   Find A− if possible. If A− does not exist, determine why. 1 − 1 1 1 1 2 0 2 1 2 2 2 1 1 2 0 −3 1 1 5 = 2 2 2  2 1 3 2   1 0− 0− 1  − −  1 2 1 2   2 3 1 9     − − 4 4 4      F.17 Exercises

2.7

1. Show the map T : Rn Rm deﬁned by T (x) = Ax where A is an m n matrix and x is an m 1 column→ vector is a linear transformation. × × This follows from matrix multiplication rules. 2. Find the matrix for the linear transformation which rotates every vector in R2 through an angle of π/3. cos (π/3) sin (π/3) 1 1 √3 − = 2 − 2 sin (π/3) cos (π/3) 1 √3 1 2 2 3. Find the matrix for the linear transformation which rotates every vector in R2 through an angle of π/4. cos (π/4) sin (π/4) 1 √2 1 √2 − = 2 − 2 sin (π/4) cos (π/4) 1 √2 1 √2 2 2 4. Find the matrix for the linear transformation which rotates every vector in R2 through an angle of π/3. − cos ( π/3) sin ( π/3) 1 1 √3 − − − = 2 2 sin ( π/3) cos ( π/3) 1 √3 1 − − − 2 2 5. Find the matrix for the linear transformation which rotates every vector in R2 through an angle of 2π/3. 2 cos (π/3) 2 sin (π/3) 1 √3 − = − 2 sin (π/3) 2 cos (π/3) √3 1 6. Find the matrix for the linear transformation which rotates every vector in R2 through an angle of π/12. Hint: Note that π/12 = π/3 π/4. −

Saylor URL: http://www.saylor.org/courses/ma212/ The Saylor Foundation 12 Exercises

cos (π/3) sin (π/3) cos ( π/4) sin ( π/4) sin (π/3)− cos (π/3) sin (−π/4)− cos ( −π/4) − − 1 √2√3 + 1 √2 1 √2 1 √2√3 = 4 4 4 − 4 1 √2√3 1 √2 1 √2√3 + 1 √2 4 − 4 4 4

7. Find the matrix for the linear transformation which rotates every vector in R2 through an angle of 2π/3 and then reflects across the x axis. 1 0 cos (2π/3) sin (2π/3) 1 1 √3 − = − 2 − 2 0 1 sin (2π/3) cos (2π/3) 1 √3 1 − − 2 2 8. Find the matrix for the linear transformation which rotates every vector in R2 through an angle of π/3 and then reflects across the x axis. 1 0 cos (π/3) sin (π/3) 1 1 √3 − = 2 − 2 0 1 sin (π/3) cos (π/3) 1 √3 1 − − 2 − 2 9. Find the matrix for the linear transformation which rotates every vector in R2 through an angle of π/4 and then reflects across the x axis. 1 0 cos (π/4) sin (π/4) 1 √2 1 √2 − = 2 − 2 0 1 sin (π/4) cos (π/4) 1 √2 1 √2 − − 2 − 2 10. Find the matrix for the linear transformation which rotates every vector in R2 through an angle of π/6 and then reflects across the x axis followed by a reflection across the y axis. 1 0 1 0 cos (π/6) sin (π/6) 1 √3 1 − − = − 2 2 0 1 0 1 sin (π/6) cos (π/6) 1 1 √3 − − 2 − 2 11. Find the matrix for the linear transformation which reflects every vector in R2 across the x axis and then rotates every vector through an angle of π/4. cos (π/4) sin (π/4) 1 0 1 √2 1 √2 − = 2 2 sin (π/4) cos (π/4) 0 1 1 √2 1 √2 − 2 − 2 12. Find the matrix for the linear transformation which rotates every vector in R2 through an angle of π/4 and next reflects every vector across the x axis. Compare with the above problem. 1 0 cos (π/4) sin (π/4) 1 √2 1 √2 − = 2 − 2 0 1 sin (π/4) cos (π/4) 1 √2 1 √2 − − 2 − 2 13. Find the matrix for the linear transformation which reflects every vector in R2 across the x axis and then rotates every vector through an angle of π/6. cos (π/6) sin (π/6) 1 0 1 √3 1 − = 2 2 sin (π/6) cos (π/6) 0 1 1 1 √3 − 2 − 2 14. Find the matrix for the linear transformation which reflects every vector in R2 across the y axis and then rotates every vector through an angle of π/6. cos (π/6) sin (π/6) 1 0 1 √3 1 − − = − 2 − 2 sin (π/6) cos (π/6) 0 1 1 1 √3 − 2 2

Saylor URL: http://www.saylor.org/courses/ma212/ The Saylor Foundation Exercises 13

15. Find the matrix for the linear transformation which rotates every vector in R2 through an angle of 5π/12. Hint: Note that 5π/12 = 2π/3 π/4. −

cos (2π/3) sin (2π/3) cos ( π/4) sin ( π/4) sin (2π/3)− cos (2π/3) sin (−π/4)− cos ( −π/4) − − 1 √2√3 1 √2 1 √2√3 1 √2 = 4 − 4 − 4 − 4 1 √2√3 + 1 √2 1 √2√3 1 √2 4 4 4 − 4

T 16. Find the matrix for proju (v) where u = (1, 2, 3) . − e (1, 2,3) T T e = i· − (1, 2, 3) i 14 − 1 2 3 1 2 4 6 14  −3− 6− 9    T 17. Find the matrix for proju (v) where u = (1, 5, 3) .

1 5 3 1 5 25 15 35  3 15 9    T 18. Find the matrix for proju (v) where u = (1, 0, 3) .

1 0 3 1 0 0 0 10  3 0 9    19. Give an example of a 2 2 matrix A which has all its entries nonzero and satisﬁes A2 = A. Such a matrix is× called idempotent. You know it can’t be invertible. So try this.

2 a a a2 + ba a2 + ba = b b b2 + ab b2 + ab Let a2 + ab = a, b2 + ab = b. A solution which yields a nonzero matrix is

2 2 1 1 − − 20. Let A be an m n matrix and let B be an n m matrix where n < m. Show that AB cannot have× an inverse. × This follows right away from Theorem 2.3.8. This theorem says there exists a vector x = 0 such that Bx = 0. Therefore, ABx = 0 also and so AB cannot be invertible. 6 21. Find ker (A) for 1 2 3 2 1 0 2 1 1 2 A = .  1 4 4 3 3   0 2 1 1 2     

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Recall ker (A) is just the set of solutions to Ax = 0. 1 2 3 2 1 0 1 0 2 1 1 0 0 2 1 1 2 0 0 1 1 1 −1 0 . After row operations, 2 2  1 4 4 3 3 0   0 0 0 0 0 0   0 2 1 1 2 0   0 0 0 0 0 0      A solution is x = 1 t 1 t t , x = 2t t + t where the t are arbitrary. 2 − 2 1 − 2 2 − 3 1 − 1 − 2 3 i 22. If A is a linear transformation, and Axp= b. Show that the general solution to the equation Ax = b is of the form xp + y where y ker (A). By this I mean to show that whenever Az = b there exists y ker (A) such∈ that x + y = z. ∈ p If Az = b, Then A (z xp) = Az Axp = b b = 0 so there exists y such that y ker (A) and x + y−= z. − − ∈ p 23. Using Problem 21, ﬁnd the general solution to the following linear system.

x 1 2 3 2 1 1 11 x 0 2 1 1 2 2 7  x  =  1 4 4 3 3  3  18   x4   0 2 1 1 2     7     x       5      2t1 t2 + t3 4 −1 t −1 t t 7/2  − 2 1 − 2 2 − 3    t1 + 0 , t F i ∈  t2   0       t   0   3    That second vector is a particular  solution. 24. Using Problem 21, ﬁnd the general solution to the following linear system.

x 1 2 3 2 1 1 6 x 0 2 1 1 2  2  7 x =  1 4 4 3 3  3  13   x4   0 2 1 1 2     7     x5            2t1 t2 + t3 1 −1 −1 − 2 t1 2 t2 t3 7/2  − −t −  +  0  , t F 1 i ∈  t2   0       t   0   3        25. Show that the function Tu deﬁned by Tu (v) v proju (v) is also a linear transformation. ≡ − This is the sum of two linear transformations so it is obviously linear.

T 26. If u = (1, 2, 3) and Tu is given in the above problem, ﬁnd the matrix Au which satisﬁes Aux = Tu (x). 13 1 3 1 0 0 1 2 3 14 7 14 1 1 −5 − 3 0 1 0 14 2 4 6 = 7 7 7  0 0 1  −  3 6 9   − 3 3 −5  − 14 − 7 14      

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27. Suppose V is a subspace of Fn and T : V Fp is a nonzero linear transformation. Show↑ that there exists a basis for Im (T ) T→(V ) ≡ T v , ,T v { 1 ··· m} and that in this situation, v , , v { 1 ··· m} is linearly independent. p Im (T ) is a subspace of F . Therefore, it has a basis T v1, ,T vm for some m p. Say { ··· } ≤ m civi = 0 i=1 X Then do T to both sides m ciT vi = T 0 = 0 i=1 X Hence each ci = 0. (T 0 =T (0 + 0) = T 0 + T 0 and so T 0 = 0) 28. In the situation of Problem 27 where V is a subspace of Fn, show that there exists ↑ z1, , zr a basis for ker (T ) . (Recall Theorem 2.4.12. Since ker (T ) is a subspace, it{ has··· a basis.)} Now for an arbitrary T v T (V ) , explain why ∈ T v = a T v + + a T v 1 1 ··· m m and why this implies v (a v + + a v ) ker (T ) . − 1 1 ··· m m ∈ Then explain why V = span (v , , v , z , , z ) . 1 ··· m 1 ··· r ker (T ) is also a subspace so it has a basis z1, , zr for some r n. Now let the basis for T (V ) be as above. Then for v an arbitrary{ ··· vector,} there exist≤ unique scalars ai such that T v = a T v + + a T v 1 1 ··· m m Then it follows that T (v (a v + + a v )) = T v (a T v + + a T v ) = 0 − 1 1 ··· m m − 1 1 ··· m m Hence the vector v (a v + + a v ) is in ker (T ) and so there exist unique scalars − 1 1 ··· m m bi such that r v (a v + + a v ) = b z − 1 1 ··· m m i i i=1 X and therefore, V = span (v , , v , z , , z ) . 1 ··· m 1 ··· r 29. In the situation of the above problem, show v1, , vm, z1, , zr is a basis for V ↑and therefore, dim (V ) = dim (ker (T )) + dim{ (T (V···)) ··· }

The claim that v1, , vm, z1, , zr is a basis will be complete if it is shown that these vectors are{ linearly··· independent.··· } Suppose then that

aizi + bj vj = 0. i j X X

Then do T to both sides. Then you have j bjT vj = 0 and so each bj = 0. Then you use the linear independence to conclude that each ai = 0. P

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30. Let A be a linear transformation from V to W and let B be a linear transformation from↑ W to U where V, W, U are all subspaces of some Fp. Explain why

A (ker (BA)) ker (B) , ker (A) ker (BA) . ⊆ ⊆

ker(BA) ker(B)

ker(A) A - A (ker(BA))

If x ker (BA) , then BAx = 0 and so Ax ker (B) . That is, BAx = 0. It follows that ∈ ∈ A (ker (BA)) ker (B) ⊆ The second inclusion is obvious because if x is sent to 0 by A, then B will send Ax to 0.

31. Let x1, , xn be a basis of ker (A) and let Ay1, ,Aym be a basis of A (ker (BA)). Let↑ z{ ker··· (BA)}. Explain why { ··· } ∈ Az span Ay , ,Ay ∈ { 1 ··· m}

and why there exist scalars ai such that

A (z (a y + + a y )) = 0 − 1 1 ··· m m and why it follows z (a y + + a y ) span x , , x . Now explain why − 1 1 ··· m m ∈ { 1 ··· n} ker (BA) span x , , x , y , , y ⊆ { 1 ··· n 1 ··· m} and so dim (ker (BA)) dim (ker (B)) + dim (ker (A)) . ≤ This important inequality is due to Sylvester. Show that equality holds if and only if A(ker BA) = ker(B).

Let Ax1, ,Axr be a basis for A (ker (BA)) . Then let y ker (BA) . Thus It follows{ that··· } ∈ Ay A (ker (BA)) . ∈ Then there are scalars ai such that r Ay = aiAxi i=1 X r Hence y i=1 aixi ker (A) . Let z1, , zm be a basis for ker (A) . This shows that − ∈ { ··· } P dim ker (BA) m + r = dim ker (A) + dim A (ker (BA)) ≤ dim ker (A) + dim ker (B) ≤

It is interesting to note that the set of vectors z1, , zm, x1, , xr is independent. To see this, say { ··· ··· }

aj zj + bixi = 0 j i X X

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Then do A to both sides and obtain each bi = 0. Then it follows that each aj is also zero because of the independence of the zj . Thus

dim ker (BA) = dim ker (A) + dim (A ker (BA))

32. Generalize the result of the previous problem to any ﬁnite product of linear mappings. dim (ker s A ) s dim ker (A ). This follows by induction. i=1 i ≤ i=1 i 33. If W VQfor W, V twoP subspaces of Fn and if dim (W ) = dim (V ) , show W = V . ⊆ Let a basis for W be w1, , wr Then if there exists v V W, you could add in v to the basis and obtain{ a linearly··· } independent set of vectors∈ of\ V which implies that the dimension of V is at least r + 1 contrary to assumption. 34. Let V be a subspace of Fnand let V , ,V be subspaces, each contained in V . Then 1 ··· m V = V V (6.27) 1 ⊕ · · · ⊕ m if every v V can be written in a unique way in the form ∈ v = v + + v 1 ··· m

where each vi Vi. This is called a direct sum. If this uniqueness condition does not hold, then one∈ writes V = V + + V 1 ··· m and this symbol means all vectors of the form

v + + v , v V for each j. 1 ··· m j ∈ j Show this is equivalent to saying that if

0 = v + + v , v V for each j, 1 ··· m j ∈ j i i then each vj = 0. Next show that in the this situation, if βi = u1, , umi is a basis for V , then β , , β is a basis for V . ··· i { 1 ··· m} span (β1, , βm) is given to equal V. It only remains to verify that β1, , βm is linearly independent.··· Suppose v V with { ··· } i ∈ i

vi = 0 i X

It is also true that i 0=0 and so each vi = 0 by assumption. Now suppose

P i cijuj = 0 i j X X Then from what was just observed, for each i,

i cijuj = 0 j X i and now, since these uj form a basis for Vi, it follows that each cij = 0 for each j for each i. Thus β , , β is a basis. { 1 ··· m}

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35. Suppose you have ﬁnitely many linear mappings L1,L2, ,Lm which map V to V ↑ n ··· where V is a subspace of F and suppose they commute. That is, LiLj = Lj Li for all i, j. Also suppose Lk is one to one on ker (Lj) whenever j = k. Letting P denote the product of these linear transformations, P = L L L , ﬁrst6 show 1 2 ··· m ker (L ) + + ker (L ) ker (P ) 1 ··· m ⊆ Next show L : ker (L ) ker (L ) . Then show j i → i ker (L ) + + ker (L ) = ker (L ) ker (L ) . 1 ··· m 1 ⊕ · · · ⊕ m Using Sylvester’s theorem, and the result of Problem 33, show

ker (P ) = ker (L ) ker (L ) 1 ⊕ · · · ⊕ m Hint: By Sylvester’s theorem and the above problem,

dim (ker (P )) dim (ker (L )) ≤ i i X = dim (ker (L ) ker (L )) dim (ker (P )) 1 ⊕ · · · ⊕ m ≤ Now consider Problem 33.

First note that, since these operators commute, it follows that Lk : ker Li ker Li. Let v ker (L ) and consider v . It is obvious, since the linear transformations→ i ∈ i i i commute that P ( vi) = 0. Thus i P P ker (L ) + + ker (L ) ker (P ) 1 ··· m ⊆ However, by Sylvester’s theorem

dim ker (P ) dim ker (L ) ≤ i i X

If vi ker (Li) and i vi = 0, then apply i=k Li to both sides. This yields ∈ 6 P Q Livk = 0 i=k Y6

Since each of these Li is one to one on ker (Lk) , it follows that vk = 0. Thus

ker (L ) + + ker (L ) = ker (L ) ker (L ) 1 ··· m 1 ⊕ · · · ⊕ m

Now it follows that a basis for ker (L1) + + ker (Lm) has i dim ker (Li) vectors in it. ··· P dim ker (P ) dim ker (L ) ≤ i i X = dim (ker (L ) + + ker (L )) 1 ··· m dim ker (P ) ≤ Thus the inequalities are all equal signs and so

ker (L ) ker (L ) = ker (P ) 1 ⊕ · · · ⊕ m

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36. Let (Fn, Fn) denote the set of all n n matrices having entries in F. With the usual operationsM of matrix addition and scalar× multiplications, explain why (Fn, Fn) can 2 M be considered as Fn . Give a basis for (Fn, Fn) . If A (Fn, Fn) , explain why there exists a monic polynomial of the formM ∈ M

λk + a λk + + a λ + a k ··· 1 0 such that Ak + a Ak + + a A + a I = 0 k ··· 1 0 The minimal polynomial of A is the polynomial like the above, for which p (A) = 0 which has smallest degree. I will discuss the uniqueness of this polynomial later. Hint: 2 Consider the matrices I, A, A2, ,An . There are n2 +1 of these matrices. Can they be linearly independent? Now consider··· all polynomials and pick one of smallest degree and then divide by the leading coefficient. n n th A basis for (F , F ) is obviously the matricies Eij where Eij has a 1 in the ij place and zerosM everywhere else. Thus the dimension of this vector space is n2. It 2 follows that the list of matrices I, A, A2, ,An is linearly dependent. Hence there exists a polynomial p (λ) which has smallest··· possible degree such that p (A) = 0 by the well ordering principle of the natural numbers. Then divide by the leading coefficient. If you insist that its leading coefficient be 1, (monic) then the polynomial is unique and it is called the minimal polynomial. It is unique thanks to the division algorithm, because if q (λ) is another one, then

q (λ) = p (λ) l (λ) + r (λ)

where the degree of r (λ) is less than the degree of p (λ) or else equals 0. If it is not zero, then r (A) = 0 and this would be a contradiction. Hence q (λ) = p (λ) l (λ) where l (λ) must be monic. Since q (λ) has smallest possible degree, this monic polynomial can only be 1. Thus q (λ) = p (λ). 37. Suppose the ﬁeld of scalars is C and A is an n n matrix. From the preceding problem,↑ and the fundamental theorem of algebra,× this minimal polynomial factors

(λ λ )r1 (λ λ )r2 (λ λ )rk − 1 − 2 ··· − k

where rj is the algebraic multiplicity of λj . Thus

(A λ I)r1 (A λ I)r2 (A λ I)rk = 0 − 1 − 2 ··· − k r1 r2 rk rj and so, letting P = (A λ1I) (A λ2I) (A λkI) and Lj = (A λj I) apply the result of Problem− 35 to verify− that ··· − −

Cn = ker (L ) ker (L ) 1 ⊕ · · · ⊕ k and that A : ker (L ) ker (L ). In this context, ker (L ) is called the generalized j → j j eigenspace for λj. You need to verify the conditions of the result of this problem hold. Let L = (A λ I)ri . Then obviously these commute since they are just polynomials i − i in A. Is Lk one to one on ker(Li)?

(A λ I)rk = (A λ I + (λ λ ) I)rk − k − i i − k

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rk rk j r j = (A λ I) (λ λ ) k− j − i i − k j=0 X rk r rk j r j = (λ λ ) k I + (A λ I) (λ λ ) k− i − k j − i i − k j=1 X

Now raise both sides to the ri power.

(A λ I)rkri = (λ λ )rkri I + g (A)(A λ I)ri − k i − k − i rk where g (A) is some polynomial in A. Let vi ker (Li) . Then suppose (A λkI) vi = 0. Then ∈ − 0 = (A λ I)rkri v = (λ λ )rkri v − k i i − k i and since λi = λk, this requires that vi = 0. Thus Lk is one to one on ker (Li) as 6 n hoped. Therefore, since C = ker i Li, it follows from the above problems that Cn = kerQ (L ) ker (L ) 1 ⊕ · · · ⊕ k Note that there was nothing sacred about C all you needed for the above to hold is that the minimal polynomial factors completely into a product of linear factors. In other words, all the above works ﬁne for Fn provided the minimal polynomial “splits”. 38. In the context of Problem 37, show there exists a nonzero vector x such that

(A λ I) x = 0. − j

This is called an eigenvector and the λj is called an eigenvalue. Hint: There must exist a vector y such that

r1 r2 r 1 r (A λ I) (A λ I) (A λ I) j − (A λ I) k y = z = 0 − 1 − 2 ··· − j ··· − k 6 Why? Now what happens if you do (A λ I) to z? − j The hint gives it away.

r1 r2 r 1 r (A λ I) z = (A λ I) (A λ I) (A λ I) (A λ I) j − (A λ I) k y − j − j − 1 − 2 ··· − j ··· − k = (A λ I)r1 (A λ I)r2 (A λ I)rj (A λ I)rk y = 0 − 1 − 2 ··· − j ··· − k 39. Suppose Q (t) is an orthogonal matrix. This means Q (t) is a real n n matrix which satisﬁes × Q (t) Q (t)T = I

T T T Suppose also the entries of Q (t) are diﬀerentiable. Show Q ′ = Q Q′Q . − This is just the product rule. T T Q′ (t) Q (t) + Q (t) Q′ (t) = 0

Hence T T T Q′ (t) = Q (t) Q′ (t) Q (t) −

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40. Remember the Coriolis force was 2Ω vB where Ω was a particular vector which came from the matrix Q (t) as described× above. Show that

i (t) i (t ) j (t) i (t ) k (t) i (t ) · 0 · 0 · 0 Q (t) = i (t) j (t0) j (t) j (t0) k (t) j (t0) .  i (t) ·k (t ) j (t) ·k (t ) k (t) ·k (t )  · 0 · 0 · 0   There will be no Coriolis force exactly when Ω = 0 which corresponds to Q′ (t) = 0. When will Q′ (t) = 0?

Recall, that letting i = e1, j = e2, k = e3 in the usual way,

Q (t) u = u1e1 (t) + u2e2 (t) + u3e3 (t) where u u e (t ) + u e (t ) + u e (t ) ≡ 1 1 0 2 2 0 3 3 0 Note that u = u e (t ) . Thus j · j 0 Q (t) u = u e (t ) e (t) · j 0 j j X So what is the rsth entry of Q (t)? It equals

e (t )T Q (t) e (t ) = e (t ) e (t ) e (t ) e (t) r 0 s 0 r 0 · s 0 · j 0 j j X = e (t ) e (t) r 0 · s which shows the desired result. 41. An illustration used in many beginning physics books is that of firing a rifle hori- zontally and dropping an identical bullet from the same height above the perfectly flat ground followed by an assertion that the two bullets will hit the ground at exactly the same time. Is this true on the rotating earth assuming the experiment takes place over a large perfectly flat field so the curvature of the earth is not an issue? Explain. What other irregularities will occur? Recall the Coriolis acceleration is 2ω [( y′ cos φ) i+ (x′ cos φ + z′ sin φ) j (y′ sin φ) k] where k points away from the center of− the earth, j points East, and i −points South. Obviously not. Because of the Coriolis force experienced by the fired bullet which is not experienced by the dropped bullet, it will not be as simple as in the physics books. For example, if the bullet is fired East, then y′ sin φ > 0 and will contribute to a force acting on the bullet which has been fired which will cause it to hit the ground faster than the one dropped. Of course at the North pole or the South pole, things should be closer to what is expected in the physics books because there sin φ = 0. Also, if you fire it North or South, there seems to be no extra force because y′ = 0.

F.18 Exercises

3.2

1. Find the determinants of the following matrices.

1 2 3 (a) 3 2 2 (The answer is 31.)  0 9 8   

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4 3 2 (b) 1 7 8 (The answer is 375.)  3 9 3  −  1 2 3 2 1 3 2 3 (c)  , (The answer is 2.) 4 1 5 0 −  1 2 1 2      1 1 2. If A− exist, what is the relationship between det (A) and det A− . Explain your answer. 1 1 1 = det AA− = det (A) det A− . 3. Let A be an n n matrix where n is odd. Suppose also that A is skew symmetric. This means AT ×= A. Show that det(A) = 0. − det (A) = det AT = det ( A) = det ( I) det (A) = ( 1)n det (A) = det (A) . − − − − 4. Is it true that det (A + B) = det (A) + det (B)? If this is so, explain why it is so and if it is not so, give a counter example. Almost anything shows that this is not true.

1 0 1 0 det + = 0 0 1 −0 1 − 1 0 1 0 det + det = 2 0 1 −0 1 − 5. Let A be an r r matrix and suppose there are r 1 rows (columns) such that all rows (columns) are× linear combinations of these r 1− rows (columns). Show det (A) = 0. − Without loss of generality, assume the last row is a linear combination of the ﬁrst r 1 rows. Then the matrix is of the form −

T r1 .  .  T rn 1  n 1−   − a rT   i=1 i i    Then from the linear property of determinants,P the determinant equals

T T r1 r1 n 1 n 1 − . − . ai det  .  = ai det  .  = 0 rT rT i=1  n 1  i=1  n 1  X  rT−  X  0−T   i        Where the ﬁrst equal sign in the above is obtained by taking 1 times a the ith row from the top and adding to the last row. − 6. Show det (aA) = an det (A) where here A is an n n matrix and a is a scalar. × Each time you take out an a from a row, you multiply by a the determinant of the matrix which remains. Since there are n rows, you do this n times, hence you get an.

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1 7. Suppose A is an upper triangular matrix. Show that A− exists if and only if all 1 elements of the main diagonal are non zero. Is it true that A− will also be upper triangular? Explain. Is everything the same for lower triangular matrices? This is obvious because the determinant of A is the product of these diagonal entries. 1 When you consider the usual process of ﬁnding the inverse, you get that A− must be upper triangular. Everything is similar for lower triangular matrices. 8. Let A and B be two n n matrices. A B (A is similar to B) means there exists an × 1∼ invertible matrix S such that A = S− BS. Show that if A B, then B A. Show also that A A and that if A B and B C, then A C.∼ ∼ ∼ ∼ ∼ ∼ 1 1 This is easy except possibly for the last claim. Say A = P − BP and B = Q− CQ. Then 1 1 1 1 A = P − BP = A = P − Q− CQP = (QP )− C (QP )

9. In the context of Problem 8 show that if A B, then det (A) = det (B) . ∼

1 1 det A = det P − BP = det P − det (B) det (P ) 1 = det (B) det P − P = det (B) . 10. Let A be an n n matrix and let x be a nonzero vector such that Ax = λx for some scalar, λ. When× this occurs, the vector, x is called an eigenvector and the scalar, λ is called an eigenvalue. It turns out that not every number is an eigenvalue. Only certain ones are. Why? Hint: Show that if Ax = λx, then (λI A) x = 0. Explain why this shows that (λI A) is not one to one and not onto. Now− use Theorem 3.1.15 to argue det(λI A) =− 0. What sort of equation is this? How many solutions does it have? − 1 If you have (λI A) x = 0 for x = 0, then (λI A)− cannot exist because if it did, you could multiply− on the left by it6 and then conclude− that x = 0. Therefore, (λI A) is not one to one and not onto. − 11. Suppose det(λI A) = 0. Show using Theorem 3.1.15 there exists x = 0 such that (λI A) x = 0. − 6 − If that determinant equals 0 then the matrix λI A has no inverse. It is not one to one and so there exists x = 0 such that (λI A−) x = 0. Also recall the process for ﬁnding the inverse. 6 − a (t) b (t) 12. Let F (t) = det . Verify c (t) d (t)

a′ (t) b′ (t) a (t) b (t) F ′ (t) = det + det . c (t) d (t) c′ (t) d′ (t) Now suppose a (t) b (t) c (t) F (t) = det d (t) e (t) f (t) .  g (t) h (t) i (t)   

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Use Laplace expansion and the ﬁrst part to verify F ′ (t) =

a′ (t) b′ (t) c′ (t) a (t) b (t) c (t) det d (t) e (t) f (t) + det d′ (t) e′ (t) f ′ (t)  g (t) h (t) i (t)   g (t) h (t) i (t)   a (t) b (t) c (t)   + det d (t) e (t) f (t) .   g′ (t) h′ (t) i′ (t)   Conjecture a general result valid for n n matrices and explain why it will be true. Can a similar thing be done with the columns?× The way to see this holds in general is to use the usual proof for the product rule and the theorem about the determinant and row operations.

a (t + h) b (t + h) c (t + h) F (t + h) F (t) = det d (t + h) e (t + h) f (t + h) −  g (t + h) h (t + h) i (t + h)   a (t) b (t) c (t)  det d (t) e (t) f (t) −  g (t) h (t) i (t)    And so this equals

a (t + h) b (t + h) c (t + h) a (t) b (t) c (t) det d (t + h) e (t + h) f (t + h) det d (t + h) e (t + h) f (t + h)  g (t + h) h (t + h) i (t + h)  −  g (t + h) h (t + h) i (t + h)   a (t) b (t) c (t)   a (t) b (t) c (t)  + det d (t + h) e (t + h) f (t + h) det d (t) e (t) f (t)  g (t + h) h (t + h) i (t + h)  −  g (t + h) h (t + h) i (t + h)   a (t) b (t) c (t)   a (t) b (t) c (t)  + det d (t) e (t) f (t) det d (t) e (t) f (t)  g (t + h) h (t + h) i (t + h)  −  g (t) h (t) i (t)      F (t+h) F (t) Now multiply by 1/h to obtain the following for the diﬀerence quotient h− .

a(t+h) a(t) b(t+h) b(t) c(t+h) c(t) a (t) b (t) c (t) h− h− h− d(t+h) d(t) e(t+h) e(t) f(t+h) f(t) det d (t + h) e (t + h) f (t + h) +det − − −    h h h  g (t + h) h (t + h) i (t + h) g (t + h) h (t + h) i (t + h)     a (t) b (t) c (t) + det d (t) e (t) f (t)  g(t+h) g(t) h(t+h) h(t) i(t+h) i(t)  h− h− h−   Now passing to a limit yields the desired formula. Obviously this holds for any size determinant. 13. Use the formula for the inverse in terms of the cofactor matrix to ﬁnd the inverse of the matrix et 0 0 A = 0 et cos t et sin t .  0 et cos t et sin t et cos t + et sin t  −  

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1 et 0 0 − 0 et cos t et sin t  0 et cos t et sin t et cos t + et sin t  −  t  e− 0 0 t t = 0 e− (cos t + sin t) (sin t) e−  t − t  0 e− (cos t sin t) (cos t) e− − −   14. Let A be an r r matrix and let B be an m m matrix such that r +m = n. Consider the following n× n block matrix × × A 0 C = . DB

where the D is an m r matrix, and the 0 is a r m matrix. Letting Ik denote the k k identity matrix,× tell why × × A 0 I 0 C = r . DI 0 B m Now explain why det (C) = det (A) det (B) . Hint: Part of this will require an expla- nation of why A 0 det = det (A) . DI m See Corollary 3.1.9. The ﬁrst follows right away from block multiplication. Now

A 0 I 0 det (C) = det det r DI 0 B m A 0 I 0 = det det r = det (A) det (B) 0 I 0 B m from expanding along the last m columns for the first one and along the first r columns for the second. 15. Suppose Q is an orthogonal matrix. This means Q is a real n n matrix which satisfies × QQT = I

Find the possible values for det (Q). You have to have det (Q) det QT = det (Q)2 =1 and so det(Q) = 1. ± 16. Suppose Q (t) is an orthogonal matrix. This means Q (t) is a real n n matrix which satisﬁes × Q (t) Q (t)T = I Suppose Q (t) is continuous for t [a, b] , some interval. Also suppose det (Q (t)) = 1. Show that it follows det (Q (t)) =∈ 1 for all t [a, b]. ∈ You have from the given equation that det (Q (t)) is always either 1 or 1. Since Q (t) is continuous, so is t det (Q (t)) and so if it starts oﬀ at 1, it cannot− jump to 1 because this would violate→ the intermediate value theorem. −

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3.6

1. Let m < n and let A be an m n matrix. Show that A is not one to one. Hint: Consider the n n matrix A which× is of the form × 1 A A 1 ≡ 0 where the 0 denotes an (n m) n matrix of zeros. Thus det A = 0 and so A is − × 1 1 not one to one. Now observe that A1x is the vector, Ax A x = 1 0 which equals zero if and only if Ax = 0. The hint gives it away. You could simply consider a vector of the form

0 a where a = 0. 6 2. Let v , , v be vectors in Fn and let M (v , , v ) denote the matrix whose ith 1 ··· n 1 ··· n column equals vi. Deﬁne

d (v , , v ) det (M (v , , v )) . 1 ··· n ≡ 1 ··· n Prove that d is linear in each variable, (multilinear), that

d (v , , v , , v , , v ) = d (v , , v , , v , , v ) , (6.28) 1 ··· i ··· j ··· n − 1 ··· j ··· i ··· n and d (e , , e ) = 1 (6.29) 1 ··· n n th where here ej is the vector in F which has a zero in every position except the j position in which it has a one. This follows from the properties of determinants which are discussed above. 3. Suppose f : Fn Fn F satisﬁes 6.28 and 6.29 and is linear in each variable. Show that f = d.× · · · × → Consider f (x , , x ) . Then by the assumptions on f it equals 1 ··· n f (x , , x ) = x x f (e e ) 1 ··· n i1 ··· in i1 ··· in i1, ,i X··· n

= x x sgn (i , , i ) f (e e ) 1i1 ··· 1in 1 ··· n 1 ··· 1 i1, ,i X··· n = x x sgn (i , , i ) d (e e ) 1i1 ··· 1in 1 ··· n 1 ··· 1 i1, ,i X··· n = x x d (e e ) = d (x , , x ) 1i1 ··· 1in i1 ··· in 1 ··· n i1, ,i X··· n

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4. Show that if you replace a row (column) of an n n matrix A with itself added to some multiple of another row (column) then the new× matrix has the same determinant as the original one. This was done above. 5. Use the result of Problem 4 to evaluate by hand the determinant

1 2 3 2 6 3 2 3 det .  −5 2 2 3   3 4 6 4      1 2 3 2 6 3 2 3 det = 5  −5 2 2 3   3 4 6 4      6. Find the inverse if it exists of the matrix et cos t sin t et sin t cos t .  et −cos t sin t  − −   1 et cos t sin t − et sin t cos t  et −cos t sin t  − −  1 t  1 t 2 e− 0 2 e− 1 1 1 1 = 2 cos t + 2 sin t sin t 2 sin t 2 cos t  1 sin t 1 cos t −cos t 1 cos t− 1 sin t  2 − 2 − 2 − 2   (n) (n 1) 7. Let Ly = y + an 1 (x) y − + + a1 (x) y′ + a0 (x) y where the ai are given continuous functions− deﬁned on a closed··· interval, (a, b) and y is some function which has n derivatives so it makes sense to write Ly. Suppose Ly = 0 for k = 1, 2, , n. k ··· The Wronskian of these functions, yi is deﬁned as

y (x) y (x) 1 ··· n y′ (x) y′ (x) 1 ··· n W (y1, , yn)(x) det  . .  ··· ≡ . .  (n 1) (n 1)   y − (x) y − (x)   1 n   ···  Show that for W (x) = W (y , , y )(x) to save space, 1 ··· n y (x) y (x) 1 ··· n y1′ (x) yn′ (x) W ′ (x) = det  . ··· .  . . .    y(n) (x) y(n) (x)   1 n   ···  Now use the differential equation, Ly = 0 which is satisfied by each of these functions, yi and properties of determinants presented above to verify that W ′ +an 1 (x) W = 0. Give an explicit solution of this linear differential equation, Abel’s formula,− and use

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your answer to verify that the Wronskian of these solutions to the equation, Ly = 0 either vanishes identically on (a, b) or never.

The last formula above follows because W ′ (x) equals the sum of determinants of matrices which have two equal rows except for the last one in the sum which is the displayed expression. Now let

(n 1) mi (x) = an 1 (x) yi − + + a1 (x) yi′ + a0 (x) yi − − ··· (n) Since each yi is a solution to Ly = 0, it follows that yi (t) = mi (t). Now from the properties of determinants, being linear in each row, y (x) y (x) 1 ··· n y′ (x) y′ (x) 1 ··· n W ′ (x) = an 1 (x) det  . .  − − . .  (n 1) (n 1)   y − (x) y − (x)   1 n   ···  = an 1 (x) W ′ (x) − − Now let A′ (x) = an 1 (x) . Then − d eA(x)W (x) = 0 dx A(x) and so W (x) = Ce− . Thus the Wronskian either vanishes for all x or for no x.

1 8. Two n n matrices, A and B, are similar if B = S− AS for some invertible n n matrix ×S. Show that if two matrices are similar, they have the same characteristic× polynomials. The characteristic polynomial of A is det (λI A) . − 1 Say A = S− BS. Then

1 det (λI A) = det λI S− BS − − 1 1 = det λS− S S− BS − 1 = det S− (λI B) S − 1 = det S− det (λI B) det (S) − 1 = det S− S det (λI B) = det (λI B) − − 9. Suppose the characteristic polynomial of an n n matrix A is of the form × n n 1 t + an 1t − + + a1t + a0 − ··· 1 and that a0 = 0. Find a formula A− in terms of powers of the matrix A. Show that 1 6 n A− exists if and only if a = 0.In fact a = ( 1) det (A). 0 6 0 − From the Cayley Hamilton theorem,

n n 1 A + an 1A − + + a1A + a0I = 0 − ··· Also the characteristic polynomial is det (tI A) − n and the constant term is ( 1) det (A) . Thus a0 = 0 if and only if det (A) = 0 if and 1 − 1 6 6 only if A− has an inverse. Thus if A− exists, it follows that

n n 1 n 1 n 2 a0I = A + an 1A − + + a1A = A A − an 1A − a1I − − ··· − − − − · · · −

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and also n 1 n 2 a0I = A − an 1A − a1I A − − − − · · · − Therefore, the inverse is

1 n 1 n 2 A − an 1A − a1I a0 − − − − · · · − 10. Letting p (t) denote the characteristic polynomial of A, show that ↑ p (t) p (t ε) ε ≡ − is the characteristic polynomial of A + εI. Then show that if det (A) = 0, it follows that det (A + εI) = 0 whenever ε is sufficiently small. 6 | | p (t) det (tI A) . Hence p (t ε) = det ((t ε) I A) = det (tI (εI + A)) and by definition,≡ this− is the characteristic− polynomial− of −εI + A. Therefore,− the constant term of this is the constant term of p (t ε). Say − n n 1 p (t) = t + an 1t − + + a1t + a0 − ··· where the smallest ai which is nonzero is ak. Then it reduces to n n 1 k t + an 1t − + + akt − ··· Then n n 1 k pε (t) = p (t ε) = (t ε) + an 1 (t ε) − + + ak (t ε) − − − − ··· − k k then the constant term of this pε (t) is ak ( 1) ε + terms having ε raised to a higher power. Hence for all ε small enough, this term− will dominate the sum of all the others and it follows that the constant term is nonzero for all ε small enough. | | 11. In constitutive modeling of the stress and strain tensors, one sometimes considers sums k of the form k∞=0 akA where A is a 3 3 matrix. Show using the Cayley Hamilton theorem that if such a thing makes any× sense, you can always obtain it as a finite sum having no moreP than n terms. Say the characteristic polynomial is q (t) . Then if n 3, ≥ tn = q (t) l (t) + r (t) where the degree of r (t) is either less than 3 or it equals zero. Thus An = q (A) l (A) + r (A) = r (A) and so all the terms An for n 3 can be replaced with some r (A) where the degree of r (t) is no more than 2. Thus,≥ assuming there are no convergence issues, the infinite 2 k sum must be of the form k=0 bkA . 12. Recall you can find the determinantP from expanding along the jth column.

det (A) = Aij (cof (A))ij i X th Think of det (A) as a function of the entries, Aij . Explain why the ij cofactor is really just ∂ det (A) . ∂Aij This follows from the formula for determinant. If you take the partial derivative, you

get (cof (A))ij.

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13. Let U be an open set in Rn and let g :U Rn be such that all the ﬁrst partial derivatives of all components of g exist and→ are continuous. Under these conditions form the matrix Dg (x) given by

∂gi (x) Dg (x)ij gi,j (x) ≡ ∂xj ≡ The best kept secret in calculus courses is that the linear transformation determined by this matrix Dg (x) is called the derivative of g and is the correct generalization of the concept of derivative of a function of one variable. Suppose the second partial derivatives also exist and are continuous. Then show that

(cof (Dg))ij,j = 0. j X Hint: First explain why

gi,k cof (Dg)ij = δjk det (Dg) i X Next diﬀerentiate with respect to xj and sum on j using the equality of mixed partial derivatives. Assume det (Dg) = 0 to prove the identity in this special case. Then explain why there exists a sequence6 ε 0 such that for g (x) g (x) + ε x, k → εk ≡ k det (Dgεk ) = 0 and so the identity holds for gεk . Then take a limit to get the desired result6 in general. This is an extremely important identity which has surprising implications. First assume det (Dg) = 0. Then from the cofactor expansion, you get 6

gi,k cof (Dg)ij = δjk det (Dg) i X If j = k you get det (Dg) on the right but if j = k, then the left is the expansion of a determinant which has two equal columns. Now6 diﬀerentiate both sides with respect to j using the above problem. You have to use the chain rule.

gi,kj cof (Dg)ij + gi,kcof (Dg)ij,j i i X X ∂ (det (Dg)) = δ g = δ cof (Dg) g jk g r,sj jk rs r,sj r,s r,s r,s X X Next sum on j

gi,kj cof (Dg)ij + gi,kcof (Dg)ij,j = δjkcof (Dg)rs gr,sj i,j i,j r,s,j X X X Of course the terms on the right are 0 unless j = k, and so

gi,kj cof (Dg)ij + gi,kcof (Dg)ij,j = cof (Dg)rs gr,sk i,j i,j r,s X X X = cof (Dg)ij gi,jk i,j X Subtract the terms at the end from each side using equality of mixed partial derivatives. This gives.

0 = g cof (Dg) = g cof (Dg) i,k ij,j i,k  ij,j i,j i j X X X  

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Since Dg is assumed invertible, this requires

cof (Dg)ij,j = 0 j X

In case det (Dg (x)) = 0 for some x, consider εk, a sequence of numbers converging to 0 with the property that det (Dg (x) + εkI) = 0. Then let gk (x) = g (x) + εkx. From the above, you have 6

cof (Dg)ij,j (x) = lim cof (Dgk)ij,j (x) = 0 k j →∞ j X X 14. A determinant of the form 1 1 1 ··· a0 a1 an ··· a2 a2 a2 0 1 ··· n ...... n 1 n 1 n 1 a − a − a − 0 1 ··· n an an an 0 1 ··· n

is called a Vandermonde determinant. Show this determinant equals

(a a ) j − i 0 i

By this is meant to take the product of all terms of the form (aj ai) such that j > i. Hint: Show it works if n = 1 so you are looking at −

1 1 a a 0 1

Then suppose it holds for n 1 and consider the case n. Consider the following polynomial. − 1 1 1 ··· a0 a1 t ··· a2 a2 t2 0 1 ··· p (t) . . . . ≡ . . . n 1 n 1 n 1 a − a − t − 0 1 ··· an an tn 0 1 ···

Explain why p (a ) = 0 for i = 0 , , n 1. Thus j ··· − n 1 − p (t) = c (t a ) . − i i=0 Y Of course c is the coeﬃcient of tn. Find this coeﬃcient from the above description of p (t) and the induction hypothesis. Then plug in t = an and observe you have the formula valid for n. 1 1 In the case of n = 1, you have = (a1 a0) so the formula holds. Now a0 a1 −

suppose it holds for n and consider p (t) as above. Then it is obvious that p (a ) = 0 j

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for i = 0, , n 1 because you have the determinant of a matrix with two equal columns. Thus··· − n 1 − p (t) = c (t a ) . − i i=0 Y because p (t) has degree n and so it has no more than n roots. So what is c? The coeﬃcient of tn is the Vandermonde determinant which is n n and by induction, it × n equals 0 i

4.6

n 1. Let u1, , un be vectors in R . The parallelepiped determined by these vectors P (u{, ···, u ) is} deﬁned as 1 ··· n n P (u1, , un) tkuk : tk [0, 1] for all k . ··· ≡ ( ∈ ) kX=1 Now let A be an n n matrix. Show that × Ax : x P (u , , u ) { ∈ 1 ··· n } is also a parallelepiped. A typical thing in Ax : x P (u , , u ) is { ∈ 1 ··· n } n t Au : t [0, 1] k k k ∈ kX=1 and so it is just P (Au , ,Au ) . 1 ··· n 2. In the context of Problem 1, draw P (e1, e2) where e1, e2 are the standard basis vectors 2 for R . Thus e1 = (1, 0) , e2 = (0, 1) . Now suppose

1 1 E = 0 1 where E is the elementary matrix which takes the third row and adds to the ﬁrst. Draw Ex : x P (e , e ) . { ∈ 1 2 }

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In other words, draw the result of doing E to the vectors in P (e1, e2). Next draw the results of doing the other elementary matrices to P (e1, e2). For E the given elementary matrix, the result is as follows.

P (e1, e2) E(P (e1, e2))

It is called a shear. Note that it does not change the area. 1 0 In caseE = , the given square P (e , e ) becomes a rectangle. If α > 0, 0 α 1 2 the square is stretched by a factor of α in the y direction. If α < 0 the rectangle is reﬂected across the x axis and in addition is stretched by a factor of α in the y α 0 | | direction. When E = the result is similar only it features changes in the x 0 1 direction. These elementary matrices do change the area unless α = 1 when the area 0 1 | | is unchanged. The permutation simply switches e and e so the result 1 0 1 2 appears to be a square just like the one you began with. 3. In the context of Problem 1, either draw or describe the result of doing elementary matrices to P (e1, e2, e3). Describe geometrically the conclusion of Corollary 4.3.7.

It is the same sort of thing. The elementary matrix either switches the ei about or it produces a shear or a magniﬁcation in one direction. 4. Consider a permutation of 1, 2, , n . This is an ordered list of numbers taken from { ··· } this list with no repeats, i1, i2, , in . Deﬁne the permutation matrix P (i1, i2, , in) as the matrix which is obtained{ ··· from} the identity matrix by placing the jth column··· of I as the ith column of P (i , i , , i ) . What does this permutation matrix do to j 1 2 ··· n the column vector (1, 2, , n)T ? ··· It produces the column (i , i , , i )T . 1 2 ··· n 5. Consider the 3 3 permutation matrices. List all of them and then determine the dimension of their× span. Recall that you can consider an m n matrix as something in Fnm. × Here they are.

1 0 0 0 0 1 0 1 0 0 1 0 , 1 0 0 , 0 0 1  0 0 1   0 1 0   1 0 0   0 1 0   1 0 0   0 0 1  1 0 0 , 0 0 1 , 0 1 0  0 0 1   0 1 0   1 0 0        So what is the dimension of the span of these? One way to systematically accomplish this is to unravel them and then use the row reduced echelon form. Unraveling these yields the column vectors

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1 0 0 0 1 0 0 0 1 1 0 0  0   1   0   0   0   1   0   1   0   1   0   0               1   0   0   0   0   1               0   0   1   0   1   0               0   0   1   0   0   1               0   1   0   0   1   0               1   0   0   1   0   0              Then arranging   these  as the columns   of a matrix yields the following along with its row reduced echelon form. 1 0 0 0 1 0 1 0 0 0 0 1 0 0 1 1 0 0 0 1 0 0 0 1  0 1 0 0 0 1   0 0 1 0 0 1   0 1 0 1 0 0   0 0 0 1 0 1     −   1 0 0 0 0 1 , row echelon form:  0 0 0 0 1 1     −   0 0 1 0 1 0   0 0 0 0 0 0       0 0 1 0 0 1   0 0 0 0 0 0       0 1 0 0 1 0   0 0 0 0 0 0       1 0 0 1 0 0   0 0 0 0 0 0      The dimension is 5.    6. Determine which matrices are in row reduced echelon form. 1 2 0 (a) 0 1 7 This one is not. 1 0 0 0 (b) 0 0 1 2  0 0 0 0  This one is.  1 1 0 0 0 5 (c) 0 0 1 2 0 4  0 0 0 0 1 3  This one is. 

7. Row reduce the following matrices to obtain the row reduced echelon form. List the pivot columns in the original matrix.

1 2 0 3 1 0 0 3 (a) 2 1 2 2 , row echelon form: 0 1 0 0  1 1 0 3   0 0 1 2  −  1 2 3  1 0 0  2 1 2 0 1 0 (b) , row echelon form:  3 0− 0   0 0 1   3 2 1   0 0 0       1 2 1 3  1 0 0 0 1 (c) 3 2 1 0 , row echelon form: 0 1 2 0  −3 2 1 1   0 0 0 1     

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8. Find the rank and nullity of the following matrices. If the rank is r, identify r columns in the original matrix which have the property that every other column may be written as a linear combination of these. 0 1 0 2 1 2 2 0 1 0 2 0 1 1 0 3 2 12 1 6 8 0 0 1 3 0 1 2 (a) , row echelon form: Rank  0 1 1 5 0 2 3   0 0 0 0 1 1 1   0 2 1 7 0 3 4   0 0 0 0 0 0 0      equals 3 nullity equals 4. A basis is columns 2,3,5  0 1 0 2 0 1 0 0 1 0 2 0 1 0 0 3 2 6 0 5 4 0 0 1 0 0 1 2 (b) , row echelon form: Rank  0 1 1 2 0 2 2   0 0 0 0 0 0 0   0 2 1 4 0 3 2   0 0 0 0 0 0 0      is 2 nullity equals 5. A basis is columns 2,3.   0 1 0 2 1 1 2 0 1 0 2 0 1 0 0 3 2 6 1 5 1 0 0 1 0 0 1 0 (c) , row echelon form: Rank  0 1 1 2 0 2 1   0 0 0 0 1 0 0   0 2 1 4 0 3 1   0 0 0 0 0 0 1      is 4 and nullity is 3. A basis is columns 2,3,5,7. 

9. Find the rank of the following matrices. If the rank is r, identify r columns in the original matrix which have the property that every other column may be written as a linear combination of these. Also ﬁnd a basis for the row and column spaces of the matrices. 1 2 0 1 0 0 3 2 1 0 1 0 (a) , row echelon form: . The rank is 3.  2 1 0   0 0 1   0 2 1   0 0 0       1 0 0   1 0 0  4 1 1 0 1 0 (b) , row echelon form: . The rank is 3  2 1 0   0 0 1   0 2 0   0 0 0       0 1 0 2 1 2 2   0 1 0 2 0 1 1 0 3 2 12 1 6 8 0 0 1 3 0 1 2 (c) , row echelon form: .  0 1 1 5 0 2 3   0 0 0 0 1 1 1   0 2 1 7 0 3 4   0 0 0 0 0 0 0      The rank is 3.    0 1 0 2 0 1 0 0 1 0 2 0 1 0 0 3 2 6 0 5 4 0 0 1 0 0 1 2 (d) , row echelon form: . The  0 1 1 2 0 2 2   0 0 0 0 0 0 0   0 2 1 4 0 3 2   0 0 0 0 0 0 0      rank is 2.    0 1 0 2 1 1 2 0 1 0 2 0 1 0 0 3 2 6 1 5 1 0 0 1 0 0 1 0 (e) , row echelon form: . The  0 1 1 2 0 2 1   0 0 0 0 1 0 0   0 2 1 4 0 3 1   0 0 0 0 0 0 1      rank is 4.   

10. Suppose A is an m n matrix. Explain why the rank of A is always no larger than min (m, n) . ×

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It is because you cannot have more than min (m, n) nonzero rows in the row reduced echelon form. Recall that the number of pivot columns is the same as the number of nonzero rows from the description of this row reduced echelon form. 11. Suppose A is an m n matrix in which m n. Suppose also that the rank of A equals × n m ≤ m. Show that A maps F onto F . Hint: The vectors e1, , em occur as columns in the row reduced echelon form for A. ···

It follows from the fact that e1, , em occur as columns in row reduced echelon form that the dimension of the column··· space of A is n and so, since this column space is A (Rn) , it follows that it equals Fm. 12. Suppose A is an m n matrix and that m > n. Show there exists b Fm such that there is no solution× to the equation ∈

Ax = b.

Since m > n the dimension of the column space of A is no more than n and so the columns of A cannot span Fm. 13. Suppose A is an m n matrix in which m n. Suppose also that the rank of A equals n. Show that ×A is one to one. Hint: If≥ not, there exists a vector, x = 0 such that Ax = 0, and this implies at least one column of A is a linear combination6 of the others. Show this would require the column rank to be less than n. The hint gives it away. The matrix has at most n independent columns. If one is a combination of the others, then its rank is less than n.

14. Explain why an n n matrix A is both one to one and onto if and only if its rank is n. × If A is one to one, then its columns are linearly independent, hence a basis and so it is also onto. Since the columns are independent, its rank is n. If its rank is n this just says that the columns are independent.

15. Suppose A is an m n matrix and w1, , wk is a linearly independent set of n ×m { ··· } vectors in A (F ) F . Suppose also that Azi wi. Show that z1, , zk is also linearly independent.⊆ − { ··· }

If i cizi = 0, apply A to both sides to obtain i ciwi = 0. By assumption, each ci = 0. P P 16. Let A, B be m n matrices. Show that rank (A + B) rank (A) + rank (B). × ≤ This is obvious from the following.

rank (A + B) = dim (span (a + b , , a + b )) 1 1 ··· n n

dim (span (a , , a , b , , b )) ≤ 1 ··· n 1 ··· n dim (span (b , , b )) + dim (span (a , , a )) ≤ 1 ··· n 1 ··· n = rank (A) + rank (B)

17. Suppose A is an m n matrix, m n and the columns of A are independent. Sup- × ≥ n pose also that z1, , zk is a linearly independent set of vectors in F . Show that Az , ,Az { is linearly··· } independent. { 1 ··· k}

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Since columns are independent, if Ax = 0, then x = 0. Therefore, the matrix is one to one. If n cj Azj = 0, j=1 X n n then A j=1 cj zj = 0 and so j=1 cj zj = 0 therefore each cj = 0. 18. Suppose PA is an m n matrix andP B is an n p matrix. Show that × × dim (ker (AB)) dim (ker (A)) + dim (ker (B)) . ≤ Hint: Consider the subspace, B (Fp) ker (A) and suppose a basis for this subspace is w , , w . Now suppose u , ∩ , u is a basis for ker (B) . Let z , , z { 1 ··· k} { 1 ··· r} { 1 ··· k} be such that Bzi = wi and argue that ker (AB) span (u , , u , z , , z ) . ⊆ 1 ··· r 1 ··· k Here is how you do this. Suppose ABx = 0. Then Bx ker (A) B (Fp) and so k ∈ ∩ Bx = i=1 aiBzi showing that P k x a z ker (B) . − i i ∈ i=1 X p Let w1, , wk be a basis for B (F ) ker (A) and suppose u1, , ur is a basis { ··· } ∩ { ··· } p for ker (B). Let Bzi = wi. Now suppose x ker (AB). Then Bx ker (A) B (F ) and so ∈ ∈ ∩ k k Bx = aiwi = aiBzi i=1 i=1 so X X k x a z ker (B) − i i ∈ i=1 X Hence k r x a z = b u − i i j j i=1 j=1 X X This shows that dim (ker (AB)) k + r dim (ker (A)) + dim (ker (B)) ≤ ≤ Note that a little more can be said. u1, , ur, z1, , zk is independent. Here is why. Suppose { ··· ··· } k aiui + bj zj = 0 i j=1 X X Then do B to both sides and conclude that

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19. Let m < n and let A be an m n matrix. Show that A is not one to one. × There are more columns than rows and at most m can be pivot columns so it follows at least one column is a linear combination of the others hence A is not one too one. 20. Let A be an m n real matrix and let b Rm. Show there exists a solution, x to the system × ∈ AT Ax = AT b

Next show that if x, x1 are two solutions, then Ax = Ax1. Hint: First show that T AT A = AT A. Next show if x ker AT A , then Ax = 0. Finally apply the Fred- T ∈ T holm alternative. Show A b ker(A A)⊥. This will give existence of a solution. ∈ T T AT A = AT AT = AT A. Also if AT Ax = 0, then

AT Ax, x = Ax 2 = 0 | | and so Ax = 0. Is AT b ker AT A ⊥? Say AT Ax = 0. Then Ax = 0. Then ∈ AT b, x = (b,Ax) = (b, 0) = 0

T T Hence A b ker(A A)⊥ and so there exists a solution to the above system. ∈ 21. Show that in the context of Problem 20 that if x is the solution there, then b Ax b Ay for every y. Thus Ax is the point of A (Rn) which is closest to|b −of every| ≤ point| − in|A (Rn). This is a solution to the least squares problem.

b Ay 2 = b Ax+Ax Ay 2 | − | | − − | = b Ax 2 + Ax Ay 2 + 2 (b Ax,A (x y)) | − | | − | − − = b Ax 2 + Ax Ay 2 + 2 AT b AT Ax, (x y) | − | | − | − − 2 2 = b Ax + Ax Ay | − | | − | and so, Ax is closest to b out of all vectors Ay. 1 0 0 1 22. Here is a point in R4 : (1, 2, 3, 4)T . Find the point in span , which  2   3   3   2      is closest to the given point.     This just says to ﬁnd the least squares solution to the equations

1 0 1 0 1 x 2 =  2 3  y  3   3 2   4          T T 1 0 1 0 1 0 1 0 1 0 1 x 0 1 2 =  2 3   2 3  y  2 3   3   3 2   3 2   3 2   4                 

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14 12 x 19 19 = , Solution is: 26 . Then the desired point is 12 14 y 19 19 26 19 1 0 26 19 19 0 19 1 26   +   =  95  26 2 26 3 26  3   2   95       26        23. Here is a point in R4 : (1, 2, 3, 4)T . Find the point on the plane described by x+ 2y 4↑z + 4w = 0 which is closest to the given point. − 2y + 4z 4w − y − The plane is of the form where y, z, w are arbitary. Thus it is  z   w    the span of the vectors   2 4 4 −1 0 −0 , ,  0   1   0   0   0   1        and so you are looking for a least squares  solution  to the system 2 4 4 1 a −1 0− 0 2 b = and to ﬁnd this, you use the method of least  0 1 0   3   c   0 0 1   4        squares.    T T 2 4 4 2 4 4 2 4 4 1 a −1 0− 0 −1 0− 0 −1 0− 0 2 b =  0 1 0   0 1 0   0 1 0   3   c   0 0 1   0 0 1   0 0 1   4                 56    5 8 8 a 0 37 − 147 8 17 16 b = 7 , Solution is: 37 The closest point is  −8 16− 17   c   0   112  − 37 then        28 2 4 4 37 − − 56 56 1 147 0 112 0 37 37   + 37   + 37   =  147  0 1 0 37  0   0   1   112         37      28     37 56 37 Check: 1 2 4 4  147  = 0 − 37  112   37    24. Suppose A, B are two invertible n n matrices. Show there exists a sequence of row operations which when done to A yield× B. Hint: Recall that every invertible matrix is a product of elementary matrices. Both A, B have row reduced echelon form equal to I. Therefore, they are row equivalent. E E A = I,F F B = I 1 ··· p 1 ··· r

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and so E E A = F F B 1 ··· p 1 ··· r Now multiply on both sides by the inverses of the Fi to obtain a product of elementary matrices which multiplied by A give B. 25. If A is invertible and n n and B is n p, show that AB has the same null space as B and also the same rank× as B. × From the above problem, AB and B have exactly the same row reduced echelon form. Thus the result follows. 26. Here are two matrices in row reduced echelon form 1 0 1 1 0 0 A = 0 1 1 ,B = 0 1 1  0 0 0   0 0 0      Does there exist a sequence of row operations which when done to A will yield B? Explain. No. The row reduced echelon form is unique. 27. Is it true that an upper triagular matrix has rank equal to the number of nonzero entries down the main diagonal? 1 0 2 0 0 1 1 7 No.  0 0 0 1   0 0 0 0      n 28. Let v1, , vn 1 be vectors in F . Describe a systematic way to obtain a vector vn which{ is··· perpendicular− } to each of these vectors. Hint: You might consider something like this e e e 1 2 ··· n v11 v12 v1n det  . . ··· .  . . .    v(n 1)1 v(n 1)2 v(n 1)n   − − ··· −  th   where vij is the j entry of the vector vi. This is a lot like the cross product. This is a good hint. You formally expand along the top row. It will work. If you replace ei with vki so that the top row becomes.

v v v k1 k2 ··· kn then the determinant will be equal to 0 because it has two equal rows. However, this determinant is also what you get when you take the dot product of the above formal determinant having the ei on the top row with vk. 29. Let A be an m n matrix. Then ker (A) is a subspace of Fn. Is it true that every subspace of Fn is× the kernel or null space of some matrix? Prove or disprove. Let M be a subspace of Fn. If it equals 0 , consider the matrix I. Otherwise, it has a basis m , , m . Consider the matrix{ } { 1 ··· k} m m 0 1 ··· k where 0 is either not there in case k = n or has n k columns. −

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30. Let A be an n n matrix and let P ij be the permutation matrix which switches the ith and jth rows of× the identity. Show that P ij AP ij produces a matrix which is similar to A which switches the ith and jth entries on the main diagonal. This is easy to see when you consider that P ij is its own inverse and that P ij multiplied on the right switches the ith and jth columns. Thus you switch the columns and then you switch the rows. This has the effect of switching Aii and Ajj . For example, 1 0 0 0 a b c d 1 0 0 0 a d c b 0 0 0 1 e f z h 0 0 0 1 n g h t =  0 0 1 0   j k l m   0 0 1 0   j m l k   0 1 0 0   n t h g   0 1 0 0   e h z f          More formally, the iith entry of P ijAPij is    ij ij ij ij Pis AspPpi = Pij Ajj Pji = Aij s,p X 31. Recall the procedure for finding the inverse of a matrix on Page 48. It was shown that the procedure, when it works, finds the inverse of the matrix. Show that whenever the matrix has an inverse, the procedure works. If A has an inverse, then it is one to one. Hence the columns are independent. There- fore, they are each pivot columns. Therefore, the row reduced echelon form of A is I. This is what was needed for the procedure to work.

F.21 Exercises

5.8 1 2 0 1 0 0 1 2 0 1. Find a LU factorization of 2 1 3 . = 2 1 0 0 3 3  1 2 3   1 0 1   0− 0 3        1 2 3 2 1 0 0 1 2 3 2 2. Find a LU factorization of 1 3 2 1 . = 1 1 0 0 1 1 1  5 0 1 3   5 10 1   0 0 −24 −17  − − −       1 2 1 1 0 0 1 0 0 1 2 1 3. Find a PLU factorization of 1 2 2 . = 0 0 1 2 1 0 0 3 1  2 1 1   0 1 0   1 0 1   0− 0− 1          1 2 1 2 1 4. Find a PLU factorization of 2 4 2 4 1 .  1 2 1 3 2  1 0 0 1 0 0  1 2 1 2 1 = 0 1 0 2 1 0 0 0 0 0 1  0 0 1   1 0 1   0 0 0 1− 1        1 2 1 1 2 2 5. Find a PLU factorization of .  2 4 1   3 2 1    1 0 0 0 1 0 0 0 1 2 1 0 0 0 1 3 1 0 0 0 4 2 =  0 0 1 0   2 0 1 0   0− 0 −1  −  0 1 0 0   1 0 1 1   0 0 0     −         

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6. Is there only one LU factorization for a given matrix? Hint: Consider the equation

0 1 1 0 0 1 = . 0 1 1 1 0 0 Apparently, LU factorization is not unique. 0 1 1 0 0 1 = 0 1 1 1 0 0 0 1 1 0 0 1 = 0 1 0 1 0 1 7. Here is a matrix and an LU factorization of it. 1 2 5 0 1 0 0 1 2 5 0 A = 1 1 4 9 = 1 1 0 0 1 1 9  0 1 2 5   0 1 1   0− 0− 1 14  −       Use this factorization to solve the system of equations

1 Ax = 2  3    1 0 0 a 1 1 1 1 0 b = 2 , Solution is: 1  0 1 1   c   3   4  −     x    24tˆ 9 1 2 5 0 1 4 − y 23tˆ4 5 0 1 1 9   = 1 , Solution is:  −  , tˆ4 R  − −  z   4 14tˆ4 ∈ 0 0 1 14 4 −  w   tˆ         4      8. Find a QR factorization for the matrix

1 2 1 3 2 1  1− 0 2   

1 √ 13 √ √ 1 √ 1 2 1 11 11 66 2 11 6 2 3 √ 5 √ √ − 1 √ 3 2 1 = 11 11 66 2 11 6 2  1− 0 2   1 √ −1 √ √ −2 √  · 11 11 33 2 11 3 2    √11 4 √11 6 √11  − 11 11 0 6 √2√11 2 √2√11  11 11  0 0 √2   9. Find a QR factorization for the matrix

1 2 1 0 3 0 1 1  1 0 2 1   

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1 2 1 0 3 0 1 1  1 0 2 1  1 √ 1 √ √  11 11 11 10 11 0 = 3 √11 3 √10√11 1 √2√5  11 − 110 − 10  1 √11 1 √10√11 3 √2√5 11 − 110 10 √ 2 √ 6 √ 4 √ 11 11 11 11 11 11 11 2 √ √ 1 √ √ 2 √ √ 0 11 10 11 22 10 11 55 10 11  1 √ √ − 1 √ √  0 0 2 2 5 5 2 5   10. If you had a QR factorization, A = QR, describe how you could use it to solve the equation Ax = b. Note, this is not how people usually solve systems of equations. You could ﬁrst solve Qy = b to obtain y = QT b. Then you would solve Rx = y which would be easy because R is upper triangular. 11. If Q is an orthogonal matrix, show the columns are an orthonormal set. That is show that for Q = q q 1 ··· n it follows that qi qj = δij. Also show that any orthonormal set of vectors is linearly independent. · n Say i=1 ciqi = 0. Then take the inner product of both sides with qk to obtain P ciδij = 0 i X Therefore, cj = 0 and so the vectors are independent. 12. Show you can’t expect uniqueness for QR factorizations. Consider

0 0 0 0 0 1  0 0 1    and verify this equals

0 1 0 0 0 √2 1 √2 0 1 √2 0 0 0  2 2    1 √2 0 1 √2 0 0 0 2 − 2     and also 1 0 0 0 0 0 0 1 0 0 0 1 .  0 0 1   0 0 1  Using Deﬁnition 5.7.4, can it be concluded  that if A is an invertible matrix it will follow there is only one QR factorization? 0 1 0 0 0 √2 0 0 0 1 √2 0 1 √2 0 0 0 = 0 0 1  2 2      1 √2 0 1 √2 0 0 0 0 0 1 2 − 2  1 0 0 0 0 0 0 0 0  0 1 0 0 0 1 = 0 0 1  0 0 1   0 0 1   0 0 1       

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1 Now if A− exists and A = QR = Q′R′ where R,R′ have positive entries down 1 T the main diagonal, then you get R (R′)− = Q Q′ U, where U is an orthogonal matrix. Thus you have an upper triangular matrix which≡ has positive entries down the diagonal equal to an orthogonal matrix. Obviously, this requires that the upper triangular matrix is the identity. Hence R = R′ and Q = Q′. Why is this so obvious? Consider the following case of a 3 3 which is orthogonal. × a b c 0 d e  0 0 f    Then this matrix times its transpose is the identity.

a b c a 0 0 a2 + b2 + c2 ce + bd cf 0 d e b d 0 = ce + bd d2 + e2 fe  0 0 f   c e f   cf fe f 2   a 0 0   a b c   a2 ab ac  b d 0 0 d e = ab b2 + d2 de + bc  c e f   0 0 f   ac de + bc c2 + f 2 + e2        Thus both of these on the right end are the identity and so b = c = 0, a = 1 since the diagonal entries are all positive. Then also d = 1 and c, e = 0 and f = 1. Now it follows that all the oﬀ diagonal entries must equal 0 since otherwise the magnitude of a column would not equal 1. 13. Suppose a , , a are linearly independent vectors in Rn and let { 1 ··· n} A = a a 1 ··· n Form a QR factorization for A.

r11 r12 r1n 0 r ··· r 22 ··· 2n a1 an = q1 qn  . .  ··· ··· . ..    0 0 r   ··· nn    Show that for each k n, ≤ span (a , , a ) = span (q , , q ) 1 ··· k 1 ··· k Prove that every subspace of Rn has an orthonormal basis. The procedure just described is similar to the Gram Schmidt procedure which will be presented later. From the way we multiply matrices,

a span (q , , q ) k ∈ 1 ··· k Hence span (a , , a ) span (q , , q ). In addition to this, each r > 0 because 1 ··· k ⊆ 1 ··· k ii the ai are a linearly independent set and if this were not so, then the triangular matrix on the right would fail to be invertible which would be a contradiction since you could express it as the product of two invertible matrices. Multiplying on the right by its inverse, you can get a similar expression for the qi and see that span(q1, , qk) span (a , , a ) . Thus ··· ⊆ 1 ··· k span (a , , a ) = span (q , , q ) 1 ··· k 1 ··· k

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If you have a subspace, let a1, , ak be a basis and then extend to a basis of n { ··· } R , a1, , an . Then form the above QR factorization and you will have that q ,{ ,···q is} an orthonormal basis for the subspace. { 1 ··· k} 14. Suppose QnRn converges to an orthogonal matrix Q where Qn is orthogonal and Rn is upper triangular having all positive entries on the diagonal. Show that then Qn converges to Q and Rn converges to the identity. Let Q = (q , , q ) ,Q = qk, , qk 1 ··· n k 1 ··· n k th where the q are the columns. Also denote by rij the ij entry of Rk. Thus

rk 11 ∗ Q R = qk, , qk .. k k 1 ··· n  .  0 rk  nn    It follows rk qk q 11 1 → 1 and so rk = rk qk 1 11 11 1 → Therefore,

qk q . 1 → 1 Next consider the second column.

rk qk + rk qk q 12 1 22 2 → 2 k Taking the inner product of both sides with q1 it follows

k k lim r12 = lim q2 q1 = (q2 q1) = 0. k k · · →∞ →∞ Therefore, k k lim r22q2 = q2 k →∞ and since rk > 0, it follows, as in the ﬁrst part that rk 1. Hence 22 22 → k lim q2 = q2. k →∞ Continuing this way, it follows k lim rij = 0 k →∞ for all i = j and 6 k k lim rjj = 1, lim qj = qj . k k →∞ →∞ Thus R I and Q Q. k → k →

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6.6

1. Maximize and minimize z = x1 2x2 + x3 subject to the constraints x1 + x2 + x3 10, x + x + x 2, and x + 2−x + x 7 if possible. All variables are nonnegative.≤ 1 2 3 ≥ 1 2 3 ≤ The constraints lead to the augmented matrix 1 1 1 1 0 0 10 1 1 1 0 1 0 2  1 2 1 0− 0 1 7    The obvious solution is not feasible. Do a row operation. 1 1 1 1 0 0 10 0 0 0 1 1 0 8  0 1 0 1 0 1 3  − −   An obvious solution is still not feasible. Do another operation couple of row operations. 1 1 1 0 1 0 2 0 0 0 1− 1 0 8  0 1 0 0 1 1 5    At this point, you can spot an obvious feasible solution. Now assemble the simplex tableau. 1 1 1 0 1 0 0 2 0 0 0 1− 1 0 0 8  0 1 0 0 1 1 0 5   1 2 1 0 0 0 1 0   − −  Now preserve the simple columns.  1 1 1 0 1 0 0 2 0 0 0 1− 1 0 0 8  0 1 0 0 1 1 0 5   0 2 0 0 1 0 1 0   −    First lets work on minimizing this. There is a +2. The ratios are then 5, 2 so the pivot is the 1 on the top of the second column. The next tableau is 1 1 1 0 1 0 0 2 0 0 0 1− 1 0 0 8  1 0 1 0 2 1 0 3  − −  2 0 2 0 1 0 1 4   − − −    There is a 1 on the bottom. The ratios of interest for that column are 3/2, 8, and so the pivot is the 2 in that column. Then the next tableau is

1 1 1 7 2 1 2 0 0 2 0 2 1 1 1 13 2 0 2 1 0 2 0 2  1 0 1 0 2− 1 0 3  − −  3 0 3 0 0 1 1 11   − 2 − 2 − 2 − 2    Now you stop because there are no more positive numbers to the left of 1 on the bottom row. The minimum is 11/2 and it occurs when x1 = x3 = x6 = 0 and x = 7/2, x = 13/2, x = 11/2.− 2 4 6 −

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Next consider maximization. The simplex tableau was

1 1 1 0 1 0 0 2 0 0 0 1− 1 0 0 8  0 1 0 0 1 1 0 5   1 2 1 0 0 0 1 0   − −    This time you work on getting rid of the negative entries. Consider the 1 in the ﬁrst column. There is only one ratio to consider so 1 is the pivot. −

1 1 1 0 1 0 0 2 0 0 0 1− 1 0 0 8  0 1 0 0 1 1 0 5   0 3 0 0 1 0 1 2   −    There remains a 1. The ratios are 5 and 8 so the next pivot is the 1 in the third row and column 5. − 1 2 1 0 0 1 0 7 0 1 0 1 0 1 0 3  0− 1 0 0 1− 1 0 5   0 4 0 0 0 1 1 7      Then no more negatives remain so the maximum is 7 and it occurs when x1 = 7, x2 = 0, x3 = 0, x4 = 3, x5 = 5, x6 = 0. 2. Maximize and minimize the following if possible. All variables are nonnegative.

(a) z = x1 2x2 subject to the constraints x1 + x2 + x3 10, x1 + x2 + x3 1, and x + 2x− + x 7 ≤ ≥ 1 2 3 ≤ an augmented matrix for the constraints is

1 1 1 1 0 0 10 1 1 1 0 1 0 1  1 2 1 0− 0 1 7    The obvious solution is not feasible. Do some row operations.

0 0 0 1 1 0 9 1 1 1 0 1 0 1  1 0 1 0− 2 1 5  − −   Now the obvious solution is feasible. Then include the objective function.

0 0 0 1 1 0 0 9 1 1 1 0 1 0 0 1  1 0 1 0− 2 1 0 5  − −  1 2 0 0 0 0 1 0   −    First preserve the simple columns.

0 0 0 1 1 0 0 9 1 1 1 0 1 0 0 1  1 0 1 0− 2 1 0 5  − −  3 0 2 0 2 0 1 2   − − −   

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Lets try to maximize ﬁrst. Begin with the ﬁrst column. The only pivot is the 1. Use it. 0 0 0 1 1 0 0 9 1 1 1 0 1 0 0 1  0 1 0 0− 1 1 0 6   0 3 1 0 1 0 1 1   −  There is still a -1 on the bottom row to the left of the 1. The ratios are 9 and 6 so the new pivot is the 1 on the third row.

0 1 0 1 0 1 0 3 1− 2 1 0 0− 1 0 7  0 1 0 0 1 1 0 6   0 4 1 0 0 1 1 7      Then the maximum is 7 when x1 = 7 and x1, x3 = 0. Next consider the minimum. 0 0 0 1 1 0 0 9 1 1 1 0 1 0 0 1  1 0 1 0− 2 1 0 5  − −  3 0 2 0 2 0 1 2   − − −    There is a positive 2 in the bottom row left of 1. The pivot in that column is the 2. 1 1 1 13 2 0 2 1 0 2 0 2 1 1 −1 7 2 1 2 0 0 2 0 2  1 0 1 0 2 1 0 5  − −  2 0 1 0 0 1 1 7   − − − −    The minimum is 7 and it happens when x = 0, x = 7/2, x = 0. − 1 2 3 (b) z = x1 2x2 3x3 subject to the constraints x1 + x2 + x3 8, x1 + x2 + 3x3 1, and x −+ x +−x 7 ≤ ≥ 1 2 3 ≤ This time, lets use artificial variables to find an initial simplex tableau. Thus you add in an artificial variable and then do a minimization procedure.

1 1 1 1 0 0 0 0 8 1 1 3 0 1 0 1 0 1  1 1 1 0− 0 1 0 0 7   0 0 0 0 0 0 1 1 0   −    First preserve the seventh column as a simple column by a row operation.

1 1 1 1 0 0 0 0 8 1 1 3 0 1 0 1 0 1  1 1 1 0− 0 1 0 0 7   1 1 3 0 1 0 0 1 1   −    Now use the third column.

2 2 1 1 23 3 3 0 1 3 0 3 0 3 1 1 3 0 1 0− 1 0 1  2 2 0 0 −1 1 1 0 20  3 3 3 − 3 3  0 0 0 0 0 0 1 1 0   −   

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It follows that a basic solution is feasible if

2 2 1 23 3 3 0 1 3 0 3 1 1 3 0 1 0 1  2 2 −1 20  3 3 0 0 3 1 3   Now assemble the simplex tableau

2 2 1 23 3 3 0 1 3 0 0 3 1 1 3 0 1 0 0 1  2 2 −1 20  3 3 0 0 3 1 0 3  1 2 3 0 0 0 1 0   −    Preserve the simple columns by doing row operations.

2 2 1 23 3 3 0 1 3 0 0 3 1 1 3 0 1 0 0 1  2 2 −1 20  3 3 0 0 3 1 0 3  2 1 0 0 1 0 1 1   − −    Lets do minimization ﬁrst. Work with the second column. 0 0 2 1 1 0 0 7 1 1− 3 0 1 0 0 1  0 0 2 0− 1 1 0 6  −  3 0 3 0 2 0 1 2   − − −    Recall how you have to pick the pivot correctly. There is still a positive number in the bottom row left of the 1. Work with that column. 0 0 0 1 0 1 0 1 1 1 1 0 0− 1 0 7  0 0 2 0 1 1 0 6  −  3 0 1 0 0 2 1 14   − − −    There is still a positive number to the left of 1 on the bottom row.

0 0 0 1 0 1 0 1 1 1 1 0 0− 1 0 7  2 2 0 0 1 3 0 20   4 1 0 0 0 3 1 21   − − − −    It follows that the minimum is 21 and it occurs when x = x = 0, x = 7. − 1 2 3 Next consider the maximum. The simplex tableau was

2 2 1 23 3 3 0 1 3 0 0 3 1 1 3 0 1 0 0 1  2 2 −1 20  3 3 0 0 3 1 0 3  2 1 0 0 1 0 1 1   − −    Use the ﬁrst column. 0 0 2 1 1 0 0 7 1 1− 3 0 1 0 0 1  0 0 2 0− 1 1 0 6  −  0 3 6 0 1 0 1 1   −   

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There is still a negative on the bottom row to the left of 1.

0 0 0 1 0 1 0 1 1 1 1 0 0− 1 0 7  0 0 2 0 1 1 0 6  −  0 3 4 0 0 1 1 7      There are no more negatives on the bottom row left of 1 so stop. The maximum is 7 and it occurs when x1 = 7, x2 = 0, x3 = 0.

(c) z = 2x1 + x2 subject to the constraints x1 x2 + x3 10, x1 + x2 + x3 1, and x + 2x + x 7. − ≤ ≥ 1 2 3 ≤ The augmented matrix for the constraints is

1 1 1 1 0 0 10 1− 1 1 0 1 0 1  1 2 1 0− 0 1 7    The basic solution is not feasible because of that 1. Lets do a row operation to change this. I used the 1 in the second column as− a pivot and zeroed out what was above and below it. Now it seems that the basic solution is feasible. 2 0 2 1 1 0 11 1 1 1 0 −1 0 1  1 0 1 0− 2 1 5  − −   Assemble the simplex tableau.

2 0 2 1 1 0 0 11 1 1 1 0 −1 0 0 1  1 0 1 0− 2 1 0 5  − −  2 1 0 0 0 0 1 0   − −    Then do a row operation to preserve the simple columns.

2 0 2 1 1 0 0 11 1 1 1 0 −1 0 0 1  1 0 1 0− 2 1 0 5  − −  1 0 1 0 1 0 1 1   − −    Lets do minimization ﬁrst. Work with the third column because there is a positive entry on the bottom.

0 2 0 1 1 0 0 9 1− 1 1 0 1 0 0 1  0 1 0 0− 1 1 0 6   2 1 0 0 0 0 1 0   − −    It follows that the minimum is 0 and it occurs when x1 = x2 = 0, x3 = 1. Now lets maximize. 2 0 2 1 1 0 0 11 1 1 1 0 −1 0 0 1  1 0 1 0− 2 1 0 5  − −  1 0 1 0 1 0 1 1   − −   

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Lets begin with the ﬁrst column.

0 2 0 1 1 0 0 9 1− 1 1 0 1 0 0 1  0 1 0 0− 1 1 0 6   0 1 2 0 2 0 1 2   −    There is still a 2 to the left of 1 in the bottom row. − 0 3 0 1 0 1 0 3 1− 2 1 0 0− 1 0 7  0 1 0 0 1 1 0 6   0 3 2 0 0 2 1 14      There are no negatives left so the maximum is 14 and it happens when x1 = 7, x2 = x3 = 0.

(d) z = x1 + 2x2 subject to the constraints x1 x2 + x3 10, x1 + x2 + x3 1, and x + 2x + x 7. − ≤ ≥ 1 2 3 ≤ The augmented matrix for the constraints is

1 1 1 1 0 0 10 1− 1 1 0 1 0 1  1 2 1 0− 0 1 7    Of course the obvious or basic solution is not feasible. Do a row operation involving a pivot in the second row to try and ﬁx this.

2 0 2 1 1 0 11 1 1 1 0 −1 0 1  1 0 1 0− 2 1 5  − −   Now all is well. Begin to assemble the simplex tableau.

2 0 2 1 1 0 0 11 1 1 1 0 −1 0 0 1  1 0 1 0− 2 1 0 5  − −  1 2 0 0 0 0 1 0   − −    Do a row operation to preserve the simple columns.

2 0 2 1 1 0 0 11 1 1 1 0 −1 0 0 1  1 0 1 0− 2 1 0 5  − −  1 0 2 0 2 0 1 2   −    Next lets maximize. There is only one negative number in the bottom left of 1.

3 3 1 27 2 0 2 1 0 2 0 2 1 1 1 7 2 1 2 0 0 2 0 2  1 0 1 0 2 1 0 5  − −  0 0 1 0 0 1 1 7      Thus the maximum is 7 and it happens when x2 = 7/2, x3 = x1 = 0.

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Next lets ﬁnd the minimum. 2 0 2 1 1 0 0 11 1 1 1 0 −1 0 0 1  1 0 1 0− 2 1 0 5  − −  1 0 2 0 2 0 1 2   −    Start with the column which has a 2. 0 2 0 1 1 0 0 9 1− 1 1 0 1 0 0 1  0 1 0 0− 1 1 0 6   1 2 0 0 0 0 1 0   − −    There are no more positive numbers so the minimum is 0 when x1 = x2 = 0, x3 = 1.

3. Consider contradictory constraints, x1 + x2 12 and x1 + 2x2 5. You know these two contradict but show they contradict using≥ the simplex algorithm.≤

You can do this by using artiﬁcial variables, x5. Thus

1 1 1 0 1 0 12 1 2− 0 1 0 0 5  0 0 0 0 1 1 0  −   Do a row operation to preserve the simple columns.

1 1 1 0 1 0 12 1 2− 0 1 0 0 5  1 1 1 0 0 1 12  −   Next start with the 1 in the ﬁrst column. 0 1 1 1 1 0 7 1− 2− 0− 1 0 0 5  0 1 1 1 0 1 7  − − −   Thus the minimum value of z = x5 is 7 but, for there to be a feasible solution, you would need to have this minimum value be 0. 4. Find a solution to the following inequalities for x, y 0 if it is possible to do so. If it is not possible, prove it is not possible. ≥

6x + 3y 4 (a) 8x + 4y ≥ 5 ≤ Use an artiﬁcial variable. Let x1 = x, x2 = y and slack variables x3, x4 with artiﬁcial variable x5. Then minimize x5 as described earlier. 6 3 1 0 1 0 4 8 4− 0 1 0 0 5  0 0 0 0 1 1 0  − Keep the simple columns. 

6 3 1 0 1 0 4 8 4− 0 1 0 0 5  6 3 1 0 0 1 4  −  

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Now proceed to minimize. 3 1 0 0 1 4 1 0 4 8 4− 0− 1 0 0 5  0 0 1 3 0 1 1  − − 4 4   It appears that the minimum value of x5 is 1/4 and so there is no solution to these inequalities with x , x 0. 1 2 ≥ 6x + 4x 11 1 3 ≤ (b) 5x1 + 4x2 + 4x3 8 6x + 6x + 5x ≥11 1 2 3 ≤ The augmented matrix is 6 0 4 1 0 0 11 5 4 4 0 1 0 8  6 6 5 0− 0 1 11    It is not clear whether there is a solution which has all variables nonnegative. However, if you do a row operation using 5 in the first column as a pivot, you get 24 4 6 7 0 5 5 1 5 0 5 5− 4− 4 0 1 0 8  6 1 −6 7  0 5 5 0 5 1 5   and so a solution is x1 = 8/5, x2 = x3 = 0. 6x + 4x 11 1 3 ≤ (c) 5x1 + 4x2 + 4x3 9 6x + 6x + 5x ≥ 9 1 2 3 ≤ The augmented matrix is 6 0 4 1 0 0 11 5 4 4 0 1 0 9  6 6 5 0− 0 1 9    Lets include an artificial variable and seek to minimize x7. 6 0 4 1 0 0 0 0 11 5 4 4 0 1 0 1 0 9  6 6 5 0− 0 1 0 0 9   0 0 0 0 0 0 1 1 0   −    Preserving the simple columns, 6 0 4 1 0 0 0 0 11 5 4 4 0 1 0 1 0 9  6 6 5 0− 0 1 0 0 9   5 4 4 0 1 0 0 1 9   −    Use the first column. 0 6 1 1 0 1 0 0 2 − −1 −5 3 0 1 6 0 1 6 1 0 2  6− 6− 5 0− 0− 1 0 0 9   0 1 1 0 1 5 0 1 3   − − 6 − − 6 2    It appears that the minimum value for x7 is 3/2 and so there will be no solution to these inequalities for which all the variables are nonnegative.

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x x + x 2 1 − 2 3 ≤ (d) x1 + 2x2 4 3x + 2x ≥ 7 1 3 ≤ The augmented matrix is

1 1 1 1 0 0 2 1− 2 0 0 1 0 4  3 0 2 0− 0 1 7    Lets add in an artiﬁcial variable and set things up to minimize this artiﬁcial variable. 1 1 1 1 0 0 0 0 2 1− 2 0 0 1 0 1 0 4  3 0 2 0− 0 1 0 0 7   0 0 0 0 0 0 1 1 0   −  Then   1 1 1 1 0 0 0 0 2 1− 2 0 0 1 0 1 0 4  3 0 2 0− 0 1 0 0 7   1 2 0 0 1 0 0 1 4   −  Work with second column. 

3 1 1 2 0 1 1 2 0 2 0 4 1 2 0 0 −1 0 1 0 4  3 0 2 0− 0 1 0 0 7   0 0 0 0 0 0 1 1 0   −    It appears the minimum value of x7 is 0 and so this will mean there is a solution when x2 = 2, x3 = 0, x1 = 0. 5x 2x + 4x 1 1 − 2 3 ≤ (e) 6x1 3x2 + 5x3 2 5x − 2x + 4x ≥ 5 1 − 2 3 ≤ The augmented matrix is

5 2 4 1 0 0 1 6 −3 5 0 1 0 2  5 −2 4 0− 0 1 5  −   lets introduce an artiﬁcial variable x7 and then minimize x7.

5 2 4 1 0 0 0 0 1 6 −3 5 0 1 0 1 0 2  5 −2 4 0− 0 1 0 0 5  −  0 0 0 0 0 0 1 1 0   −    Then, preserving the simple columns,

5 2 4 1 0 0 0 0 1 6 −3 5 0 1 0 1 0 2  5 −2 4 0− 0 1 0 0 5  −  6 3 5 0 1 0 0 1 2   − −   

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work with the ﬁrst column. 5 2 4 1 0 0 0 0 1 −3 1 6 4 0 5 5 5 1 0 1 0 5  0− 0 0 −1− 0 1 0 0 4  −  0 3 1 6 1 0 0 1 4   − 5 5 − 5 − 5    There is still a positive entry.

5 2 4 1 0 0 0 0 1 1 −1 5 3 4 2 0 4 1 0 1 0 4  −0− 0 0 −1− 0 1 0 0 4  −  1 1 0 5 1 0 0 1 3   − 4 − 2 − 4 − 4    It appears that there is no solution to this system of inequalities because the minimum value of x7 is not 0.

5. Minimize z = x1 + x2 subject to x1 + x2 2, x1 + 3x2 20, x1 + x2 18. Change to a maximization problem and solve as follows:≥ Let y =≤M x . Formulate≤ in terms i − i of y1, y2. You could ﬁnd the maximum of 2M x x for the given constraints and this would − 1 − 2 happen when x1 +x2 is as small as possible. Thus you would maximize y1 +y2 subject to the constraints

M y + M y 2 − 1 − 2 ≥ M y + 3 (M y ) 20 − 1 − 2 ≤ M y + M y 18 − 1 − 2 ≤ To simplify, this would be

2M 2 y + y − ≥ 1 2 4M 20 y + 3y − ≤ 1 2 2M 18 y + y − ≤ 1 2

You could simply regard M as large enough that yi 0 and use the techniques just developed. The augmented matrix for the constraints≥ is then

1 1 1 0 0 2M 2 1 3 0 1 0 4M −20  1 1 0− 0 1 2M − 18  − −   Here M is large. Use the 3 as a pivot to zero out above and below it.

2 1 2 14 3 0 1 3 0 3 M + 3 1 3 0 1 0 4M 20  2 0 0 −1 1 2 M − 34  3 3 − 3 − 3 Then it is still the case that the basic solution is not feasible. Lets use the bottom row and pick the 2/3 as a pivot.

0 0 1 0 1 16 3 3 0 3 0 2 2 3M 3  2 0 0 −1 1 2 M − 34  3 3 − 3 − 3  

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Now it appears that the basic solution is feasible provided M is large. Then assemble the simplex tableau.

0 0 1 0 1 0 16 0 3 0 3 3 0 3M 3  2 2  2 0 0 −1 1 0 2 M − 34 3 3 − 3 − 3  1 1 0 0 0 1 0   − −    Do row operations to preserve the simple columns.

0 0 1 0 1 0 16 3 3 0 3 0 2 2 0 3M 3  2 0 0 −1 1 0 2 M − 34  3 3 − 3 − 3  0 0 0 0 1 1 2M 18   − −    There is a negative number to the left of the 1 on the bottom row and we want to maximize so work with this column. Assume M is very large. Then the pivot should be the top entry in this column.

0 0 1 0 1 0 16 3 3 0 3 2 2 0 0 3M 27  2 − −1 2 − 14  3 0 1 3 0 0 3 M + 3  0 0 1 0 0 1 2M 2   −    It follows that the maximum of y1 +y2 is 2M 2 and it happens when y1 = M +7, y2 = M 9, y = 0. Thus the minimum of x + x−is − 3 1 2 M y + M y = M (M + 7) + M (M 9) = 2 − 1 − 2 − − − F.23 Exercises

7.3

1. If A is the matrix of a linear transformation which rotates all vectors in R2 through 30◦, explain why A cannot have any real eigenvalues. Because the vectors which result are not parallel to the vector you begin with. 2. If A is an n n matrix and c is a nonzero constant, compare the eigenvalues of A and cA. × λ cλ → 1 3. If A is an invertible n n matrix, compare the eigenvalues of A and A− . More generally, for m an arbitrary× integer, compare the eigenvalues of A and Am. 1 m λ λ− and λ λ . → → 4. Let A, B be invertible n n matrices which commute. That is, AB = BA. Suppose x is an eigenvector of B.× Show that then Ax must also be an eigenvector for B. Say Bx = λx. Then B (Ax) = A (Bx) = Aλx = λ (Ax)

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5. Suppose A is an n n matrix and it satisﬁes Am = A for some m a positive integer larger than 1. Show× that if λ is an eigenvalue of A then λ equals either 0 or 1. | | Let x be the eigenvector. Then Amx = λmx,Amx = Ax = λx and so λm = λ Hence if λ = 0, then 6 m 1 λ − = 1 and so λ = 1. | | 6. Show that if Ax = λx and Ay = λy, then whenever a, b are scalars, A (ax + by) = λ (ax + by) . Does this imply that ax + by is an eigenvector? Explain. The formula is obvious from properties of matrix multiplications. However, this vector might not be an eigenvector because it might equal 0 and

Eigenvectors are never 0

7. Find the eigenvalues and eigenvectors of the matrix 1 1 7 − − 1 0 4 .  −1 1 5  − −   Determine whether the matrix is defective. 1 1 7 3 2 −1− 0 4 , eigenvectors: 1 1, 1 2. This is a defective matrix.      −1 1 5  1 ↔ 1 ↔ − −       8. Find the eigenvalues and eigenvectors  of the matrix  3 7 19 − − 2 1 8 .  −2 −3 10  − −   Determine whether the matrix is defective. 3 7 19 3 1 2 −2 −1 8 , eigenvectors: 1 1, 2 2, 1 3        −2 −3 10   1  ↔  1  ↔  1  ↔ − −       This matrix has distinct eigenvalues so it is not defective.          9. Find the eigenvalues and eigenvectors of the matrix 7 12 30 − − 3 7 15 .  −3 −6 14  − −   Determine whether the matrix is defective. 7 12 30 2 5 2 −3 − 7 15 , eigenvectors: −1 , 0 1, 1 2      −3 −6 14   0   1  ↔ −  1  ↔ − −     This matrix is not defective because, even though  λ = 1 is a repeated eigenvalue, it     has a 2 dimensional eigenspace.

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10. Find the eigenvalues and eigenvectors of the matrix 7 2 0 8 −1 0 .  2− 4 6  −   Determine whether the matrix is defective. 7 2 0 1 0 − 2 8 1 0 , eigenvectors: −1 3, 0 6      2− 4 6   −1  ↔  1  ↔ −     This matrix is defective.         11. Find the eigenvalues and eigenvectors of the matrix

3 2 1 0− 5− 1 .  0 2 4    Determine whether the matrix is defective. 3 2 1 1 0 1 − − − 0 5 1 , eigenvectors: 0 , 1 3, 1 6      − 2  ↔   ↔ 0 2 4  0 1   1  This matrix is not defective.           12. Find the eigenvalues and eigenvectors of the matrix 6 8 23 4 5 −16  3 4 −12  −   Determine whether the matrix is defective. 13. Find the eigenvalues and eigenvectors of the matrix 5 2 5 − 12 3 10 .  12 4 −11  −   Determine whether the matrix is defective. 1 5 5 2 5 3 6 12 3 −10 , eigenvectors: −1 , 0 1    12 4 −11   0   1  ↔ − −   This matrix is defective. In this case, there is only one eigenvalue, 1 of multiplicity   3 but the dimension of the eigenspace is only 2. − 14. Find the eigenvalues and eigenvectors of the matrix

20 9 18 6 5 − 6 .  30 14 −27  −   Determine whether the matrix is defective. 20 9 18 6 5 − 6 , eigenvectors:  30 14 −27  −  

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3 1 9 4 2 13 1 1, 1 2, 3 3  4  ↔ −   ↔  13  ↔ −  1   1   1  Not defective.           15. Find the eigenvalues and eigenvectors of the matrix

1 26 17 4 4− 4 .  9 −18 9  − −   Determine whether the matrix is defective. 16. Find the eigenvalues and eigenvectors of the matrix

3 1 2 − − 11 3 9 .  8 0 −6  −   Determine whether the matrix is defective. 3 3 1 2 4 11− 3 −9 , eigenvectors: 1 0 This one is defective.  −   4  8 0 6 1 ↔ −       17. Find the eigenvalues and eigenvectors of the matrix 2 1 2 −11 2 9 .  − 8− 0 7  −   Determine whether the matrix is defective. 3 2 1 2 4 −11 2 9 , eigenvectors: 1 1  4   − 8− 0 7   1  ↔ −   This is defective.     18. Find the eigenvalues and eigenvectors of the matrix

2 1 1 2 3 −2 .  2 2 −1  −   Determine whether the matrix is defective. 2 1 1 1 1 1 − − 2 2 3 2 , eigenvectors: 1 , 0 1, 1 2      2 2 −1   0   1  ↔  1  ↔ −     This is non defective.           19. Find the complex eigenvalues and eigenvectors of the matrix

4 2 2 − − 0 2 2 .  2 0− 2   

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4 2 2 − − 0 2 2 , eigenvectors:  2 0− 2   1  i i 1 4, −i 2 2i, i 2 + 2i  −  ↔  −  ↔ −   ↔  1   1   1        20. Find the eigenvalues  and eigenvectors of the matrix 9 6 3 − 0 6 0 .  3 6 9  − −   Determine whether the matrix is defective. 9 6 3 2 1 1 − − − 0 6 0 , eigenvectors: 1 , 0 6, 0 12      3 6 9   0   1  ↔  1  ↔ − −     This is nondefective.           4 2 2 21. Find the complex eigenvalues and eigenvectors of the matrix 0− 2 −2 . De-  2 0− 2  termine whether the matrix is defective.   4 2 2 − − 0 2 2 , eigenvectors:  2 0− 2   1  i i 1 4, −i 2 2i, i 2 + 2i  −  ↔  −  ↔ −   ↔  1   1   1              4 2 0 − 22. Find the complex eigenvalues and eigenvectors of the matrix 2 4 0 .  2− 2 2  − − Determine whether the matrix is defective.   4 2 0 − 2 4 0 , eigenvectors:  2− 2 2  − −  0 1  1 0 , 1 2, 1 6     ↔ −  −  ↔ −  1 0   1  This is not defective.       1 1 6 − 23. Find the complex eigenvalues and eigenvectors of the matrix 7 5 6 .  1− 7− 2  − Determine whether the matrix is defective.   1 1 6 7 5 −6 , eigenvectors:  1− 7− 2  −  1  i i 1 6, −i 2 6i, i 2 + 6i  −  ↔ −  −  ↔ −   ↔  1   1   1             

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This is not defective. 4 2 0 24. Find the complex eigenvalues and eigenvectors of the matrix 2 4 0 . Deter-  −2 2 6  − mine whether the matrix is defective.   4 2 0 2 4 0 , eigenvectors:  −2 2 6  −  0  1 1 0 6, i 4 2i, i 4 + 2i   ↔  −  ↔ −   ↔  1   1   1  This is not defective.          25. Here is a matrix. 1 a 0 0 0 1 b 0  0 0 2 c   0 0 0 2    Find values of a, b, c for which the matrix is defective and values of a, b, c for which it is nondefective. First consider the eigenvalue λ = 1

0 a 0 0 0 0 0 b 0 0  0 0 1 c 0   0 0 0 1 0      Then you have ax2 = 0, bx3 = 0. If neither a nor b = 0 then λ = 1 would be a defective eigenvalue and the matrix would be defective. If a = 0, then the dimension of the eigenspace is clearly 2 and so the matrix would be nondefective. If b = 0 but a = 0, then you would have a defective matrix because the eigenspace would have dimension6 less than 2. If c = 0, then the matrix is defective. If c = 0 and a = 0, then it is non defective. Basically,6 if a, c = 0, then the matrix is defective. 6 26. Here is a matrix. a 1 0 0 b 1  0 0 c  where a, b, c are numbers. Show this is sometimes defective depending on the choice of a, b, c. What is an easy case which will ensure it is not defective? An easy case which will ensure that it is not defective is for a, b, c to be distinct. If you have any repeats, then this will be defective. 27. Suppose A is an n n matrix consisting entirely of real entries but a + ib is a complex eigenvalue having× the eigenvector, x + iy. Here x and y are real vectors. Show that then a ib is also an eigenvalue with the eigenvector, x iy. Hint: You should remember− that the conjugate of a product of complex numbers− equals the product of the conjugates. Here a + ib is a complex number whose conjugate equals a ib. − A (x + iy) = (a + ib)(x + iy) . Now just take complex conjugates of both sides.

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28. Recall an n n matrix is said to be symmetric if it has all real entries and if A = AT . Show the eigenvalues× of a real symmetric matrix are real and for each eigenvalue, it has a real eigenvector. Let λ be an eigenvalue which corresponds to x = 0. Then λxT = xT AT . Then 6 λxT ¯x = xT AT ¯x = xT A¯x = xT ¯xλ¯

which shows that λ¯ = λ. We have A¯x = λ¯x and Ax = λx so

A (¯x + x) = λ (¯x + x)

Hence it has a real eigenvector. 29. Recall an n n matrix is said to be skew symmetric if it has all real entries and if A = AT . Show× that any nonzero eigenvalues must be of the form ib where i2 = 1. In words,− the eigenvalues are either 0 or pure imaginary. − Let A be skew symmetric. Then if x is an eigenvector for λ,

λxT ¯x = xT AT ¯x = xT A¯x = xT ¯xλ¯ − − and so λ = λ¯. Thus a + ib = (a ib) and so a = 0. − − − 30. Is it possible for a nonzero matrix to have only 0 as an eigenvalue? 12 16 Sure. Try this one. −9− 12 31. Show that the eigenvalues and eigenvectors of a real matrix occur in conjugate pairs. This follows from the observation that if Ax = λx, then Ax = λx 32. Suppose A is an n n matrix having all real eigenvalues which are distinct. Show × 1 there exists S such that S− AS = D, a diagonal matrix. If

λ1 0 . D =  ..  0 λ  n    define eD by eλ1 0 eD .. ≡  .  0 eλn   and define   A D 1 e Se S− . ≡ Next show that if A is as just described, so is tA where t is a real number and the eigenvalues of At are tλk. If you differentiate a matrix of functions entry by entry so th th that for the ij entry of A′ (t) you get aij′ (t) where aij (t) is the ij entry of A (t) , show d eAt = AeAt dt Next show det eAt = 0. This is called the matrix exponential. Note I have only defined it for the case6 where the eigenvalues of A are real, but the same procedure will work even for complex eigenvalues. All you have to do is to define what is meant by

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ea+ib. It is clear that the eigenvalues of tA are tλ where λ is an eigenvalue of A. Thus eAt is of the form etλ1 0 . 1 S  ..  S− 0 etλn   And so  

tλ1 λ1e 0 d At . 1 e = S . S− dt  .  0 λ etλn  n   tλ1 λ1 0 e 0 . . 1 = S  ..   ..  S− 0 λ 0 etλn  n       

tλ1 λ1 0 e 0 . 1 . 1 = S  ..  S− S  ..  S− 0 λ 0 etλn  n    = AetA   

7 1 1 12 4 6 1 −7 1 33. Find the principle directions determined by the matrix 4 12 6 . The  −1 1 −2  6 6 3 1 1 − eigenvalues are 3 , 1, and 2 listed according to multiplicity.  1 1 2 1 −1 1 1 1 , 1 , 2 , 2 , 1 , 3  −1    1    0            34. Find the principle directions determined by the matrix 5 1 1 3 3 3 1 −7 −1 3 6 6 The eigenvalues are 1, 2, and 1. What is the physical interpreta-  − 1 1 7  − 3 6 6 tion of the repeated eigenvalue? There are apparently two directions in which there is no stretching. 1 1 2 2 2 − 1 , 0 1, 1 2  0   1  ↔  1  ↔       35. Find oscillatory solutions to the system of diﬀerential equations, x′′ = Ax where A = 3 1 1 − − − 1 2 0 The eigenvalues are 1, 4, and 2.  −1− 0 2  − − − − −  

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The eigenvalues are given and so there are oscillatory solutions of the form

1 −1 (a cos (t) + b sin (t)) ,  1   0  1 c sin √2t + d cos √2t  −  1  2  1 (e cos (2t) + f sin (2t))  1    where a,b,c,d,e,f are scalars. 36. Let A and B be n n matrices and let the columns of B be × b , , b 1 ··· n and the rows of A are aT , , aT . 1 ··· n Show the columns of AB are Ab Ab 1 ··· n and the rows of AB are aT B aT B. 1 ··· n th th T The i entry of the j column of AB is ai bj . Therefore, these columns are as th T T indicated. Also the i row of AB will be ai b1 ai bn . Thus this row is just aT B. ··· i 37. Let M be an n n matrix. Then deﬁne the adjoint of M, denoted by M ∗ to be the transpose of the× conjugate of M. For example,

2 i ∗ 2 1 i = . 1 + i 3 i −3 −

A matrix M, is self adjoint if M ∗ = M. Show the eigenvalues of a self adjoint matrix are all real.

First note that (AB)∗ = B∗A∗. Say Mx = λx, x = 0. Then 6 2 λ x = λx∗x = (λx)∗ x = (Mx)∗ x = x∗M ∗x | | 2 = x∗Mx = x∗λx = λ x | | Hence λ = λ.

38. Let M be an n n matrix and suppose x1, , xn are n eigenvectors which form a linearly independent× set. Form the matrix S···by making the columns these vectors. 1 1 Show that S− exists and that S− MS is a diagonal matrix (one having zeros everywhere except on the main diagonal) having the eigenvalues of M on the main diagonal. When this can be done the matrix is said to be diagonalizable.

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Since the vectors are linearly independent, the matrix S has an inverse. Denoting this inverse by T w1 1 . S− =  .  wT  n  it follows by deﬁnition that   T wi xj = δij . Therefore,

T w1 1 1 . S− MS = S− (Mx , ,Mx ) = . (λ x , , λ x ) 1 ··· n  .  1 1 ··· n n wT  n    λ1 0 . =  ..  0 λ  n    39. Show that a n n matrix M is diagonalizable if and only if Fn has a basis of eigenvectors. Hint: The× ﬁrst part is done in Problem 38. It only remains to show that if the 1 matrix can be diagonalized by some matrix S giving D = S− MS for D a diagonal matrix, then it has a basis of eigenvectors. Try using the columns of the matrix S. The formula says that MS = SD th Letting xk denote the k column of S, it follows from the way we multiply matrices that Mxk = λkxk th where λk is the k diagonal entry on D. 40. Let 1 2 2 3 4 0 A =   0 1 3   and let   0 1 1 1 B =   2 1     1 2 Multiply AB verifying the block multiplication formula. Here A = ,A = 11 3 4 12 2 ,A = 0 1 and A = (3) . 0 21 22 1 2 2 0 1 6 5 AB = 3 4 0 1 1 = 4 7  0 1 3   2 1   7 4       

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Now consider doing it by block multiplication. This leads to

1 2 0 1 2 + 2 1 3 4 1 1 0   0 1 0 1 + 3 2 1  1 1     6 5 6 5  = 4 7 = 4 7  7 4   7 4      41. Suppose A, B are n n matrices and λ is a nonzero eigenvalue of AB. Show that then it is also an eigenvalue× of BA. Hint: Use the deﬁnition of what it means for λ to be an eigenvalue. That is, ABx = λx where x = 0. Maybe you should multiply both sides by B. 6 B (ABx) = λBx Hence (BA)(Bx) = λ (Bx) . Is Bx = 0? No, it can’t be since λ = 0. If Bx = 0, then the left side of the given equation would be 0 and the right wouldn’t6 be. 42. Using the above problem show that if A, B are n n matrices, it is not possible that AB BA = aI for any a = 0. Hint: First show× that if A is a matrix, then the eigenvalues− of A aI are λ 6 a where λ is an eigenvalue of A. − − If it is true that AB BA = aI for a = 0, Then you get AB aI = BA. If − 6 − λ1, , λn are the eigenvalues of AB, then the eigenvalues of AB aI = BA are {λ ···a, }, λ a which cannot be the same set, contrary to the above− problem. { 1 − ··· n − } 43. Consider the following matrix.

0 0 a ··· − 0 1 0 a1 C =  . . −.  .. .. .    0 1 an 1   − −    n 1 n Show det (λI C) = a0 + λa1 + an 1λ − + λ . This matrix is called a companion matrix for the− given polynomial.··· − λI C is of the form − λ 0 a ··· 0 1 λ a1  − . . .  .. .. .    0 1 λ + an 1   − −  Expand along the bottom row. 

λ 0 1 λ  − . .  (λ + an 1) det .. .. −    1 λ   −   0 1 λ   −   

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λ 0 0 a ··· 0 1 λ a1  − 1 λ a  +1 det − 2  . . .   .. .. .     0 1 an 2   − −  That last determinant is of the form  λ 0 0 a ··· 0 1 λ a1  − 1 λ a  det − 2  . . .   .. .. .     0 1 λ + (an 2 λ)   − − −    By induction, this is of the form

n 1 n 2 λ − + (an 2 λ) λ − + + a1λ + a0 − − ··· and so, the sum of the two is of the form

n 1 n 1 n 2 λ − (λ + an 1) + λ − + (an 2 λ) λ − + + a1λ + a0 − − n n 1 n 2 − ··· = λ + an 1λ − + an 2λ − + + a1λ + a0 − − ··· It is routine to verify that when n = 2 this determinant gives the right polynomial. 44. A discreet dynamical system is of the form

x (k + 1) = Ax (k) , x (0) = x0

where A is an n n matrix and x (k) is a vector in Rn. Show ﬁrst that × k x (k) = A x0

for all k 1. If A is nondefective so that it has a basis of eigenvectors, v1, , vn where ≥ { ··· } Avj = λj vj

you can write the initial condition x0 in a unique way as a linear combination of these eigenvectors. Thus n x0 = aj vj j=1 X Now explain why n n k k x (k) = aj A vj = aj λj vj j=1 j=1 X X which gives a formula for x (k) , the solution of the dynamical system. The ﬁrst formula is obvious from induction. Thus

n n k k k A x0 = aj A vj = ajλj vj j=1 j=1 X X because if Av = λv, then if Akx = λkx, do A to both sides. Thus Ak+1x = λk+1x.

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45. Suppose A is an n n matrix and let v be an eigenvector such that Av = λv. Also suppose the characteristic× polynomial of A is

n n 1 det (λI A) = λ + an 1λ − + + a1λ + a0 − − ··· Explain why n n 1 A + an 1A − + + a1A + a0I v = 0 − ··· If A is nondefective, give a very easy proof of the Cayley Hamilton theorem based on this. Recall this theorem says A satisﬁes its characteristic equation,

n n 1 A + an 1A − + + a1A + a0I = 0. − ··· Let v be any vector in a basis of eigenvectors of A. Since A is nondefective, such a basis exists. Then

n n 1 n n 1 A + an 1A − + + a1A + a0I v = λ + an 1λ − + + a1λ + a0 v = 0 − ··· − ··· because λ is by deﬁnition a solution to the characteristic polynomial. Since the above holds for vectors in a basis, it holds for any v and so

n n 1 A + an 1A − + + a1A + a0I = 0 − ··· 46. Suppose an n n nondefective matrix A has only 1 and 1 as eigenvalues. Find A12. × − Let x be some vector. Let x = j aj vj where vj is an eigenvector. Then

P n n 12 12 A x = aj λj vj = ajvj = x j=1 j=1 X X It follows that A12 = I. 47. Suppose the characteristic polynomial of an n n matrix A is 1 λn. Find Amn where m is an integer. Hint: Note ﬁrst that A is nondefective.× Why?− The eigenvalues are distinct because they are the nth roots of 1. Hence from the above formula, if x is a given vector with

n x = aj vj j=1 X then n n n nm nm nm A x = A aj vj = aj A vj = aj vj = x j=1 j=1 j=1 X X X so Anm = I. 48. Sometimes sequences come in terms of a recursion formula. An example is the Fi- bonacci sequence. x0 = 1 = x1, xn+1 = xn + xn 1 − Show this can be considered as a discreet dynamical system as follows.

x 1 1 x x 1 n+1 = n , 1 = xn 1 0 xn 1 x0 1 −

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Now use the technique of Problem 44 to ﬁnd a formula for xn. What are the eigenvalues and eigenvectors of this matrix? 1 1 , eigenvectors: 1 0 1 1 √5 1 1 1 √5 + 1 1 1 2 − 2 √5, 2 2 √5 + 1 ↔ 2 − 2 1 ↔ 2 2 Now also 1 1 1 1 √5 1 1 1 √5 + 1 1 √5 2 − 2 + √5 + 2 2 = 2 − 10 1 10 2 1 1 Therefore, the solution is of the form

x n+1 = x n

n 1 1 1 1 1 1 √5 √5 √5 2 − 2 2 − 10 2 − 2 1 n 1 1 1 1 1 √5 + 1 + √5 + √5 + 2 2 10 2 2 2 1 In particular,

1 1 1 1 n 1 1 1 1 n x = √5 √5 + √5 + √5 + n 2 − 10 2 − 2 10 2 2 2 49. Let A be an n n matrix having characteristic polynomial × n n 1 det (λI A) = λ + an 1λ − + + a1λ + a0 − − ··· Show that a = ( 1)n det (A). 0 − The characteristic polynomial equals det (λI A) . To get the constant term, you plug in λ = 0 and obtain det ( A) = ( 1)n det (A−). − − F.24 Exercises

7.10

1. Explain why it is typically impossible to compute the upper triangular matrix whose existence is guaranteed by Schur’s theorem. To get it, you must be able to get the eigenvalues and this is typically not possible. 2. Now recall the QR factorization of Theorem 5.7.5 on Page 133. The QR algorithm is a technique which does compute the upper triangular matrix in Schur’s theorem. There is much more to the QR algorithm than will be presented here. In fact, what I am about to show you is not the way it is done in practice. One ﬁrst obtains what is called a Hessenburg matrix for which the algorithm will work better. However, the idea is as follows. Start with A an n n matrix having real eigenvalues. Form A = QR where Q is orthogonal and R is upper× triangular. (Right triangular.) This

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can be done using the technique of Theorem 5.7.5 using Householder matrices. Next take A RQ. Show that A = QA QT . In other words these two matrices, A, A are 1 ≡ 1 1 similar. Explain why they have the same eigenvalues. Continue by letting A1 play the role of A. Thus the algorithm is of the form An = QRn and An+1 = Rn+1Q. Explain T why A = QnAnQn for some Qn orthogonal. Thus An is a sequence of matrices each similar to A. The remarkable thing is that often these matrices converge to an upper T triangular matrix T and A = QTQ for some orthogonal matrix, the limit of the Qn where the limit means the entries converge. Then the process computes the upper triangular Schur form of the matrix A. Thus the eigenvalues of A appear on the diagonal of T. You will see approximately what these are as the process continues.

Suppose you have found An. Then An = QnRn and then

T T T An+1 RnQn = Qn AnQn = Qn Qn 1An 1Qn 1Qn ≡ − − − ··· Now the product of orthogonal matrices is orthogonal and so this shows by induction that each An is orthogonally similar to A. 3. Try the QR algorithm on 1 2 −6− 6 which has eigenvalues 3 and 2. I suggest you use a computer algebra system to do the computations. 1 2 1 √37 6 √37 √37 38 √37 − − = − 37 − 37 37 6 6 6 √37 1 √37 0 6 √37 37 − 37 37 √ 38 √ 1 √ 6 √ 191 260 37 37 37 37 37 37 37 37 37 A1 = − − = − 0 6 √37 6 √37 1 √37 36 6 37 37 − 37 37 − 37 191 260 37 37 36 − 6 = 37 − 37 191 √37√1021 36 √37√1021 1 √37√1021 1348 √37√1021 37 777 − 37 777 37 − 37 777 36 √37√1021 191 √37√1021 0 6 √37√1021 37 777 37 777 1021 3959 7952 1021 1021 A2 = 216 −1146 1021 1021 3959 7952 1021 1021 −1146 = 0.211 56 1021 2 0.998 51 5. 447 9 10− 3. 883 3 7. 715 7 2 − × − 5. 447 9 10− 0.998 51 0 1. 545 1 × 2 3. 883 3 7. 715 7 0.998 51 5. 447 9 10− A3 = − 2 − × = 0 1. 545 1 5. 447 9 10− 0.998 51 × 3. 457 2 7. 915 8 2 − 8. 417 6 10− 1. 542 8 × You can keep on going. You see that it appears to be converging to an upper triangular matrix having 3, 1 down the diagonal. 4. Now try the QR algorithm on 0 1 2− 0 Show that the algorithm cannot converge for this example. Hint: Try a few iterations of the algorithm.

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0 1 0 1 2 0 = 2− 0 1− 0 0 1 2 0 0 1 0 2 A = = 1 0 1 1− 0 1− 0 0 2 0 1 1 0 = 1− 0 1− 0 0 2 1 0 0 1 0 1 A = = . Now it is back to where you started. 2 0 2 1− 0 2− 0 0 1 0 2 Thus the algorithm merely bounces between the two matrices and 2− 0 1− 0 and so it can’t possibly converge. 0 1 0 2 5. Show the two matrices A − and B − are similar; that is ≡ 4 0 ≡ 2 0 1 there exists a matrix S such that A = S− BS but there is no orthogonal matrix Q such that QT BQ = A. Show the QR algorithm does converge for the matrix B although it fails to do so for A. Both matrices have distinct eigenvalues 2 and so they are both similar to a diagonal matrix having these eigenvalues on the± diagonal. Therefore, the matrices are indeed similar. Lets try the QR algorithm on the second. 0 2 0 1 2 0 = 2− 0 1− 0 0 2 2 0 0 1 0 2 A = = and so the algorithm keeps on returning 1 0 2 1− 0 2− 0 the same matrix in the case of B. Now consider the matrix A. 0 1 0 1 4 0 = 4− 0 1− 0 0 1 4 0 0 1 0 4 A = = 1 0 1 1− 0 1− 0 0 4 0 1 1 0 = 1− 0 1− 0 0 4 1 0 0 1 0 1 A = = 2 0 4 1− 0 4− 0 0 1 0 4 In this case, the algorithm bounces between and . 4− 0 1− 0 6. Let F be an m n matrix. Show that F ∗F has all real eigenvalues and furthermore, they are all nonnegative.×

The eigenvalues are real because the matrix F ∗F is Hermitian. If λ is one of them with eigenvector x, λ (x, x) = (F ∗F x, x) = (F x,F x) 0 ≥ 7. If A is a real n n matrix and λ is a complex eigenvalue of A having eigenvector z + iw, show that× w = 0. 6 A (z + iw) = (a + ib)(z + iw) A (z iw) = (a ib)(z iw) − − −

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Now subtract to obtain A (2iw) = 2iaw + 2ibz. Now if w = 0, then you would have 0 =ibz and this requires b = 0 since otherwise, the eigenvector was equal to 0 which it isn’t. Hence λ was not complex after all. 8. Suppose A = QT DQ where Q is an orthogonal matrix and all the matrices are real. Also D is a diagonal matrix. Show that A must be symmetric. AT = QT DT Q = QT DQ = A. 9. Suppose A is an n n matrix and there exists a unitary matrix U such that ×

A = U ∗DU

where D is a diagonal matrix. Explain why A must be normal.

A∗A = U ∗D∗UU ∗DU = U ∗D∗DU

AA∗ = U ∗DUU ∗D∗U = U ∗DD∗U. But D∗D = DD∗ and both are equal to the diagonal matrix which has the squares of the absolute values of the diagonal entries of D down the diagonal. 10. If A is Hermitian, show that det (A) must be real.

A = U ∗DU where D is a real diagonal matrix. Therefore, 1 det (A) = det U − det (U) det (D) = det (D) R. ∈ 11. Show that every unitary matrix preserves distance. That is, if U is unitary,

Ux = x . | | | |

2 2 Ux = (Ux,Ux) = (U ∗Ux, x) = (x, x) = x . | | | | 12. Show that if a matrix does preserve distances, then it must be unitary. Let U preserve distances. Then

Re (Ux,Uy) Re (x, y) = 0 − Re (U ∗Ux x, y) = 0 −

Let θ = 1 and θ (U ∗Ux x, y) = (U ∗Ux x, y) . Then replace y in the above with ¯θy. Then| | you get − | − |

Re U ∗Ux x,θ¯y = Re θ (U ∗Ux x, y) = (U ∗Ux x, y) = 0 − − | − | and since y is arbitrary, (U ∗Ux x, y) = 0 − for all y. In particular, this holds for y =U ∗Ux x and so U ∗U = I. −

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13. Show that a complex normal matrix A is unitary if and only if its eigenvalues have magnitude equal to 1. Since A is normal, it follows that there is a unitary matrix U such that

AU = UD

where D is a diagonal matrix. Then from the above problems, A is unitary if and only if it preserves distances. Let the columns of U be the orthonormal set u1, , un . Then we have Au = λ u . For A to preserve distances, you must have{ ···λ =} 1. k k k | k| Conversely, if λk = 1 for all k, then λkuk is also an orthonormal set and so A preserves distances| | and is therefore normal.{ } 14. Suppose A is an n n matrix which is diagonally dominant. Recall this means × a < a | ij| | ii| j=i X6 1 show A− must exist. This follows from Gerschgorin’s theorem. The condition implies that 0 is not an eigenvalue and so the matrix is one to one and hence invertible. 15. Give some disks in the complex plane whose union contains all the eigenvalues of the matrix 1 + 2i 4 2 0 i 3  5 6 7    B (1 + 2i, 6) ,B (i, 3) ,B (7, 11) 16. Show a square matrix is invertible if and only if it has no zero eigenvalues. To say the matrix has no zero eigenvalues is to say that it is one to one because it maps no nonzero vector to 0. 17. Using Schur’s theorem, show the trace of an n n matrix equals the sum of the eigenvalues and the determinant of an n n matrix× is the product of the eigenvalues. × Let A be an n n matrix. By Schur, there exists a unitary matrix U and an upper triangular matrix× T such that A = U ∗TU Then det (A) = det (T ) = the product of the diagonal entries of T. These are the eigenvalues of T and both A and T have the same eigenvalues because they have the same characteristic polynomial. Thus the determinant of A equals the product of the eigenvalues. Since A and T are similar, they have the same trace also. However, the trace of T is just the sum of the eigenvalues of A. 18. Using Schur’s theorem, show that if A is any complex n n matrix having eigenvalues 2 ×n 2 λi listed according to multiplicity, then i,j Aij i=1 λi . Show that equality holds{ } if and only if A is normal. | | ≥ | | P P By Schur’s theorem, there is a unitary U and upper triangular T such that U ∗AU = T. Then 2 A = trace (AA∗) = trace (UTU ∗UT ∗U ∗) | ij | i,j X

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n 2 = trace (UTT ∗U ∗) = trace (TT ∗) λ ≥ | i| i=1 X Equality holds if and only if there are no nonzero oﬀ diagonal terms. This happens if and only if T is a diagonal matrix. But if this is so, then A must be normal because U ∗AU = D, a diagonal matrix and this implies that A is normal. 19. Here is a matrix. 1234 6 5 3 0 654 9 123  98− 123 10, 000 11   56 78 98 400    I know this matrix has an inverse before doing any computations. How do I know? Gerschgorin’s theorem shows that there are no zero eigenvalues and so the matrix is invertible. 20. Show the critical points of the following function are

1 (0, 3, 0) , (2, 3, 0) , and 1, 3, − − − −3 and classify them as local minima, local maxima or saddle points. f (x, y, z) = 3 x4 + 6x3 6x2 + zx2 2zx 2y2 12y 18 3 z2. − 2 − − − − − − 2 0 = 3 x4 + 6x3 6x2 + zx2 2zx 2y2 12y 18 3 z2 . ∇ − 2 − − − − − − 2 2xz 2z 12x + 18x2 6x3 = 0 − − 4y 12 = 0− After much fuss, you ﬁnd the above solutions  x2− 2x− 3z = 0  − − to these systems of equations.  18x2 + 36x + 2z 12 0 2x 2 − − − Hessian is 0 4 0  2x 2− 0 3  − − (0, 3, 0) ,   − 12 0 2 −0 4− 0  2− 0 3  − − The eigenvalues are all negative and so this point is a local maximum. (2, 3, 0) − 12 0 2 −0 4 0  2− 0 3  − The eigenvalues are all negative so this is a local maximum. 1, 3, 1 − − 3 16 0 0 3 0 4 0  0− 0 3  − This is a saddle point.  

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21. Here is a function of three variables.

f (x, y, z) = 13x2 + 2xy + 8xz + 13y2 + 8yz + 10z2

change the variables so that in the new variables there are no mixed terms, terms involving xy, yz etc. Two eigenvalues are 12 and 18. The function is T x 13 1 4 x f (x, y, z) = y 1 13 4 y  z   4 4 10   z  13 1 4      1 13 4 , eigenvectors:  4 4 10  1  2  1 1 − 1 6, −1 12, 1 18  − 2  ↔   ↔   ↔  1   0   1        2 2 2 Then in terms of the new variables, the quadratic form is 6x′ + 12y′ + 18z′ . Note how I only needed to ﬁnd the eigenvalues. This changes in the next problems. 22. Here is a function of three variables.

f (x, y, z) = 2x2 4x + 2 + 9yx 9y 3zx + 3z + 5y2 9zy 7z2 − − − − − change the variables so that in the new variables there are no mixed terms, terms involving xy, yz etc. The function is f (x, y, z) = T x 2 9/2 3/2 x x − y 9/2 5 9/2 y + 4 9 3 y + 2  z   3/2 9/2 − 7   z  − −  z  − − −  2  9/2 3/2      − 9/2 5 9/2 , eigenvectors:  3/2 9/2 − 7  − − −  5 0 2 3 1, 1 17 , −3 19  −  ↔ −  3  ↔ − 2  −  ↔ 2  1   1   1      1√  1 √     7 35 0 7 14 3 √ 1 √ − 3 √ Appropriate orthogonal matrix: Q = 35 35 10 10 14 14  −1 √ 3 √ −1 √  35 35 10 10 14 14 1 0 0  1 0 0  T − 17 − 17 T Q AQ = 0 2 0 ,A = Q 0 2 0 Q Thus you need to have  − 19   − 19  0 0 2 0 0 2 the new variables be    1 √ 1 √ T 7 35 0 7 14 x x′ 3 √ 1 √ − 3 √ 35 35 10 10 14 14 y = y′  −1 √ 3 √ −1 √   z   z  35 35 10 10 14 14 ′       Then in terms of the new variables, the quadratic form is

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T x′ 1 0 0 x′ − 17 y′ 0 2 0 y′ +    − 19    z′ 0 0 2 z′    1 √    1 √ 7 35 0 7 14 x′ 3 √ 1 √ − 3 √ 4 9 3 35 35 10 10 14 14 y′ + 2 − −  −1 3 −1    √35 √10 √14 z′ 35 10 14 2 2  17 2 19 2 19    = (x′) + √5√7x′ (y′) + (z′) + √2√7z′ + 2 − 7 − 2 2 7 23. Here is a function of three variables.

f (x, y, z) = x2 + 2xy + 2xz y2 + 2yz z2 + x − − − change the variables so that in the new variables there are no mixed terms, terms involving xy, yz etc. The function is T x 1 1 1 x x − y 1 1 1 y + 1 0 0 y  z   1− 1 1   z   z  −  1 1 1    1  1 1 −1 1 1 , eigenvectors: 1 1, −1 , −1 2      1− 1 1   1  ↔  0   −2  ↔ − −        1 √ 1 √  1 √    3 3 2 2 6 6  1 √ −1 √ − 1 √ Appropriate orthogonal matrix: Q = 3 3 2 2 6 6 .  1 √ −1 √  3 3 0 3 6 New variables,  

1 √ 1 √ 1 √ T x′ 3 3 2 2 6 6 x 1 √ −1 √ − 1 √ y′ = 3 3 2 2 6 6 y  z   1 √ −1 √   z  ′ 3 3 0 3 6       Quadratic form in the new variables: T x′ 1 0 0 x′ y′ 0 2 0 y′    −    z′ 0 0 2 z′ −    1 √ 1 √ 1 √ 3 3 2 2 6 6 x′ 1 √ −1 √ − 1 √ + 1 0 0 3 3 2 2 6 6 y′  1 −1    √3 0 √6 z′ 3 3 2 1  2 1 2 1  =(x′) + √3x′ 2(y′) √2y′ 2(z′) √6z′ 3 − − 2 − − 6 24. Show the critical points of the function,

f (x, y, z) = 2yx2 6yx 4zx2 12zx + y2 + 2yz. − − − − are points of the form,

(x, y, z) = t, 2t2 + 6t, t2 3t − − for t R and classify them as local minima, local maxima or saddle points. ∈

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6y 12z 4xy 8xz − − − − 2yx2 6yx 4zx2 12zx + y2 + 2yz = 2x2 6x + 2y + 2z ∇ − − − −  − 4x−2 12x + 2y  − − 6y 12z 4xy 8xz = 0   − 2x−2 6x−+ 2y −+ 2z = 0 Let x = t. Then y = 2t2 + 6t. Now place these into the − 4x−2 12x + 2y = 0 − − middle equation and you then ﬁnd that 2t2 6t + 2 2t2 + 6t + 2z = 0, Solution is: z = t2 3t. This also works in the top− equation.− − − 6 2t2 + 6t 12 t2 3t 4t 2t2 + 6t 8t t2 3t = 0 − − − − − − − − 2yx2 6yx 4zx2 12zx + y2 + 2yz, Hessian is − − − − 4y 8z 4x 6 8x 12 − 4x− 6− 2− − 2−  −8x −12 2 0  − − Plug in the critical points.  0 4t 6 8t 12 − − − − 4t 6 2 2 You need to consider the sign of the eigenvalues.  −8t −12 2 0  − −  λ 4t 6 8t 12 det 4−t 6− 2 −λ − 2− = 80t2λ + 240tλ λ3 + 2λ2 + 184λ. Thus you  −8t −12− 2 λ  − − − − need to be looking at the solutions to λ3 2λ2 80t2λ 240tλ 184λ = 0 − − − − You get an eigenvalue which equals 0. Then you need to solve λ2 2λ 184 + 80t2 + 240t = 0 − − 2 3 184 + 80t + 240t has a minimum value when t = 2 and at this point, the value of 3 2 3 − this function is 184 + 80 2 + 240 2 = 4. Therefore, from the quadratic formula, there is a positive eigenvalue and a negative− eigenvalue for any value of t. Thus this function always has a direction in which it appears to have a local min. and a direction in which it appears to have a local max. for each of the critical points. 25. Show the critical points of the function 1 1 f (x, y, z) = x4 4x3 + 8x2 3zx2 + 12zx + 2y2 + 4y + 2 + z2. 2 − − 2 are (0, 1, 0) , (4, 1, 0) , and (2, 1, 12) and classify them as local minima, local maxima− or saddle− points. − − 6x2 24x 6z + 16 0 12 6x Hessian: − 0− 4− 0  12 6x 0 1  − (0, 1, 0)  − 16 0 12 0 4 0  12 0 1  This is a saddle point because it has a negative and two positive eigenvalues. (4, 1, 0) − 22 0 12 0 4− 0  12 0 1  −  

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This is also a saddle point because it has a negative and two positive eigenvalues. (2, 1, 12) − − 64 0 0 0 4 0  0 0 1  This is a local minimum because all eigenvalues are positive. 26. Let f (x, y) = 3x4 24x2 + 48 yx2 + 4y. Find and classify the critical points using the second derivative− test. − 12x3 2xy 48x 3x4 24x2 + 48 yx2 + 4y = − − ∇ − − 4 x2 − Critical points: (2, 0) , ( 2, 0) . − 36x2 2y 48 2x 3x4 24x2 + 48 yx2 + 4y, Hessian is − − − − − 2x 0 − (2, 0) 96 4 4− 0 − Clearly it has eigenvalues of diﬀerent sign because the determinant is negative. There- fore, this is a saddle point. ( 2, 0) works out the same way. To have some fun, graph this function of two variables− and see if you can see the critical points are saddle points from the picture. 27. Let f (x, y) = 3x4 5x2 + 2 y2x2 + y2. Find and classify the critical points using the second derivative test.− − 12x3 2xy2 10x 3x4 5x2 + 2 y2x2 + y2 = − − ∇ − − 2y 2x2y − Critical points: (1, 1) , ( 1, 1), (1, 1) , ( 1, 1) , (0, 0) , 1 √5√6, 0 , 1 √5√6, 0 − − − − − 6 6 36x2 2y2 10 4xy 3x4 5x2 + 2 y2x2 + y2, Hessian is − − − − − 4xy 2 2x2 − − (1, 1) 24 4 4− 0 − Saddle point. The eigenvalues have opposite sign. ( 1, 1) , saddle point also. (1, 1) saddle point ( 1, 1) saddle point. − − − − 1 √5√6, 0 − 6 20 0 0 1 3 1 √ √ Local minimum. 6 5 6, 0 also a local minimum. 28. Let f (x, y) = 5x4 7x2 2 3y2x2 + 11y2 4y4. Find and classify the critical points using the second derivative− − − test. − 20x3 6xy2 14x 5x4 7x2 2 3y2x2 + 11y2 4y4 = − − ∇ − − − − 6x2y 16y3 + 22y − − Critical points (1, 1) , ( 1, 1) , (0, 0) , 1 √7√10, 0 , 1 √7√10, 0 − − 10 10 5x4 7x2 2 3y2x2 + 11y2 4y4, − − − −

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60x2 6y2 14 12xy Hessian is −12xy− 6x2 − 48y2 + 22 − − − (1, 1) 40 12 12 −32 − − The determinant is negative and so this is a saddle point. ( 1, 1) − 40 12 12 32 − The determinant is negative and so this is also a saddle point. (0, 0) 14 0 −0 22 This is a saddle point. 1 √ √ 10 7 10, 0 28 0 0 89 5 This is a local minimum. So is 1 √7√10, 0 . − 10 29. Let f (x, y, z) = 2x4 3yx2 + 3x2 + 5x2z + 3y2 6y + 3 3zy + 3z + z2. Find and classify the critical− points− using the second derivative− test.− 2x4 3yx2 + 3x2 + 5x2z + 3y2 6y + 3 3zy + 3z + z2 ∇ − − − − 6x 6xy + 10xz 8x3 = −3x2 + 6y 3z− 6  −5x2 3y +− 2z +− 3  − Critical points: (0, 1, 0)  24x2 6y + 10z + 6 6x 10x Hessian is − − 6x −6 3 At the point of interest this is  −10x 3− 2  −   54 0 0 −0 6 3  0 3− 2  −   It has two positive eigenvalues and one negative eigenvalue so this is a saddle point. 30. Let f (x, y, z) = 3yx2 3x2 x2z y2 + 2y 1 + 3zy 3z 3z2. Find and classify the critical points using− the second− − derivative− test. − − 3yx2 3x2 x2z y2 + 2y 1 + 3zy 3z 3z2 ∇ − − − − − − 6xy 6x 2xz = 3x2 −2y +− 3z + 2  x2 −+ 3y 6z 3  − − − Critical points: (0, 1, 0) 

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6y 2z 6 6x 2x − − − Hessian is 6x 2 3 At the critical point, this is  2x −3 6  − −   0 0 0 0 2 3  0− 3 6  −   eigenvalues: √13 4, √13 4, 0 − − − This has two directions in which it appears to have a local maximum. However, the test fails because of the zero eigenvalue. 31. Let Q be orthogonal. Find the possible values of det (Q) . 1 ± 32. Let U be unitary. Find the possible values of det (U) . U is unitary which means it preserves length. Therefore, if Ux = λx, it must be the case that λ = 1. Hence λ = eiθ for some θ. In other words, λ is a point on the unit circle. | | 33. If a matrix is nonzero can it have only zero for eigenvalues? 1 1 Deﬁnitely yes. Consider this one. −1 1 − 34. A matrix A is called nilpotent if Ak = 0 for some positive integer k. Suppose A is a nilpotent matrix. Show it has only 0 for an eigenvalue. Say λ is an eigenvalue. Then for some x = 0, 6 Ax = λx Then λk is an eigenvalue for Ak and so λk = 0. Hence λ = 0. 35. If A is a nonzero nilpotent matrix, show it must be defective. If it is not defective, then there exists S such that 1 S− AS = 0 But then A = 0. Hence, if A = 0 is nilpotent, then A is defective. 6 36. Suppose A is a nondefective n n matrix and its eigenvalues are all either 0 or 1. Show A2 = A. Could you say anything× interesting if the eigenvalues were all either 0,1,or 1? By DeMoivre’s theorem, an nth root of unity is of the form − 2kπ 2kπ cos + i sin n n Could you generalize the sort of thing just described to get An = A? Hint: Since A 1 is nondefective, there exists S such that S− AS = D where D is a diagonal matrix. 1 1 2 2 1 2 S− AS = D and so S− A S = D = D = S− AS and so, in the ﬁrst case, A = A. If the eigenvaluse are 0,1, 1, then all even powers of A are equal and all odd powers are equal. − 2kπ 2kπ In the last case, you could have the eigenvalues of A equal to cos n + i sin n . Then 1 n S− A S = I and so An = I.

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37. This and the following problems will present most of a diﬀerential equations course. To begin with, consider the scalar initial value problem

y′ = ay, y (t0) = y0

a(t t0) When a is real, show the unique solution to this problem is y = y0e − . Next suppose y′ = (a + ib) y, y (t0) = y0 (6.30) where y (t) = u (t) + iv (t) . Show there exists a unique solution and it is

a(t t0) y (t) = y e − (cos b (t t ) + i sin b (t t )) 0 − 0 − 0 (a+ib)(t t0) e − y . (6.31) ≡ 0 Next show that for a real or complex there exists a unique solution to the initial value problem y′ = ay + f, y (t0) = y0 and it is given by t a(t t0) at as y (t) = e − y0 + e e− f (s) ds. Zt0 at Hint: For the ﬁrst part write as y′ ay = 0 and multiply both sides by e− . Then explain why you get − d at e− y (t) = 0, y (t ) = 0. dt 0 Now you ﬁnish the argument. To show uniqueness in the second part, suppose

y′ = (a + ib) y, y (0) = 0

and verify this requires y (t) = 0. To do this, note

y′ = (a ib) y, y (0) = 0 − and that

d 2 y (t) = y′ (t) y (t) + y′ (t) y (t) dt | | = (a + ib) y (t) y (t) + (a ib) y (t) y (t) − = 2a y (t) 2 , y 2 (t ) = 0 | | | | 0 2 2at Thus from the ﬁrst part y (t) = 0e− = 0. Finally observe by a simple computation that 6.30 is solved by 6.31.| For| the last part, write the equation as

y′ ay = f − at and multiply both sides by e− and then integrate from t0 to t using the initial condition. 38. Now consider A an n n matrix. By Schur’s theorem there exists unitary Q such that × 1 Q− AQ = T

where T is upper triangular. Now consider the ﬁrst order initial value problem

x′ = Ax, x (t0) = x0.

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1 Show there exists a unique solution to this ﬁrst order system. Hint: Let y = Q− x and so the system becomes

1 y′ = T y, y (t0) = Q− x0 (6.32)

Now letting y = (y , , y )T , the bottom equation becomes 1 ··· n 1x yn′ = tnnyn, yn (t0) = Q− 0 n . Then use the solution you get in this to get the solution to the initial value problem which occurs one level up, namely

1x yn′ 1 = t(n 1)(n 1)yn 1 + t(n 1)nyn, yn 1 (t0) = Q− 0 n 1 − − − − − − − Continue doing this to obtain a unique solution to 6.32. 39. Now suppose Φ(t) is an n n matrix of the form × Φ(t) = x (t) x (t) (6.33) 1 ··· n where xk′ (t) = Axk (t) . Explain why Φ′ (t) = AΦ(t) if and only if Φ (t) is given in the form of 6.33. Also explain why if c Fn, ∈ y (t) Φ(t) c ≡ solves the equation y′ (t) = Ay (t) .

This is because y′ (t) = Φ′ (t) c = AΦ(t) c = Ay (t) . 40. In the above problem, consider the question whether all solutions to

x′ = Ax (6.34)

are obtained in the form Φ (t) c for some choice of c Fn. In other words, is the general solution to this equation Φ (t) c for c Fn? Prove∈ the following theorem using linear algebra. ∈

Theorem F.24.1 Suppose Φ(t) is an n n matrix which satisﬁes ×

Φ′ (t) = AΦ(t) .

1 Then the general solution to 6.34 is Φ(t) c if and only if Φ(t)− exists for some t. 1 1 Furthermore, if Φ′ (t) = AΦ(t) , then either Φ(t)− exists for all t or Φ(t)− never exists for any t.

(det (Φ (t)) is called the Wronskian and this theorem is sometimes called the Wronskian alternative.) Hint: Suppose ﬁrst the general solution is of the form Φ(t) c where c is an arbitrary n 1 constant vector in F . You need to verify Φ (t)− exists for some t. In fact, show

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1 1 Φ(t)− exists for every t. Suppose then that Φ (t0)− does not exist. Explain why there exists c Fn such that there is no solution x to ∈

c = Φ (t0) x

By the existence part of Problem 38 there exists a solution to

x′ = Ax, x (t0) = c

1 but this cannot be in the form Φ (t) c. Thus for every t,Φ(t)− exists. Next suppose 1 for some t0, Φ(t0)− exists. Let z′ = Az and choose c such that

z (t0) = Φ (t0) c

Then both z (t) , Φ(t) c solve

x′ = Ax, x (t0) = z (t0)

Apply uniqueness to conclude z = Φ (t) c. Finally, consider that Φ (t) c for c Fn ∈ 1 either is the general solution or it is not the general solution. If it is, then Φ (t)− 1 exists for all t. If it is not, then Φ (t)− cannot exist for any t from what was just shown.

1 41. Let Φ′ (t) = AΦ(t) . Then Φ (t) is called a fundamental matrix if Φ (t)− exists for all t. Show there exists a unique solution to the equation

x′ = Ax + f, x (t0) = x0 (6.35)

and it is given by the formula

t 1 1 x (t) = Φ (t)Φ(t0)− x0 + Φ (t) Φ(s)− f (s) ds Zt0 Now these few problems have done virtually everything of significance in an entire un- dergraduate differential equations course, illustrating the superiority of linear algebra. The above formula is called the variation of constants formula. Hint: Uniquenss is easy. If x , x are two solutions then let u (t) = x (t) x (t) and 1 2 1 − 2 argue u′ = Au, u (t0) = 0. Then use Problem 38. To verify there exists a solution, you could just differentiate the above formula using the fundamental theorem of calculus and verify it works. Another way is to assume the solution in the form

x (t) = Φ (t) c (t)

and ﬁnd c (t) to make it all work out. This is called the method of variation of parameters. For the method of variation of parameters,

x′ (t) = AΦ(t) c (t) + Φ (t) c′ (t)

x′ (t) = AΦ(t) c (t) + f (t)

and so t 1 c (t) = Φ(s)− f (s) ds + k Zt0

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Then the solution desired is

t 1 x (t) = Φ (t) k + Φ (t) Φ(s)− f (s) ds Zt0 1 Plug in t = t0. This yields x0 = Φ (t0) k and so k = Φ (t0)− x0 and so the solution is

t 1 1 x (t) = Φ (t)Φ(t0)− x0 + Φ (t) Φ(s)− f (s) ds Zt0 Note that by approximating with Riemann sums and passing to a limit, it follows that the constant Φ (t) can be taken into the integral to write

t 1 1 x (t) = Φ (t)Φ(t0)− x0 + Φ(t)Φ(s)− f (s) ds Zt0

42. Show there exists a special Φ such that Φ′ (t) = AΦ(t) , Φ (0) = I, and suppose 1 Φ(t)− exists for all t. Show using uniqueness that

1 Φ( t) = Φ (t)− − and that for all t, s R ∈ Φ(t + s) = Φ (t)Φ(s) Explain why with this special Φ, the solution to 6.35 can be written as

t x (t) = Φ (t t ) x + Φ(t s) f (s) ds. − 0 0 − Zt0

th Hint: Let Φ (t) be such that the j column is xj (t) where

xj′ = Axj , xj (0) = ej. Use uniqueness as required. Just follow the hint. Such a Φ (t) clearly exists. By the Wronskian alternative in one 1 of the above problems Φ (t)− exists for all t. Consider AΦ(t) Φ(t) A = Ψ (t) −

Ψ′ (t) = AΦ′ (t) Φ′ (t) A − = AAΦ(t) AΦ(t) A − = A (Ψ (t))

and also Ψ (0) = 0. By uniqueness given above, Ψ (t) = 0 for all t. Therefore,

(Φ (t)Φ( t))′ = AΦ(t)Φ( t) Φ(t) AΦ( t) − − − − = Φ (t) AΦ( t) Φ(t) AΦ( t) = 0 − − − and so Φ(t)Φ( t) = C − a constant. However, when t = 0,C = I and so Φ (t)Φ( t) = I. Also ﬁxing s, − Ψ(t) Φ(t + s) Φ(t)Φ(s) ≡ −

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Ψ′ (t) = Φ′ (t + s) Φ′ (t)Φ(s) − = AΦ(t + s) AΦ(t)Φ(s) − = AΨ(t)

and Ψ (0) = 0. Therefore, Ψ(t) = 0 for all t by uniqueness. It follows that for all t, s,

Φ(t + s) = Φ (t)Φ(s)

Therefore, in the variation of constants formula, it reduces to

t x (t) = Φ (t t ) x + Φ(t s) f (s) ds − 0 0 − Zt0

One thing might not be clear. If Ψ′ (t) = 0, why is Ψ (t) a constant? This works for scalar valued functions by a use of the mean value theorem. However, this is a matrix valued function. Nevertheless this is true. You just consider the ijth entry which is

T ei Ψ(t) ej

Its derivative equals 0 and so it is constant. 43. You can see more on this problem and the next one in the latest version of Horn and Johnson, [16]. Two n n matrices A, B are said to be congruent if there is an invertible P such that × B = P AP ∗

Let A be a Hermitian matrix. Thus it has all real eigenvalues. Let n+ be the number of positive eigenvalues, n , the number of negative eigenvalues and n0 the number of zero eigenvalues. For k a− positive integer, let I denote the k k identity matrix and k × Ok the k k zero matrix. Then the inertia matrix of A is the following block diagonal n n matrix.× × In+

I −  n  On0 Show that A is congruent to its inertia matrix. Next show that congruence is an equivalence relation. Finally, show that if two Hermitian matrices have the same inertia matrix, then they must be congruent. Hint: First recall that there is a unitary matrix, U such that

Dn+

U ∗AU = D −  n  On0   where the Dn+ is a diagonal matrix having the positive eigenvalues of A, Dn− being

deﬁned similarly. Now let Dn− denote the diagonal matrix which replaces each entry

of D − with its absolute value. Consider the two diagonal matrices n

1/2 Dn−+ 1/2 D = D∗ = −  Dn−  In  0    Now consider D∗U ∗AUD.

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The hint gives it away. When you block multiply, the thing which results is the inertia matrix. Congruance is obviously an equivalence relation. Consider the transitive law for example. Say A = PBP ∗,B = QCQ∗. Then A = PQCQ∗P ∗ = PQC (PQ)∗ . Now if you have two Hermitian matrices which have the same inertia matrix, then since congruance is an equivalence relation, it follows that the two matrices are congruant. 44. Show that if A, B are two congruent Hermitian matrices, then they have the same inertia matrix. Hint: Let A = SBS∗ where S is invertible. Show that A, B have the same rank and this implies that they are each unitarily similar to a diagonal matrix which has the same number of zero entries on the main diagonal. Therefore, letting VA be the span of the eigenvectors associated with positive eigenvalues of A and VB being deﬁned similarly, it suﬃces to show that these have the same dimensions. Show that (Ax, x) > 0 for all x V . Next consider S∗V . For x V , explain why ∈ A A ∈ A 1 1 (BS∗x,S∗x) = S− A (S∗)− S∗x,S∗x

1 1 = S− Ax,S∗x = Ax, S− ∗ S∗x = (Ax, x) > 0 Next explain why this shows that S∗VA is a subspace ofVB and so the dimension of VB is at least as large as the dimension of VA. Hence there are at least as many positive eigenvalues for B as there are for A. Switching A, B you can turn the inequality around. Thus the two have the same inertia matrix. The above manipulation is straight forward. Now an orthonormal basis exists for Fn which consists of eigenvectors of B. Say you have v , , v , w , , w , z , , z { 1 ··· q 1 ··· r 1 ··· s} where the vi correspond to the positive eigenvalues, the wi to the negative eigenvalues and the z to the eigenvalue 0. We know that B is positive on span (v , , v ) . Could i 1 ··· q it be positive on any larger subspace? Could it be positive on span (v1, , vq, u) where ··· u span (w , , w , z , , z )? ∈ 1 ··· r 1 ··· s No, it couldn’t because you could then consider (Bu, u) and it would end up being no larger than 0. It follows that S∗V V = span (v , , v ) and so the dimension A ⊆ V 1 ··· q of S∗VA, which is the same as the dimension of VA which is the same as the number of positive eigenvalues, counted according to multiplicity, is no larger than q. Now reverse the argument letting A B. Hence the two have the same number of positive eigenvalues. Since they have←→ the same number of zero eigenvalues, this implies they have the same inertia matrix. 45. Let A be an m n matrix. Then if you unraveled it, you could consider it as a vector in Cnm. The Frobenius× inner product on the vector space of m n matrices is deﬁned as × (A, B) trace (AB∗) ≡ Show that this really does satisfy the axioms of an inner product space and that it also amounts to nothing more than considering m n matrices as vectors in Cnm. × First note that

(A, B) = Aij Bij = Aij Bij = (B,A) i,j i,j X X (A, A) = A A = A 2 ij ij | ij | i,j i,j X X Thus (A, A) = 0 if and only if A = 0. Clearly for a, b scalars, (aA + bB, C) = a (A, C) + b (B,C)

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and so this does satisfy the axioms of the inner product. Also, this inner product coincides with the standard inner product on Cnm. 46. Consider the n n unitary matrices. Show that whenever U is such a matrix, it ↑follows that ×

U nn = √n | |C Next explain why if U is any sequence of unitary matrices, there exists a subse- { k} quence Uk ∞ such that limm Uk = U where U is unitary. Here the limit { m }m=1 →∞ m takes place in the sense that the entries of Ukm converge to the corresponding entries of U. From the above and the fact that U is unitary,

2 U n2 = trace (UU ∗) = trace (I) = n. | |C n2 Let Uk be a sequence of unitary matrices. Since these, considered as vectors in C , are contained{ } in a closed and bounded set, it follows that there is a subsequence which converges to U. Why is U a unitary matrix? You have

I = Uk U ∗ UU ∗ m km → and so U is unitary. 47. Let A, B be two n n matrices. Denote by σ (A) the set of eigenvalues of A. Deﬁne ↑ × dist (σ (A) , σ (B)) = max min λ µ : µ σ (B) λ σ(A) {| − | ∈ } ∈ Explain why dist (σ (A) , σ (B)) is small if and only if every eigenvalue of A is close to some eigenvalue of B. Now prove the following theorem using the above problem and Schur’s theorem. This theorem says roughly that if A is close to B then the eigenvalues of A are close to those of B in the sense that every eigenvalue of A is close to an eigenvalue of B.

Theorem F.24.2 Suppose limk Ak = A. Then →∞

lim dist (σ (Ak) , σ (A)) = 0 k →∞

By Schur’s theorem, there exist unitary matrices Uk such that

Uk∗AkUk = Tk,

where Tk is upper triangular. I claim that there exists a subsequence Tkm m∞=1 such that { } T T km → Where T is unitarily similar to A.

Using the above problem, there exists a subsequence km such that

lim Uk = U m m →∞ a unitary matrix. It follows then that

lim Tk = lim U ∗ Ak Uk = U ∗AU T n m m km m m →∞ →∞ ≡

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Since Uk∗m Akm Ukm is upper triangular, it follows that so is T . This shows the claim. Next suppose that the conclusion does not hold. Then there exists ε > 0 and a subsequence, still called k such that

dist (σ (A ) , σ (A)) ε k ≥

It follows that for every choice of Uk such that Uk∗AkUk = Tk, a triangular matrix and for every choice of unitary U such that U ∗AU = T, a triangular matrix,

T T n2 dist (σ (A ) , σ (A)) ε | k − |C ≥ k ≥ Recall from the proof of Schur’s theorem, you change the order of the eigenvalues on the diagonal by adjusting the unitary matrix U. No matter how you permute the entries on the diagonals, the two are always far apart. Now pick Uk for each k, such that Uk∗AkUk = Tk a triangular matrix. By the ﬁrst part of the proof, there

exists a subsequence km such that Tkm T where T is unitarily similar to A. This contradicts the above.{ } → a b 48. Let A = be a 2 2 matrix which is not a multiple of the identity. Show c d × that A is similar to a 2 2 matrix which has one entry equal to 0. Hint: First note that there exists a vector×a such that Aa is not a multiple of a. Then consider

1 B = a Aa − A a Aa

Show B has a zero on the main diagonal. p p Let the vector be such that this vector and A are linearly independent. q q Then by brute force you do the following computation.

1 cp + dq ap bq a b p ap + bq − − D q p c d q cp + dq − Then you do the multiplication. This yields

1 0 (ad bc) bq2 cp2 + apq dpq − − − D cp2 bq2 apq + dpq (a + d) bq2 cp2 + apq dpq − − − − − and since the trace equals 0 the term in the bottom right is also equal to 0. 49. Let A be a complex n n matrix which has trace equal to 0. Show that A is similar to↑ a matrix which has× all zeros on the main diagonal. Hint: Use Problem 30 on Page 122 to argue that you can say that a given matrix is similar to one which has the diagonal entries permuted in any order desired. Then use the above problem and block multiplication to show that if the A has k nonzero entries, then it is similar to a matrix which has k 1 nonzero entries. Finally, when A is similar to one which has at most one nonzero entry,− this one must also be zero because of the condition on the trace. Suppose A has k nonzero diagonal entries. First of all, we can assume k 2 because the trace equals 0. Then from Problem 30 on Page 122 it is similar to a block≥ matrix of the form PQ RS

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where the diagonal entries are just re ordered and P is a 2 2 matrix which which is not a multiple of the identity and has both diagonal entries× nonzero. Otherwise, the trace could not equal zero if all the diagonal entries were equal. Then from the above 1 problem, there exists L such that L− PL has at least one zero on the diagonal.

L 1 0 PQ L 0 L 1PLL 1Q − = − − 0 I RS 0 I RLS Thus this new matrix is similar to the one above and it has at least one fewer nonzero entries on the main diagonal. Continue this process obtaining a sequence of similar matrices, each having fewer nonzero diagonal entries than the preceding one till there is at most one nonzero entry. It follows that this one must be zero also because the trace equals 0. 50. An n n matrix X is a comutator if there are n n matrices A, B such that X = AB↑ BA.× Show that the trace of any comutator is× 0. Next show that if a complex matrix− X has trace equal to 0, then it is in fact a comutator. Hint: Use the above problem to show that it suﬃces to consider X having all zero entries on the main diagonal. Then deﬁne

1 0 2 Xij   i j if i = j A = . ,Bij = − 6 .. ( 0 if i = j    0 n      The ﬁrst part is easy because the trace of a product is the same in either order. Suppose then that X has zero trace. Then by the above problem, it is similar to a matrix Y which has a zero diagonal. Suppose you can show that Y = AB BA. Then 1 − if Y = S− XS, you could write

1 S− XS = AB BA, − 1 1 1 1 1 X = S (AB BA) S− = SAS− SBS− SBS− SAS− − − Thus X will be a comutator. It suﬃces then to show that whenever a matrix has all zero diagonal entries, it must be a comutator. Now use the deﬁnition of A, B in the hint. (AB BA) A B B A − ij ≡ is sj − ir rj s r X X Since A is a diagonal matrix, this reduces for i = j to 6 iX X iB B j = ij j ij = X ij − ij i j − i j ij − − while if i = j, you have iB B i = 0 ii − ii Thus it yields Xij .

F.25 Exercises

8.4

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1 1 1 0 1. Let H denote span 2 , 4 , 3 , 1 . Find the dimension of H  0   0   1   1  and determine a basis.        1 1 1 0 1 0 0 1 2 4 3 1 , row echelon form: 0 1 0− 0 . The ﬁrst three vectors form  0 0 1 1   0 0 1 1  a basis and the dimension is 3.   2. Let M = u = (u , u , u , u ) R4 : u = u = 0 . Is M a subspace? Explain. 1 2 3 4 ∈ 3 1 Sure. It is closed with respect to linear combinations. 3. Let M = u = (u , u , u , u ) R4 : u u . Is M a subspace? Explain. 1 2 3 4 ∈ 3 ≥ 1 No. Not a subspace. Consider (0, 0, 1, 0) and multiply by 1. − 4 4 4. Let w R and let M = u = (u1, u2, u3, u4) R : w u = 0 . Is M a subspace? Explain.∈ ∈ · Of course. It is the kernel of a linear transformation.

4 5. Let M = u = (u1, u2, u3, u4) R : ui 0 for each i = 1, 2, 3, 4 . Is M a subspace? Explain. ∈ ≥ NO. Multiply something by 1. − 4 6. Let w, w1 be given vectors in R and deﬁne

M = u = (u , u , u , u ) R4 : w u = 0 and w u = 0 . 1 2 3 4 ∈ · 1 · Is M a subspace? Explain. Yes. It is the intersection of two subspaces. 7. Let M = u = (u , u , u , u ) R4 : u 4 . Is M a subspace? Explain. 1 2 3 4 ∈ | 1| ≤ No. Take something nonzero in M where say u1 = 1. Now multiply by 100. 8. Let M = u = (u , u , u , u ) R4 : sin (u ) = 1 . Is M a subspace? Explain. 1 2 3 4 ∈ 1 No. Let u1 = π/2 and multiply by 2. 9. Suppose x , , x is a set of vectors from Fn. Show that 0 is in span (x , , x ) . { 1 ··· k} 1 ··· k 0 = i 0xi 10. ConsiderP the vectors of the form 2t + 3s s t : s, t R .  −  ∈   t + s    Is this set of vectors a subspace of R3? If so, explain why, give a basis for the subspace and ﬁnd its dimension. 2 3 Yes. A basis is 1 , 1  −1   1     

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11. Consider the vectors of the form 2t + 3s + u s t  −  : s, t, u R . t + s ∈    u        Is this set of vectors a subspace of R4? If so, explain why, give a basis for the subspace and ﬁnd its dimension. It is a subspace. It is spanned by

3 2 1 1 1 0 , ,  1   1   0   0   0   1              It will be a basis if it is a linearly independent set. 3 2 1 1 0 0 1 1 0 0 1 0 , row echelon form: . Thus the vectors are linearly inde-  1 1 0   0 0 1   0 0 1   0 0 0      pendent.    12. Consider the vectors of the form 2t + u + 1 t + 3u   : s, t, u, v R . t + s + v ∈    u        Is this set of vectors a subspace of R4? If so, explain why, give a basis for the subspace and find its dimension. Because of that 1 this is not a subspace. 13. Let V denote the set of functions defined on [0, 1]. Vector addition is defined as (f + g)(x) f (x) + g (x) and scalar multiplication is defined as (αf)(x) α (f (x)). Verify V is≡ a vector space. What is its dimension, finite or infinite? ≡

Pick n points x1, , xn . Then let ei (x) = 0 unless x = xi when it equals 1. Then e n is linearly{ independent,··· } this for any n. { i}i=1 14. Let V denote the set of polynomial functions defined on [0, 1]. Vector addition is defined as (f + g)(x) f (x) + g (x) and scalar multiplication is defined as (αf)(x) α (f (x)). Verify V is≡ a vector space. What is its dimension, finite or infinite? ≡ This is clearly a subspace of the set of all functions defined on [0, 1] . Its dimension is still infinite. For example, you can see that 1, x, x2, , xn is linearly independent for each n. You could use Theorem 8.2.9 to show this,··· for example. 15. Let V be the set of polynomials defined on R having degree no more than 4. Give a basis for this vector space. 1, x, x2, x3, x4

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16. Let the vectors be of the form a + b√2 where a, b are rational numbers and let the field of scalars be F = Q, the rational numbers. Show directly this is a vector space. What is its dimension? What is a basis for this vector space? A basis is 1, √2 . In fact it is actually a field. 1 √ − a b√2 a + b 2 = a 2− 2b2 . Since a, b are rational, it follows that the denominator is not 0. Thus every element− has an inverse so it is actually a field. It is certainly a vector space over the rational numbers.

17. Let V be a vector space with field of scalars F and suppose v1, , vn is a basis for V . Now let W also be a vector space with field of scalars F{. Let···L : v} , , v { 1 ··· n} → W be a function such that Lvj = wj . Explain how L can be extended to a linear transformation mapping V to W in a unique way. L ( n c v ) n c w i=1 i i ≡ i=1 i i 18. If youP have 5 vectorsP in F5 and the vectors are linearly independent, can it always be concluded they span F5? Explain. Yes. If not, you could add in a vector not in the span and obtain a linearly independent set of 6 vectors which is not possible. 19. If you have 6 vectors in F5, is it possible they are linearly independent? Explain. No. There is a spanning set having 5 vectors and this would need to be as long as the linearly independent set. 20. Suppose V,W are subspaces of Fn. Show V W defined to be all vectors which are in both V and W is a subspace also. ∩ If x, y V W then αa+βb V and is also in W because these are subspaces. Hence V W∈is a∩ subspace. ∈ ∩ 21. Suppose V and W both have dimension equal to 7 and they are subspaces of a vector space of dimension 10. What are the possibilities for the dimension of V W ? Hint: Remember that a linear independent set can be extended to form a basis.∩ See the next problem.

22. Suppose V has dimension p and W has dimension q and they are each contained in a subspace, U which has dimension equal to n where n > max (p, q) . What are the possibilities for the dimension of V W ? Hint: Remember that a linear independent set can be extended to form a basis.∩

Let x1, , xk be a basis for V W. Then there is a basis for V and W which are respectively{ ··· } ∩

x , , x , y , , y , x , , x , z , , z { 1 ··· k k+1 ··· p} { 1 ··· k k+1 ··· q} It follows that you must have k + p k + q k n and so you must have − − ≤ p + q n k − ≤ 23. If b = 0, can the solution set of Ax = b be a plane through the origin? Explain. 6 No. It can’t. It does not contain 0.

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24. Suppose a system of equations has fewer equations than variables and you have found a solution to this system of equations. Is it possible that your solution is the only one? Explain. No. There must then be infinitely many solutions. If the system is Ax = b, then there are infinitely many solutions to Ax = 0 and so the solutions to Ax = b are a particular solution to Ax = b added to the solutions to Ax = 0 of which there are infinitely many. 25. Suppose a system of linear equations has a 2 4 augmented matrix and the last column is a pivot column. Could the system of linear× equations be consistent? Explain. No. This would lead to 0 = 1. 26. Suppose the coefficient matrix of a system of n equations with n variables has the property that every column is a pivot column. Does it follow that the system of equations must have a solution? If so, must the solution be unique? Explain. Yes. It has a unique solution. 27. Suppose there is a unique solution to a system of linear equations. What must be true of the pivot columns in the augmented matrix. The last one must not be a pivot column and the ones to the left must each be pivot columns. 28. State whether each of the following sets of data are possible for the matrix equation Ax = b. If possible, describe the solution set. That is, tell whether there exists a unique solution no solution or infinitely many solutions.

(a) A is a 5 6 matrix, rank (A) = 4 and rank(A b) = 4. Hint: This says b is in the span× of four of the columns. Thus the columns| are not independent. Inﬁnite solution set. (b) A is a 3 4 matrix, rank (A)=3 and rank(A b) = 2. × | This surely can’t happen. If you add in another column, the rank does not get smaller. (c) A is a 4 2 matrix, rank (A) = 4 and rank(A b) = 4. Hint: This says b is in the span× of the columns and the columns must| be independent. You can’t have the rank equal 4 if you only have two columns. (d) A is a 5 5 matrix, rank (A) = 4 and rank(A b) = 5. Hint: This says b is not in the span× of the columns. | In this case, there is no solution to the system of equations represented by the augmented matrix. (e) A is a 4 2 matrix, rank (A)=2 and rank(A b) = 2. × | In this case, there is a unique solution since the columns of A are independent.

29. Suppose A is an m n matrix in which m n. Suppose also that the rank of A equals × n m ≤ m. Show that A maps F onto F . Hint: The vectors e1, , em occur as columns in the row reduced echelon form for A. ··· This says that the columns of A have a subset of m vectors which are linearly independent. Therefore, this set of vectors is a basis for Fm. It follows that the span of the columns is all of Fm. Thus A is onto.

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30. Suppose A is an m n matrix in which m n. Suppose also that the rank of A equals n. Show that A is one× to one. Hint: If not,≥ there exists a vector, x such that Ax = 0, and this implies at least one column of A is a linear combination of the others. Show this would require the column rank to be less than n. The columns are independent. Therefore, A is one to one. 31. Explain why an n n matrix A is both one to one and onto if and only if its rank is n. × The rank is n is the same as saying the columns are independent which is the same as saying A is one to one which is the same as saying the columns are a basis. Thus the span of the columns of A is all of Fn and so A is onto. If A is onto, then the columns must be linearly independent since otherwise the span of these columns would have dimension less than n and so the dimension of Fn would be less than n. 32. If you have not done this problem already, here it is again. It is a very important result. Suppose A is an m n matrix and B is an n p matrix. Show that × × dim (ker (AB)) dim (ker (A)) + dim (ker (B)) . ≤ p Let w1, , wk be a basis for B (F ) ker (A) and suppose u1, , ur is a basis { ··· } ∩ { ··· } p for ker (B). Let Bzi = wi. Now suppose x ker (AB). Then Bx ker (A) B (F ) and so ∈ ∈ ∩ k k Bx = aiwi = aiBzi i=1 i=1 X X so k x a z ker (B) − i i ∈ i=1 X Hence k r x a z = b u − i i j j i=1 j=1 X X This shows that dim (ker (AB)) k + r dim (ker (A)) + dim (ker (B)) ≤ ≤ Note that a little more can be said. u1, , ur, z1, , zk is independent. Here is why. Suppose { ··· ··· } k aiui + bj zj = 0 i j=1 X X Then do B to both sides and conclude that

bjBzj = 0 j X which forces each bj = 0. Then the independence of the ui implies each ai = 0. It follows that in fact, dim (ker (AB)) = k + r dim (ker (A)) + dim (ker (B)) ≤ and the remaining inequality is an equality if and only if B (Fp) ker (A). ⊇ This is because B (Fp) ker (A) ker (A) and these are equal exactly when they have the same dimension. ∩ ⊆

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33. Recall that every positive integer can be factored into a product of primes in a unique way. Show there must be infinitely many primes. Hint: Show that if you have any finite set of primes and you multiply them and then add 1, the result cannot be divisible by any of the primes in your finite set. This idea in the hint is due to Euclid who lived about 300 B.C. Consider p p p + 1. Then if 1 2 ··· n p p p + 1 = lp 1 2 ··· n k you would have 1 + m = l pk

where m is an integer, the product of the other primes different than pk. This is a contradiction. Hence there must be another prime to divide p1p2 pn + 1 and so there are infinitely many primes. ··· 34. There are lots of fields. This will give an example of a finite field. Let Z denote the set of integers. Thus Z = , 3, 2, 1, 0, 1, 2, 3, . Also let p be a prime number. We will say that two integers,{· · · − a,− b are− equivalent···} and write a b if a b is divisible by p. Thus they are equivalent if a b = px for some integer∼ x. First− show that a a. Next show that if a b then−b a. Finally show that if a b and b c then∼ a c. For a an integer,∼ denote by [∼a] the set of all integers which∼ is equivalent∼ to a, the∼ equivalence class of a. Show first that is suffices to consider only [a] for a = 0, 1, 2, , p 1 and that for 0 a < b p 1, [a] = [b]. That is, [a] = [r] where r 0, 1, 2···, , p− 1 . Thus there≤ are exactly≤ p−of these6 equivalence classes. Hint: Recall∈ { the Euclidean··· − } algorithm. For a > 0, a = mp + r where r < p. Next define the following operations. [a] + [b] [a + b] ≡ [a][b] [ab] ≡ Show these operations are well defined. That is, if [a] = [a′] and [b] = [b′] , then [a] + [b] = [a′] + [b′] with a similar conclusion holding for multiplication. Thus for addition you need to verify [a + b] = [a′ + b′] and for multiplication you need to verify [ab] = [a′b′]. For example, if p = 5 you have [3] = [8] and [2] = [7] . Is [2 3] = [8 7]? Is [2 + 3] = [8 + 7]? Clearly so in this example because when you subtract,× the result× is divisible by 5. So why is this so in general? Now verify that [0] , [1] , , [p 1] with these operations is a Field. This is called the integers modulo{ a prime··· and− is} written Zp. Since there are infinitely many primes p, it follows there are infinitely many of these finite fields. Hint: Most of the axioms are easy once you have shown the operations are well defined. The only two which are tricky are the ones which give the existence of the additive inverse and the multiplicative inverse. Of these, the first is not hard. [x] = [ x]. Since p is prime, there exist integers x, y such that 1 = px+ky and so− 1 ky =−px which says 1 ky and so [1] = [ky] . Now you finish the argument. What is the− multiplicative identity∼ in this collection of equivalence classes? Of course you could now consider field extensions based on these fields.

The only substantive issue is why Zp is a ﬁeld. Let [x] Zp where [x] = [0]. Thus x is not a multiple of p. Then x and p are relatively prime.∈ Hence from Theorem6 1.9.3, there exists a, b such that 1 = ap + bx Then [1 bx] = [ap] = 0 −

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and it follows that [b][x] = [1] 1 so [b] = [x]− .

35. Suppose the field of scalars is Z2 described above. Show that 0 1 0 0 0 0 0 1 1 0 = 0 0 1 0 − 1 0 0 0 0 1 Thus the identity is a comutator. Compare this with Problem 50 on Page 198. 1 0 1 0 This is easy. The left side equals = 0 1 0 1 − 36. Suppose V is a vector space with field of scalars F. Let T (V,W ) , the space of linear transformations mapping V onto W where W is another∈ L vector space. Define an equivalence relation on V as follows. v w means v w ker (T ) . Recall that ker (T ) v : T v = 0 . Show this is an equivalence∼ relation.− ∈ Now for [v] an equiv- ≡ { } alence class define T ′ [v] T v. Show this is well defined. Also show that with the operations ≡

[v] + [w] [v + w] ≡ α [v] [αv] ≡ this set of equivalence classes, denoted by V/ ker (T ) is a vector space. Show next that T ′ : V/ ker (T ) W is one to one, linear, and onto. This new vector space, V/ ker (T ) is called a quotient→ space. Show its dimension equals the diﬀerence between the dimension of V and the dimension of ker (T ). It is obviously an equivalence relation. Is the operation of addition well deﬁned? If [v] = [v′] , similar for w, w′, is it the case that [v + w] = [v′ + w′]? Is

v + w (v′ + w′) ker (T )? − ∈ Of course so. It is just the sum of two things in ker (T ) . Similarly the operation of scalar multiplication is well defined. Is T ′ well defined? If v w, is it true that T v =T w? Of course so. ∼ T (v w) = 0 − Is T ′ linear? This is obvious. Is T ′ one to one? If T ′ [v] = 0, then it says that T v = 0 and so v ker (T ) . Hence [v] = [0] . Thus T ′ is certainly one to one. If T is onto, it is ∈ obvious that T ′ is onto. Just pick w W . Then T v = w for some v. Then T ′ [v] = w. ∈ 37. Let V be an n dimensional vector space and let W be a subspace. Generalize the above problem to define and give properties of V/W . What is its dimension? What is a basis? Here you define an equivalence relation by v u if v u W. This is clearly an equivalence relation. Define [v] + [u] [u∼+ v] and −α [u] ∈ [αu]. Then by a repeat of the above, everything is well defined.≡ It is clear that≡ V/W , the set of equivalence classes is a vector space with these operations as just defined. What is its dimension? Let w , , w be a basis for W . Now extend to get a basis for { 1 ··· r} V, w1, , wr, v1, , vn r . Then if { ··· ··· − } n r − ci [vi] = [0] , i=1 X

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n r it follows that − c v W and so there exist scalars d such that i=1 i i ∈ i P r n r − diwi + civi = 0 i=1 i=1 X X

Now, since w1, , wr, v1, , vn r is a basis, it follows that all the di and ci are { ··· ··· − } equal to 0. Therefore, in particular the vectors [vi] form a linearly independent set in V/W. Do these vectors span V/W ? Suppose [v] V/W . Then ∈ r n r − v = diwi + civi i=1 i=1 X X for suitable scalars ci and di. Now it follows that

n r − [v] = ci [vi] i=1 X and so the dimension of V/W equals n r where r is the dimension of W . − 38. If F and G are two fields and F G, can you consider G as a vector space with field of scalars F? Explain. ⊆ Yes. This is obvious. 39. Let A denote the algebraic numbers, those numbers which are roots of polynomials having rational coefficients which are in R. Show A can be considered a vector space with field of scalars Q. What is the dimension of this vector space, finite or infinite? As in the above problem, since A is a field, it can be considered as a vector space over Q. So what is the dimension? It is clearly infinite. To see this, consider xn 7. By the rational root theorem, this polynomial is irreducible. Therefore, the minimal− n n n 2 n 1 polynomial for √7 is x 7 and so for α = √7, 1, α, α , , α − are linearly independent since otherwise, −xn 7 would not be the minimal··· polynomial of α and would end up not being irreducible.− Since n is arbitrary, this shows that the dimension of A is infinite.

αi n 40. As mentioned, for distinct algebraic numbers αi, the complex numbers e i=1 are linearly independent over the field of scalars A where A denotes the algebraic{ numbers,} those which are roots of a polynomial having integer (rational) coefficients. What is the dimension of the vector space C with field of scalars A, finite or infinite? If the field of scalars were C instead of A, would this change? What if the field of scalars were R? How many distinct algebraic numbers can you get? Clearly as many as desired. Just consider the nth roots of some complex number. If you have n distinct algebraic αi n numbers αi, then from the Lindemann Weierstrass theorem mentioned above, e i=1 is linearly independent. Of course if the field of scalars were C this would{ change} completely. Then the dimension of C over C is clearly 1. 41. Suppose F is a countable field and let A be the algebraic numbers, those numbers which are roots of a polynomial having coefficients in F which are in G, some other field containing F. Show A is also countable. You consider each polynomial of degree n which has coefficients in the countable field F. There are countably many of these. Each has finitely many roots in A. Therefore,

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there are countably many things in A which correspond to polynomials of degree n The totality of A is obtained then from taking a countable union of countable sets, and this is countable. 42. This problem is on partial fractions. Suppose you have

p (x) R (x) = , degree of p (x) < degree of denominator. q (x) q (x) 1 ··· m

where the polynomials qi (x) are relatively prime and all the polynomials p (x) and qi (x) have coefficients in a field of scalars F. Thus there exist polynomials ai (x) having coefficients in F such that

m 1 = ai (x) qi (x) i=1 X Explain why m p (x) m a (x) q (x) a (x) p (x) R (x) = i=1 i i = i q (x) q (x) q (x) 1 m i=1 j=i j P ··· X 6 This is because you just did Q

p (x) p (x) m a (x) q (x) R (x) = = i=1 i i q (x) q (x) q (x) q (x) 1 ··· m 1P ··· m Then simplifying this yields

m a (x) p (x) R (x) = i q (x) i=1 j=i j X 6 Q Now continue doing this on each term in the above sum till ﬁnally you obtain an expression of the form m bi (x) q (x) i=1 i X Using the Euclidean algorithm for polynomials, explain why the above is of the form

m r (x) M (x) + i q (x) i=1 i X

where the degree of each ri (x) is less than the degree of qi (x) and M (x) is a polynomial. You just note that bi (x) = li (x) qi (x) + ri (x)

where the degree of ri (x) is less than the degree of qi (x) . Then replacing each bi (x) by this expression, you obtain the above expression. Now argue that M (x) = 0. You have m r (x) p (x) M (x) + i = q (x) q (x) q (x) i=1 i 1 m X ···

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Now multiply both sides by q1 (x) qm (x) . If M (x) is not zero, then you would have the degree of the left is larger than··· the degree on the right in the equation which results. Hence M (x) = 0. From this explain why the usual partial fractions expansion of calculus must be true. You can use the fact that every polynomial having real coeﬃcients factors into a product of irreducible quadratic polynomials and linear polynomials having real coeﬃcients. This follows from the fundamental theorem of algebra in the appendix. n What if some qi (x) = ki (x) ? In this case, you can go further. You have the partial fractions expansion described above and you have a term of the form a (x) , degree of a (x) < degree k (x)n k (x)n I claim this can be written as n ai (x) i i=1 k (x) X where the degree of ai (x) < degree of k (x). This can be proved by induction. If n = 1, there is nothing to show. Suppose then it is true for n 1. − a (x) 1 a (x) n = n 1 k (x) k (x) k (x) − !

n 1 If the degree of a (x) is less than the degree of k (x) − , then there is nothing to show. Induction takes care of it. If this is not the case, you can write

n 1 a (x) 1 a (x) 1 b (x) k (x) − + r (x) n = n 1 = n 1 k (x) k (x) k (x) − ! k (x) k (x) − !

n 1 where the degree of r (x) is less than the degree of k (x) − . Then this reduces to a (x) b (x) r (x) 1 n = + n 1 k (x) k (x) k (x) − k (x) The second term gives what is desired by induction. I claim that the degree of b (x) must be less than the degree of k (x) . If not, you could multiply both sides by k (x)n and obtain n 1 a (x) = b (x) k (x) − + r (x) and the degree of the left is less than the degree of k (x)n while the degree of the right is greater than the degree of k (x)n. Therefore, the desired decomposition has been obtained. Now the usual procedure for partial fractions follows. If you have a rational function p (x) /q (x) where the degree of q (x) is larger than the degree of p (x) , there exist factors of q (x) into products of linear and irreducible over R quadratic factors by using the fundamental theorem of algebra. Then you just apply the above decomposition result. 43. Suppose f , , f is an independent set of smooth functions deﬁned on some inter- { 1 ··· n} val (a, b). Now let A be an invertible n n matrix. Deﬁne new functions g1, , gn as follows. × { ··· } g1 f1 . .  .  = A  .  g f  n   n     

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Is it the case that g1, , gn is also independent? Explain why. n { ··· } Suppose i=1 aigi = 0. Then P 0 = ai Aij fj = fj Aij ai i j j i X X X X It follows that Aij ai = 0 i X T for each j. Therefore, since A is invertible, it follows that each ai = 0. Hence the functions gi are linearly independent.

F.26 Exercises

9.5

1. If A, B, and C are each n n matrices and ABC is invertible, why are each of A, B, and C invertible? × This is because ABC is one to one. 2. Give an example of a 3 2 matrix with the property that the linear transformation determined by this matrix× is one to one but not onto.

1 0 0 1  0 0    3. Explain why Ax = 0 always has a solution whenever A is a linear transformation. 4. Review problem: Suppose det (A λI) = 0. Show using Theorem 3.1.15 there exists x = 0 such that (A λI) x = 0. − 6 − Since the matrix (A λI) is not invertible, it follows that there exists x which it sends to 0. − 5. How does the minimal polynomial of an algebraic number relate to the minimal polynomial of a linear transformation? Can an algebraic number be thought of as a linear transformation? How? They are pretty much the same thing. If you have an algebraic number α it can be thought of as a linear transformation which maps A to A according to the rule α (β) = αβ. 6 6. Recall the fact from algebra that if p (λ) and q (λ) are polynomials, then there exists l (λ) , a polynomial such that

q (λ) = p (λ) l (λ) + r (λ)

where the degree of r (λ) is less than the degree of p (λ) or else r (λ) = 0. With this in mind, why must the minimal polynomial always divide the characteristic polynomial? That is, why does there always exist a polynomial l (λ) such that p (λ) l (λ) = q (λ)? Can you give conditions which imply the minimal polynomial equals the characteristic polynomial? Go ahead and use the Cayley Hamilton theorem.

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If q (λ) is the characteristic polynomial and p (λ) is the minimal polynomial, then from the above equation, 0 = r (A) and this cannot happen unless r (λ) = 0 because if not, p (λ) was not the minimal polynomial after all. 7. In the following examples, a linear transformation, T is given by specifying its action on a basis β. Find its matrix with respect to this basis. 1 1 1 1 1 (a) T = 2 + 1 ,T = 2 2 −1 −1 −1 1 1 The basis is , . Formally, 2 −1 1 1 1 1 1 2 + 1 , = , A 2 −1 −1 2 −1 1 1 1 1 = A 5− 1 2− 1 1 1 1 − 1 1 2 0 A = = 2− 1 5− 1 1 1 0 0 1 1 0 (b) T = 2 + 1 ,T = 1 1 −1 −1 1 0 1 0 0 1 2 + 1 , = , A 1 −1 1 1 −1 2 1 A = 1 0 1 1 1 1 1 1 (c) T = 2 + 1 ,T = 1 0 2 0 2 0 − 2 1 1 1 1 1 1 2 + 1 , 1 = , A 2 0 0 − 2 0 2 1 1 A = 2 1 − 8. Let β = u , , u be a basis for Fn and let T : Fn Fn be deﬁned as follows. { 1 ··· n} → n n T akuk = akbkuk ! Xk=1 Xk=1 First show that T is a linear transformation. Next show that the matrix of T with

respect to this basis is [T ]β =

b1 .  ..  b  n  Show that the above deﬁnition is equivalent to simply specifying T on the basis vectors of β by T (uk) = bkuk. Formally, (b u , b u , , b u ) = (u , u , , u )[T ] 1 1 2 2 ··· n n 1 2 ··· n β Thus [T ]β is the diagonal matrix just described.

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9. In the situation of the above problem, let γ = e1, , en be the standard basis for ↑ n th{ ··· } F where ek is the vector which has 1 in the k entry and zeros elsewhere. Show that [T ]γ = 1 u u [T ] u u − (6.36) 1 ··· n β 1 ··· n We have T ui = biui and so we know [T ]β. The problem is to ﬁnd [T ]γ where γ is the good basis. Consider the diagram

[T ] Fn γ Fn → ↓ ↓ Fn T Fn →

↑[T ] ↑ Fn β Fn → 1 On the left side going down you get x x u1 un − x. Going up on the → → ··· right, you need to have x u1 un x u1 un x. Then you need → ··· → ··· 1 [T ] = u u [T ] u u − γ 1 ··· n β 1 ··· n 10. Generalize the above problem to the situation where T is given by specifying its action↑ on the vectors of a basis β = u , , u as follows. { 1 ··· n} n T uk = ajkuj . j=1 X Letting A = (a ) , verify that for γ = e , , e , 6.36 still holds and that [T ] = A. ij { 1 ··· n} β This isn’t really any diﬀerent. As explained earlier, the matrix of T with respect to

the given basis is just A where it is as described. Thus A = [T ]β and by a repeat of the above argument,

1 [T ] = u u [T ] u u − γ 1 ··· n β 1 ··· n 11. Let P3 denote the set of real polynomials of degree no more than 3, defined on an interval [a, b]. Show that P3 is a subspace of the vector space of all functions defined 2 3 on this interval. Show that a basis for P3 is 1, x, x , x . Now let D denote the differentiation operator which sends a function to its derivative. Show D is a linear transformation which sends P3 to P3. Find the matrix of this linear transformation with respect to the given basis. 0, 1, 2x, 3x2 = 1, x, x2, x3 A where A is the desired matrix. Then clearly

0 1 0 0 0 0 2 0 A =  0 0 0 3   0 0 0 0      12. Generalize the above problem to Pn, the space of polynomials of degree no more than n with basis 1, x, , xn . { ··· } The pattern seems pretty clear. A will be an (n + 1) (n + 1) matrix which has zeros everywhere except on the super diagonal where you have,× starting at the top and going towards the bottom, the numbers in the following order. 1, 2, 3, , n. ···

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13. In the situation of the above problem, let the linear transformation be T = D2 + 1, defined as T f = f ′′ + f. Find the matrix of this linear transformation with respect to the given basis 1, x, , xn . { ··· } First consider the operator D2. 0 0 2 6x 12x2 = 1 x x2 x3 x4 A ··· ··· To see the pattern, let n = 4. Then you would have 0 0 2 0 0 0 0 0 6 0   A = 0 0 0 0 12  0 0 0 0 0     0 0 0 0 0      Then the matrix will be this one added to the identity. Hence it is the matrix 1 0 2 0 0 0 1 0 6 0  0 0 1 0 12   0 0 0 1 0     0 0 0 0 1      14. In calculus, the following situation is encountered. There exists a vector valued function f :U Rm where U is an open subset of Rn. Such a function is said to have a derivative→ or to be differentiable at x U if there exists a linear transformation T : Rn Rm such that ∈ → f (x + v) f (x) T v lim | − − | = 0. v 0 v → | | First show that this linear transformation, if it exists, must be unique. Next show that for β = e , , e ,, the standard basis, the kth column of [T ] is { 1 ··· n} β ∂f (x) . ∂xk Actually, the result of this problem is a well kept secret. People typically don’t see this in calculus. It is seen for the first time in advanced calculus if then.

Suppose that T ′ also works. Then letting t > 0,

f (x + tv) f (x) tT v f (x + tv) f (x) tT ′v lim | − − | = lim | − − | = 0 t 0 t t 0 t → → In particular, t (T v T ′v) lim − = 0 t 0 → t which says that T v = T ′v and since v is arbitrary, this shows T = T ′.

1 15. Recall that A is similar to B if there exists a matrix P such that A = P − BP. Show that if A and B are similar, then they have the same determinant. Give an example of two matrices which are not similar but have the same determinant. These two are not similar but have the same determinant. 1 0 1 1 , 0 1 0 1

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You can see these are not similar by noticing that the second has an eigenspace of dimension equal to 1 so it is not similar to any diagonal matrix which is what the first one is. 16. Suppose A (V,W ) where dim (V ) > dim (W ) . Show ker (A) = 0 . That is, show there exist nonzero∈ L vectors v V such that Av = 0. 6 { } ∈ If not, then A would be one to one and so it would take a basis to a linearly independent set which is impossible because W has smaller dimension than V . 17. A vector v is in the convex hull of a nonempty set S if there are finitely many vectors of S, v , , v and nonnegative scalars t , , t such that { 1 ··· m} { 1 ··· m} m m v = tkvk, tk = 1. Xk=1 kX=1 Such a linear combination is called a convex combination. Suppose now that S V, m ⊆ a vector space of dimension n. Show that if v = k=1 tkvk is a vector in the convex hull for m > n + 1, then there exist other scalars t′ such that P{ k} m 1 − v = tk′ vk. Xk=1 Thus every vector in the convex hull of S can be obtained as a convex combination of at most n + 1 points of S. This incredible result is in Rudin [23]. Hint: Consider L : Rm V R defined by → × m m L (a) akvk, ak ≡ ! kX=1 Xk=1 Explain why ker (L) = 0 . Next, letting a ker (L) 0 and λ R, note that λa ker (L) . Thus for6 all{ λ} R, ∈ \{ } ∈ ∈ ∈ m v = (tk + λak) vk. Xk=1 Now vary λ till some t + λa = 0 for some a = 0. k k k 6 If any of the tk = 0, there is nothing to prove. Hence you can assume each of these tk > 0. Suppose then that m > n + 1 and consider

m m L (a) akvk, ak ≡ ! kX=1 Xk=1 a mapping from Rm to V R a vector space of dimension n+1. Then since m > n+1, this mapping L cannot be× one to one. Hence there exists a = 0 such that La = 0. Thus also λa ker (L). By assumption, v = m t v . Then6 for each λ R, ∈ k=1 k k ∈ m P v = (tk + λak) vk Xk=1 m because k=1 akvk = 0. Now (tk + λak) = 1 because ak = 0. Also, for λ = 0, each of the tk + λak > 0. Adjust λ so that each of these is still 0 but one vanishes. Then youP have gotten v as a convexP combination of fewerP than ≥m vectors.

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18. For those who know about compactness, use Problem 17 to show that if S Rn and S is compact, then so is its convex hull. ⊆ Suppose that the convex hull of S is not compact. Then there exists n+1 v , v , , v ∞ , each in S and s C s : s = 1, s [0, 1] 1k 2k ··· (n+1)k k=1 k ∈ ≡ i=1 i i ∈ such that n o n+1 ∞ P sikvik ( i=1 ) X k=1 has no convergent subsequence. However, there exists a further subsequence, still denoted as k such that for each i, limk vik = vi. This is because S is given to be →∞ compact. Also, taking yet another subsequence, you can assume that limk sik = si because of the compactness of C. But this yields an obvious contradiction→∞ to the assertion that the above has no convergent subsequence. 19. Suppose Ax = b has a solution. Explain why the solution is unique precisely when Ax = 0 has only the trivial (zero) solution.

This is because the general solution is yp +y where Ayp = b and Ay = 0. Now A0 = 0 and so the solution is unique precisely when this is the only solution y to Ay = 0. 20. Let A be an n n matrix of elements of F. There are two cases. In the first case, × F contains a splitting field of pA (λ) so that p (λ) factors into a product of linear polynomials having coefficients in F. It is the second case which of interest here where pA (λ) does not factor into linear factors having coefficients in F. Let G be a splitting field of pA (λ) and let qA (λ) be the minimal polynomial of A with respect to the field G. Explain why qA (λ) must divide pA (λ). Now why must qA (λ) factor completely into linear factors? In the field G pA (λ) = qA (λ) l (λ) + r (λ)

where r (λ) either equals 0 or has degree less than the degree of qA (λ). Then it follows that r (A) = 0 and so it must be the case that r (λ) = 0. Hence qA (λ) must divide pA (λ). Now G was the splitting ﬁeld of pA (λ) and this new polynomial divides this one. Therefore, this new polynomial also must factor completely into linear polynomials having coeﬃcients in G.

pA (λ) = qA (λ) l (λ)

and all roots of pA (λ) are in G so the same is true of the roots of qA (λ) because these are all roots of pA (λ). 21. In Lemma 9.2.2 verify that L is linear.

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n n

L a αivi + b βivi i=1 i=1 ! nX X

= L (aαi + bβi) vi i=1 ! n X

= (aαi + bβi) Lvi i=1 Xn n

= a αiLvi + b βiLvi i=1 i=1 X n X n

= aL αivi + bL βivi i=1 ! i=1 ! X X F.27 Exercises

10.6

1. In the discussion of Nilpotent transformations, it was asserted that if two n n matrices A, B are similar, then Ak is also similar to Bk. Why is this so? If two matrices× are similar, why must they have the same rank? 1 k 1 k Say A = S− BS. Then A = S− B S and so these are also similar. You can write

1 1 1 1 S− BSS− BSS− BS S− BS ··· 1 and note that the inside SS− cancels. Therefore, the result of the above is as claimed. Hence Ak is similar to Bk. n 1 n Now the dimension of A (F ) is the dimension of S− BS (F ) which is the same as the 1 n n 1 dimension of S− B (F ) which is the same as the dimension of B (F ) because S, S− are one to one and onto. Hence they each take a basis to a basis. 2. If A, B are both invertible, then they are both row equivalent to the identity matrix. Are they necessarily similar? Explain. 1 1 1 0 Obviously not. Consider , . These are both in Jordan form. 0 1 0 1 3. Suppose you have two nilpotent matrices A, B and Ak and Bk both have the same rank for all k. Does it follow that A, B are similar? What if it is not known that A, B are nilpotent? Does it follow then? 1 1 Yes, this is so. You have A = S− JS and B = T − J ′T where J, J ′ are both Jordan 1 1 k k canonical form. Then you have that J = SAS− and J ′ = TBT − and so J and J ′ have the same rank for all k. Therefore, J = J ′ because these matrices are nilpotent. It follows that 1 1 SAS− = TBT − and consequently, 1 1 1 1 1 B = T − SAS− T = S− T − AS− T so that A is similar to B. In case they are not" nilpotent, the above problem gives a counter example.

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4. When we say a polynomial equals zero, we mean that all the coeﬃcients equal 0. If we assign a diﬀerent meaning to it which says that a polynomial

n k p (λ) = akλ = 0, kX=0 when the value of the polynomial equals zero whenever a particular value of λ F is placed in the formula for p (λ) , can the same conclusion be drawn? Is there∈ any difference in the two definitions for ordinary fields like Q? Hint: Consider Z2, the integers mod 2. This sort of thing where it equals 0 if it sends everything to 0 works fine when we think of polynomials as functions. However, consider λ2 + λ where the coefficients come from Z2. It sends everything to 0 but we don’t want to consider this polynomial to be the zero polynomial because its coefficients are not all 0. 5. Let A (V,V ) where V is a finite dimensional vector space with field of scalars F. Let p (λ∈) L be the minimal polynomial and suppose φ (λ) is any nonzero polynomial such that φ (A) is not one to one and φ (λ) has smallest possible degree such that φ (A) is nonzero and not one to one. Show φ (λ) must divide p (λ). Say φ (A) x = 0 where x = 0. Consider all monic polynomials φ (λ) such that φ (A) x = 0. One such is the minimal6 polynomial p (λ). This condition is less severe than saying that φ (A) = 0. Therefore, there are more polynomials considered. If you take the one of least degree, its degree must be no larger than the degree of p (λ) . By assumption there is a polynomial φ (λ) for which φ (A) = 0 but such that φ (A) x = 0. You can do the usual thing. 6 p (λ) = l (λ) φ (λ) + r (λ) where r (λ) = 0 or else it has smaller degree than φ (λ) . Then

0 = l (A) φ (A) x + r (A) x

It follows that r (A) x = 0. If r (λ) = 0 this would be a contradiction to the definition of φ (λ). Therefore, φ (λ) divides p (6 λ) . 6. Let A (V,V ) where V is a finite dimensional vector space with field of scalars F. Let p (∈λ) L be the minimal polynomial and suppose φ (λ) is an irreducible polynomial with the property that φ (A) x = 0 for some specific x = 0. Show that φ (λ) must divide p (λ) . Hint: First write 6

p (λ) = φ (λ) g (λ) + r (λ)

where r (λ) is either 0 or has degree smaller than the degree of φ (λ). If r (λ) = 0 you are done. Suppose it is not 0. Let η (λ) be the monic polynomial of smallest degree with the property that η (A) x = 0. Now use the Euclidean algorithm to divide φ (λ) by η (λ) . Contradict the irreducibility of φ (λ) . Say φ (A) x = 0 where x = 0. Consider all monic polynomials φ (λ) such that φ (A) x = 0. One such is the minimal6 polynomial p (λ). This condition is less severe than saying that φ (A) = 0. Therefore, there are more polynomials considered. If you take the one of least degree, η its degree must be no larger than the degree of p (λ) . You can do the usual thing. p (λ) = g (λ) η (λ) + r (λ)

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where r (λ) = 0 or else it has smaller degree than η (λ) . Then

0 = g (A) η (A) x + r (A) x

It follows that r (A) x = 0. If r (λ) = 0 this would be a contradiction to the deﬁnition of η (λ). Therefore, η (λ) divides p (6 λ) . Also

φ (λ) = η (λ) l (λ) + r (λ)

where the degree of r (λ) is less than the degree of η (λ) or else r (λ) = 0. If r (λ) = 0, then r (A) x = 0 and it would have degree too small. Hence r (λ) = 0 and so η6 (λ) must divide φ (λ) . But this requires that the two must be scalar multiples of each other. Hence φ (λ) must also divide p (λ). 7. Suppose A is a linear transformation and let the characteristic polynomial be

q det (λI A) = φ (λ)nj − j j=1 Y

where the φj (λ) are irreducible. Explain using Corollary 8.3.11 why the irreducible factors of the minimal polynomial are φj (λ) and why the minimal polynomial is of the form q rj φj (λ) j=1 Y where r n . You can use the Cayley Hamilton theorem if you like. j ≤ j This follows right away from the Cayley Hamilton theorem and that corollary. Since the minimal polynomial divides the characteristic polynomial, it must be of the form just mentioned. 8. Let 1 0 0 A = 0 0 1  0 1− 0  Find the minimal polynomial for A.   You know that this satisﬁes a polynomial of degree no more than 3. Consider 2 3 1 0 0 1 0 0 1 0 0 1 0 0 0 1 0 , 0 0 1 , 0 0 1 , 0 0 1 These are  0 0 1   0 1− 0   0 1− 0   0 1− 0   1 0 0   1 0 0   1 0 0  1 0 0  0 1 0 , 0 0 1 , 0 1 0 , 0 0 1  0 0 1   0 1− 0   0− 0 1   0 1 0  − − Thus it is desired  to ﬁnd a linear  combination of these which equals 0 and out of all of them to pick the one using as few of them as possible starting with the left side. An easy way to do this is to use the row reduced echelon from. Write them as column vectors and then do row operations.

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1 1 1 1 1 1 1 1 1 0 0 1 0 0 0 0 0 0 0 0 0 1 0 1 0 0 0 0 0 0 0 0 0 0 1− 1 0 0 0 0 0 0 0 0 0 0 0 0 1 0 1 0 , 1 0 1 0 , row echelon form: 0 0 0 0 0 1 −0 1 0 1− 0 1 0 0 0 0 0 −0 0 0 0− 0 0 0 0 0 0 0 0 1 0 1 0 1 0 1 0 0 0 0 1 0 1 −0 1 0 1− 0 0 0 0 0 − − Then it follows that A3 = I A + A2 and so the minimal polynomial is, − λ3 λ2 + λ 1 − − Does it work? 3 2 1 0 0 1 0 0 0 0 1 0 0 1 +  0 1− 0  −  0 1− 0      1 0 0 1 0 0 0 0 0 0 0 1 0 1 0 = 0 0 0  0 1− 0  −  0 0 1   0 0 0        9. Suppose A is an n n matrix and let v be a vector. Consider the A cyclic set of × m 1 m vectors v,Av, ,A − v where this is an independent set of vectors but A v is a linear combination··· of the preceding vectors in the list. Show how to obtain a monic polynomial of smallest degree, m, φv (λ) such that

φv (A) v = 0

Now let w , , w be a basis and let φ (λ) be the least common multiple of the { 1 ··· n} φwk (λ) . Explain why this must be the minimal polynomial of A. Give a reasonably easy algorithm for computing φv (λ). You could simply form the cyclic set of vectors described, make them the columns of a matrix and then do the row reduced echelon form to reveal linear relations. This is how you can compute φv (λ). Then, having found this, for each vector in a basis, and letting φ (λ) be the least common multiple, it follows that φ (A) v = 0 for each v in a basis and so φ (A) = 0. Also, if η (A) = 0, then η (λ) must be divisible by each of the φv (λ) for v in the given basis. Here is why.

η (λ) = l (λ) φv (λ) + r (λ) , degree of r (λ) < degree of φv (λ) or r (λ) = 0

Then if r (λ) = 0, you could contradict the fact that φv (λ) has smallest degree out of 6 all those φ (λ) for which φ (A) v = 0. Therefore, η (λ) is divisible by each φv (λ) for each v in the given basis. The polynomial of smallest such degree is by deﬁnition the least common multiple. 10. Here is a matrix. 7 1 1 −21 −3 −3  −70− 10− 10  Using the process of Problem 9 ﬁnd the minimal polynomial of this matrix. It turns out the characteristic polynomial is λ3.

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1 0 0 1 7 1 1 1 0 0 0 1 0 1 0 0 −21 −3 −3 0 0 0 0 0  0 0 1   0   −70− 10− 10   0   0 0 0   0   1 7 0 1 7 0     1 0 0 − − 0 21 0 , 0 21 0 , row echelon form: 0 1 0 It fol-  0   −70   0   0− 70 0   0 0 0  2 lows that  φe1 (λ) , λ = 0 + 0λ. The same sort of thing will happen withe2, e3. Therefore, the minimal polynomial is just λ2. Note that

2 7 1 1 0 0 0 −21 −3 −3 = 0 0 0  −70− 10− 10   0 0 0      11. Find the minimal polynomial for 1 2 3 A = 2 1 4  3 2 1  −   by the above technique. Is what you found also the characteristic polynomial?

First here is the string of vectors which comes from e1 1 1 4 26 0 2 −8 −24  0   3   −2   − 6  − − −  1 1 4  26    1 0 0 14 − − − 0 2 8 24 , row echelon form: 0 1 0 0  0 3 −2 − 6   0 0 1 3  − − −  2 3   φe (λ): 14 + 3λ = λ 1 − The minimal polynomial is then λ3 3λ2 +14. We know it can’t have any larger degree and so this must be it. − 3 2 1 2 3 1 2 3 2 1 4 3 2 1 4  3 2 1  −  3 2 1  − −     0 1 2 3 0 0 0 +14 2 1 4 = 0 0 0  3 2 1   0 0 0  − This is also the characteristic equation. Note that no computation involving determinants was required. 12. Let A be an n n matrix with ﬁeld of scalars C. Letting λ be an eigenvalue, show the dimension of× the eigenspace equals the number of Jordan blocks in the Jordan canonical form which are associated with λ. Recall the eigenspace is ker (λI A) . − Let the blocks associated with λ in the Jordan form be

J1 (λ) J2 (λ)  .  ..    J (λ)   r   

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Thus this is an upper triangular matrix which has λ down the main diagonal. If you take away λI from it, you get a matrix of the form

N1 (λ) N2 (λ)  .  ..    N (λ)   r    where 0 1 0 . 0 0 .. N (λ) =   i .  .. 1     0 0      How many columns of zeros will there be? Exactly r where r is the number of those blocks. You get one at the beginning of each block. Hence there will be exactly r free variables and so the dimension of the eigenspace will equal to the number of blocks. 13. For any n n matrix, why is the dimension of the eigenspace always less than or equal to the× algebraic multiplicity of the eigenvalue as a root of the characteristic equation? Hint: Note the algebraic multiplicity is the size of the appropriate block in the Jordan form. This is obvious from the above problem. The dimension of the eigenspace equals the number of blocks which must be no larger than the length of the diagonal. 14. Give an example of two nilpotent matrices which are not similar but have the same minimal polynomial if possible. 2 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 = ,  0 0 0 0   0 0 0 0   0 0 0 0   0 0 0 0       2   0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 =  0 0 0 1   0 0 0 0   0 0 0 0   0 0 0 0      These are two which work.  15. Use the existence of the Jordan canonical form for a linear transformation whose minimal polynomial factors completely to give a proof of the Cayley Hamilton theorem which is valid for any field of scalars. Hint: First assume the minimal polynomial factors completely into linear factors. If this does not happen, consider a splitting field of the minimal polynomial. Then consider the minimal polynomial with respect to this larger field. How will the two minimal polynomials be related? Show the minimal polynomial always divides the characteristic polynomial. First suppose the minimal polynomial factors. Then the linear transformation has a Jordan form. J (λ1) . J =  ..  J (λ )  r   

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where each block is as described in the description of the Jordan form. Say J (λi) is ri ri in size. Then the characteristic polynomial is the determinant of λI J a matrix× of the form −

(λ λ1) I + N (λ1) − .  ..  (λ λ ) I + N (λ )  − r r    where the blocks N (λi) have some 1’s on the super diagonal and zeros elsewhere. Thus the characteristic polynomial is− the determinant of this matrix which is

r q (λ) = (λ λ )ri − i i=1 Y Then q (J) is given by r (J λ I)ri − i i=1 Y The matrix J λ I is of the form − i J (λ λ ) 1 − i ..  .  J (0)    .   ..     J (λ λ )   r − i    ri ri where the block J (0) is nilpotent with J (0) = 0. It follows that (J λiI) is of the form − J (λ λ )ri 1 − i ..  .  0    .   ..     J (λ λ )ri   r − i  Therefore, the product r (J λ I)ri = 0 from block multiplication. i=1 − i Now in the general caseQ where the minimal polynomial pF (λ) does not necessarily factor, consider a splitting field for the minimal polynomial. The new minimal polynomial divides the old one. Therefore, the new minimal polynomial with respect to this splitting field G can be completely factored. Hence the Jordan form exists with respect to this new splitting field and new minimal polynomial pG (λ) . Then from the above, it follows that pG (J) = 0. However, pG (λ) divides pF (λ) and so it is also the case that pF (J) = 0. Since J is a matrix of T with respect to a suitable basis, it follows that pF (T ) = 0 also, where T is the original linear transformation. 16. Here is a matrix. Find its Jordan canonical form by directly finding the eigenvectors and generalized eigenvectors based on these to find a basis which will yield the Jordan form. The eigenvalues are 1 and 2. 3 2 5 3 −1− 0 1 2  −4 3 6 4  − −  1 1 1 3   − −   

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Why is it typically impossible to ﬁnd the Jordan canonical form? 3 2 5 3 4 1 −1− 0 1 2 1 0  − , eigenvectors:   1,   2 4 3 6 4 3 ↔ 1 ↔ − −      1 1 1 3   1   0   − −      Now it is necessary to ﬁnd generalized eigenvectors.       3 2 5 3 1 0 0 0 x1 4 −1− 0 1 2 0 1 0 0 x 1  −     2  =   4 3 6 4 − 0 0 1 0 x3 3 − −  1 1 1 3   0 0 0 1   x4   1   − −         4 2 5 3  x 4    4tˆ 4 1 4 − −1 −1 1 2 x 1 tˆ + 1  − −   2  =  , Solution is:  4  4 3 5 4 x3 3 3tˆ 2 − − 4 −  1 1 1 2   x4   1   tˆ   − −       4  A generalized eigenvector  is      4 −1  2  −  0      3 2 5 3 1 0 0 0 x1 1 −1− 0 1 2 0 1 0 0 x 0  −  2    2  =   4 3 6 4 − 0 0 1 0 x3 1 − −  1 1 1 3   0 0 0 1   x4   0   − −                5 2 5 3 x1 1 tˆ3 −1 −2 1 2 x 0 1  − −   2  =  , Solution is:   4 3 4 4 x 1 tˆ − − 3 3  1 1 1 1   x4   0   1   − −        A generalized eigenvector  is      0 1  0   1    So let the matrix be as follows   4 4 1 0 4 4 1 0 1 −1 0 1 1− 1 0 1 ,  3   2   1   0   3 2 1 0  − −  1   0   0   1   1 0 0 1                      4 4 1 0 1− 1 0 1 S =  3 2 1 0  −  1 0 0 1      1 Now to get the Jordan form you take S− AS, but of course we know the Jordan form at this point. There will be two blocks, one for 1 and one for 2. Nevertheless, here it

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is 1 4 4 1 0 − 3 2 5 3 4 4 1 0 1− 1 0 1 −1− 0 1 2 1− 1 0 1  3 2 1 0   −4 3 6 4   3 2 1 0  − − − −  1 0 0 1   1 1 1 3   1 0 0 1     − −          1 1 0 0 0 1 0 0 =  0 0 2 1   0 0 0 2      17. People like to consider the solutions of ﬁrst order linear systems of equations which are of the form x′ (t) = Ax (t) where here A is an n n matrix. From the theorem on the Jordan canonical form, ×1 1 there exist S and S− such that A = SJS− where J is a Jordan form. Deﬁne 1 y (t) S− x (t) . Show y′ = Jy. ≡ This is easy. You have 1 x′ = SJS− x and so 1 1 S− x ′ = J S− x , y′ = Jy

Now suppose Ψ (t) is an n n matrix whose columns are solutions of the above diﬀerential equation. Thus × Ψ′ = AΨ 1 Now let Φ be deﬁned by SΦS− = Ψ. Show

Φ′ = JΦ.

You have 1 Φ = S− ΨS and so

1 1 1 1 Φ′ = S− Ψ′S = S− AΨS = S− ASΦS− S = JΦ

18. In the above Problem show that

det (Ψ)′ = trace (A) det (Ψ)

and so det (Ψ (t)) = Cetrace(A)t This is called Abel’s formula and det (Ψ (t)) is called the Wronskian. trace(J)t It is easiest to consider Φ′ = JΦ and verify that det (Φ (t)) = Ce . This will do it because J and A are similar and so are Φ and Ψ.

n det (Φ)′ = det Φi i=1 X

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th where Φi has the i row diﬀerentiated and the other rows of Φ left unchanged. Thus, th th letting φi denote the i row of Φ, you have that the j entry of φi′ is given by

φij′ (t) = JikΦkj = λiΦij + aiΦ(i+1)j Xk Thus φi′ (t) = λiφi (t) + aiφi+1 (t)

where ai either is 0 or 1. Then since the determinant of a matrix having two equal rows equals 0, it follows that

n (det (Φ))′ = λi detΦ = trace(J) det (Φ) i=1 X and so det (Φ) = Cetrace(J)t

19. Let A be an n n matrix and let J be its Jordan canonical form. Recall J is a block × diagonal matrix having blocks Jk (λ) down the diagonal. Each of these blocks is of the form λ 1 0 . λ .. J (λ) =   k .  .. 1     0 λ      Now for ε > 0 given, let the diagonal matrix Dε be given by

1 0 ε Dε =  .  ..    0 εk 1   −    1 Show that Dε− Jk (λ) Dε has the same form as Jk (λ) but instead of ones down the super diagonal, there is ε down the super diagonal. That is Jk (λ) is replaced with

λ ε 0 ..  λ .  .  .. ε     0 λ      Now show that for A an n n matrix, it is similar to one which is just like the Jordan canonical form except instead× of the blocks having 1 down the super diagonal, it has ε. 1 That D− J (λ) D is of the right form is easy to see by considering the 4 4 case. ε k ε × 1 0 0 0 λ 1 0 0 1 0 0 0 1 0 ε− 0 0 0 λ 1 0 0 ε 0 0  2     2  0 0 ε− 0 0 0 λ 1 0 0 ε 0 3 3  0 0 0 ε−   0 0 0 λ   0 0 0 ε             

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1 0 0 0 λ ε 0 0 1 2 0 ε− 0 0 0 λε ε 0 =  2   2 3  0 0 ε− 0 0 0 λε ε 3 3  0 0 0 ε−   0 0 0 λε          λ ε 0 0 0 λ ε 0 =  0 0 λ ε   0 0 0 λ    Finally, the Jordan form is of the form 

J1 0 .  ..  0 J  m    where the blocks Jk are as just described. So multiply on the right by

Dε1 0 .  ..  0 D  εm    and on the left by the inverse of this matrix. Using block mutiplication you get the desired modification. Note that this shows that every matrix in n n matrix having entries in C is similar to one which is close to a diagonal matrix. × 20. Let A be in (V,V ) and suppose that Apx = 0 for some x = 0. Show that Ape = 0 L 6 6 k 6 for some ek e1, , en , a basis for V . If you have a matrix which is nilpotent, (Am = 0 for∈ some { m···) will} it always be possible to find its Jordan form? Describe how to do it if this is the case. Hint: First explain why all the eigenvalues are 0. Then consider the way the Jordan form for nilpotent transformations was constructed in the above. Yes, it is possible to compute the Jordan form. In fact, the proof of existence tells how to do it. A nilpotent matrix has all eigenvalues equal to 0. This is because if Ax =λx, then Akx = λkx and so for large k, you get λkx = 0. Finding the eigenvalues is always the difficulty. 21. Suppose A is an n n matrix and that it has n distinct eigenvalues. How do the minimal polynomial and× characteristic polynomials compare? Determine other conditions based on the Jordan Canonical form which will cause the minimal and characteristic polynomials to be different. If the eigenvalues are distinct, then the two polynomials are obviously the same. More generally, A has a Jordan form

J (λ1) .  ..  J (λ )  r    where each J (λ ) is an r r block diagonal matrix consisting of Jordan blocks, the k k × k λk being the distinct eigenvalues. Corresponding to one of these J (λk) you get the term (λ λ )rk − k

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in the characteristic equation. If J (λk) is of the form

λk 1 ..  λk .  .  .. 1     λ   k    then J λ I is of the form − i J (λ λ ) 1 − i ..  .  J (0)    .   ..     J (λ λ )   r − i    where J (0)ri = 0 and no smaller exponent will work. Therefore, the characteristic polynomial and the minimal polynomial will coincide. However, if J (λk) is composed of many blocks, the extreme case being when it is a diagonal matrix having λk down the diagonal, a smaller exponsne will work and so there will be a diﬀerence between the characteristic and minimal polynomials. For example, consider the minimal polynomial for the matrix

0 1 0 0 0 0 0 0 N =  0 0 0 1   0 0 0 0      The characteristic polynomial is just λ4 but N 2 = 0. If you had an unbroken string of ones down the diagonal, then the minimal polynomial would also be λ4. Getting diﬀerence between the two polynomials involves having lots of repeated eigenvalues and lots of blocks for the Jordan form for each eigenvalue. 22. Suppose A is a 3 3 matrix and it has at least two distinct eigenvalues. Is it possible that the minimal× polynomial is diﬀerent than the characteristic polynomial? You might have distinct eigenvalues in which case the two polynomials are the same or you would have the following Jordan forms.

a 1 0 a 0 0 0 a 0 or 0 a 0  0 0 b   0 0 b      In the ﬁrst case, the two polynomials are the same. The characteristic polynomial is

(λ a)2 (λ b) − − In the ﬁrst case a 1 0 a 0 0 a 1 0 b 0 0 0 a 0 0 a 0 0 a 0 0 b 0  0 0 b  −  0 0 a   0 0 b  −  0 0 b         

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0 a b 0 − = 0 0 0 = 0  0 0 0  6 and so the characteristic polynomial and minimal polynomial are the same. In the second case however,

a 0 0 a 0 0 a 0 0 b 0 0 0 a 0 0 a 0 0 a 0 0 b 0  0 0 b  −  0 0 a   0 0 b  −  0 0 b          0 0 0 = 0 0 0  0 0 0  and so the minimal polynomial is different than the characteristic polynomial 23. If A is an n n matrix of entries from a field of scalars and if the minimal polynomial of A splits over× this field of scalars, does it follow that the characteristic polynomial of A also splits? Explain why or why not. If the minimal polynomial splits, then the matrix has a Jordan form with respect to this field of scalars. Thus both the characteristic polynomial and minimal polynomial are of the form m (λ λ )ri − i i=1 Y Hence the characteristic polynomial also splits. 24. In proving the uniqueness of the Jordan canonical form, it was asserted that if two n n matrices A, B are similar, then they have the same minimal polynomial and also that× if this minimal polynomial is of the form

s ri p (λ) = φi (λ) i=1 Y ri ri where the φi (λ) are irreducible, then ker (φi (A) ) has the same dimension as ker (φi (B) ) . Why is this so? This was what was responsible for the blocks corresponding to an eigenvalue being of the same size. Since the two are similar, they come from the same linear transformation T . Therefore, 1 this is pretty obvious. Also, if B = S− AS, then

ri 1 ri 1 ri ker (φi (B) ) = ker φi S− AS = ker S− (φi (A) ) S ri This clearly has the same dimension as ker (φi (A) ) . 25. Show that a given matrix is non defective (diagonalizable) if and only if the minimal polynomial has no repeated roots. If the matrix is non defective, then its Jordan form is

J (λ1) .  ..  J (λ )  r   

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where each J (λk) is a diagonal matrix, J (λk) of size rk rk. Then the minimal polynomial is × (λ λ ) . − i i Y Conversely, suppose the minimal polynomial is of this form. Also suppose that some J (λi) say J (λi) is not diagonal. Then consider

(J λ I) − i i Y The ith term in the product is

J (λ λ ) 1 1 − i ..  .  J (0)  i   .   ..     J (λ λ )   r r − i    where for some i, Ji (0) is not the zero matrix. It follows that the above product cannot be zero either. In fact, you could let x Fri be such that J (0) x = y = 0. Then ∈ 6 0 0 . .  .   .  (J λ I) x = z , z = 0 − i     6 i  .   .  Y  .   .       0   0          because J (λ λ ) is one to one on Fri whenever k = i. It follows that each of the i i − k 6 J (λk) must be a diagonal matrix. 26. Describe a straight forward way to determine the minimal polynomial of an n n × matrix using row operations. Next show that if p (λ) and p′ (λ) are relatively prime, then p (λ) has no repeated roots. With the above problem, explain how this gives a way to determine whether a matrix is non defective. You take the matrix A and consider the sequence

I, A, A2,A3, ,An ··· There is some linear combination of these which equals 0. You just want to ﬁnd the one which involves the lowest exponents. Thus you string these out to make them 2 vectors in Fn and then make them the columns of a n2 n matrix. Find the row reduced echelon form of this matrix. This will allow you to× identify a linear relation which gives the minimal polynomial.

If p (λ) and p′ (λ) are relatively prime, then you would need to have

p (λ) = (λ λ ) − i i Y the λi being distinct. Thus the minimal polynomial would have no repeated roots and so the matrix is non defective by the above problem.

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27. In Theorem 10.3.4 show that the cyclic sets can be arranged in such a way that the

length of βxi+1 divides the length of βxi . You make this condition a part of the induction hypothesis. Then at the end when m 1 you extend to φ (A) − (V ) ker (φ (A)) , you note that, thanks to Lemma 10.3.2, ⊆ m 1 each of the β for y ker φ (A) − has length equal to a multiple of d, the degree yi i ∈ m 1 m 1 of φ (λ) while the β for x φ (A) − (V ) ker φ (A) − each have length equal to x ∈ \ d thanks to the assumption that φ (λ) is irreducible over the ﬁeld of scalars and each of these is in ker (φ (A)). 28. Show that if A is a complex n n matrix, then A and AT are similar. Hint: Consider a Jordan block. Note that × 0 0 1 λ 1 0 0 0 1 λ 0 0 0 1 0 0 λ 1 0 1 0 = 1 λ 0  1 0 0   0 0 λ   1 0 0   0 1 λ          Such a backwards identity will work for any size Jordan block. Therefore, J is similar to J T because you can use a similarity matrix which is block diagonal, each block being appropriate for producing the transpose of the corresponding Jordan block. Thus there is a matrix S such that

1 T J = S− J S

1 Now also J = P − AP for a suitable P. Therefore,

1 1 1 T P − AP = S− P − AP S

and so

1 T T T 1 1 A = PS− P A P − SP − 1 T 1 1 − T T 1 1 = P − SP − A P − SP − 29. Let A be a linear transformation deﬁned on a ﬁnite dimensional vector space V . Let the minimal polynomial be q mi φi (λ) i=1 Y i i i i and let βvi , , βvi be the cyclic sets such that βvi , , βvi is a basis for 1 ··· ri 1 ··· ri mi i ker (φi (A) ). Let v = i j vj . Now let q (λ) be anyn polynomial ando suppose that P P q (A) v = 0

Show that it follows q (A) = 0. n 1 First suppose that V is a vector space with a cyclic basis x, Ax, ,A − x such that Anx is a linear combination of the vectors just listed. Let φ···(λ) be the monic polynomial of degree n which results from this. Thus φ (A) x = 0. Then actually, φ (A) = 0. This is because if v is arbitrary, then

n 1 − i v = aiA x. i=0 X

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Then n 1 − i φ (A) v = aiA φ (A) x = 0 i=0 X Now suppose that q (A) x = 0. Does it follow that q (A) = 0?

q (λ) = φ (λ) l (λ) + r (λ)

where r (λ) either has smaller degree than φ (λ) or r (λ) = 0. Suppose that r (λ) = 0. Then 6 r (A) x = 0 and the degree of r (λ) is less than n which contradicts the linear independence of

n 1 x, Ax, ,A − x ··· Hence r (λ) = 0. It follows that φ(λ) divides q (λ) and so q (A) = 0. i Now consider the problem of interest. Since each span β i is invariant, it follows vj i i i that q (A) vj span βvi . Therefore, 0 = i,j q (A) vj . By independence, each vector ∈ j in the sum equals 0 and so from the ﬁrst part of the argument q (A) restricted to mi P ker (φi (A) ) equals 0. Since the whole space is a direct sum of these, it follows that q (A) = 0.

F.28 Exercises

10.9

1. Letting A be a complex n n matrix, in obtaining the rational canonical form, one obtains the Cn as a direct sum× of the form

span βx span βx 1 ⊕ · · · ⊕ r m 1 m where βx is an ordered cyclic set of vectors, x,Ax, ,A − x such that A x is in the span of the previous vectors. Now apply the Gram··· Schmidt process to the ordered

basis βx1 , βx2 , , βxr , the vectors in each βxi listed according to increasing power of A, thus obtaining··· an ordered basis (q , , q ) . Letting Q be the unitary matrix 1 ··· n which has these vectors as columns, show that Q∗AQ equals a matrix B which satisﬁes Bij = 0 if i j 2. Such a matrix is called an upper Hessinberg matrix and this shows that every n− ≥n matrix is orthogonally similar to an upper Hessinberg matrix. × You note that qk is in the span of

βx1 βx2 x , ,Am1 x , x , ,Am2 x , , x ,Ax , ,Aj x 1 ··· 1 2 ··· 2 ··· ··· l l ··· l z }| { z }| { q where there are k vectors in the above list. Since each span βxi is A invariant, A k is in the span of

βx1 βx2 x , ,Am1 x , x , ,Am2 x , , x ,Ax , ,Aj x ,Aj+1x 1 ··· 1 2 ··· 2 ··· ··· l l ··· l l and by thez Gram}| Schmidt{ z process,}| this{ is the same as

span (q , , q ) 1 ··· k+1

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It follows that if p k + 2, then ≥ qp∗Aqk = 0 th Thus the pk entry of Q∗AQ equals 0 if p k 2. That is, this matrix is upper Hessinberg. − ≥

2. In the argument for Theorem 10.2.4 it was shown that m (A) φl (A) v = v whenever rk rk v ker (φk (A) ) . Show that m (A) restricted to ker (φk (A) ) is the inverse of the ∈ rk linear transformation φl (A) on ker (φk (A) ) . In this theorem you had polynomials m (λ) , n (λ) such that

rk m (λ) φl (λ) + n (λ) φk (λ) = 1

Thus you also have rk m (A) φl (A) + n (A) φk (A) = I If x ker (φ (A)rk ) , then if follows that ∈ k =0

rk m (A) φl (A) x + n (A) φk (A) x = x z }| { and so m (A) is indeed the inverse of φl (A). 3. Suppose A is a linear transformation and let the characteristic polynomial be

q det (λI A) = φ (λ)nj − j j=1 Y

1 2 3 A = 2 1 4  3 2 1  −   by the above technique with the ﬁeld of scalars being the rational numbers. Is what you found also the characteristic polynomial? 1 2 3 2 1 4 , characteristic polynomial: X3 3X2 + 14  3 2 1  − − It is irreducible over the rationals by the rational root theorem. Therefore, the above must be the minimal polynomial as well.

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5. Show, using the rational root theorem, the minimal polynomial for A in the above problem is irreducible with respect to Q. Letting the field of scalars be Q find the rational canonical form and a similarity transformation which will produce it. To find the rational canonical form, you need to look for cyclic sets. You have a single irreducible polynomial, λ3 3λ + 14 = φ (λ). Also φ (A) sends every vector to 0. − Therefore, you can simply start with a vector. Take e1.

2 1 1 2 3 1 1 2 3 1 0 , 2 1 4 0 , 2 1 4 0  0   3 2 1   0   3 2 1   0  − −           3 1 2 3 1 2 1 4 0  3 2 1   0  − These vectors are     1 1 4 26 − − 0 , 2 , 8 , 24  0   3   −2   − 6  − − − The last is definitely a linear combination   of the first  three because the first three are an independent set. Hence, you ought to have these as columns. Then you have

1 1 1 4 − 1 2 3 1 1 4 0 2 −8 2 1 4 0 2 −8  0 3 −2   3 2 1   0 3 −2  − − − − −       0 0 14 − = 1 0 0  0 1 3  and this is the rational canonical form for this matrix. 6. Find the rational canonical form for the matrix 1 2 1 1 2 3 0− 2  1 3 2 4   1 2 1 2      This was just pulled out of the air. To ﬁnd the rational canonical form, look for cyclic sets.

2 1 0 0 0 1 1 2 1 1 1 1 2 1 1 1 0 1 0 0 0 2 3 0− 2 0 2 3 0− 2 0 , ,  0 0 1 0   0   1 3 2 4   0   1 3 2 4   0   0 0 0 1   0   1 2 1 2   0   1 2 1 2   0                          3 4 1 2 1 1 1 1 2 1 1 1 2 3 0− 2 0 2 3 0− 2 0 ,  1 3 2 4   0   1 3 2 4   0   1 2 1 2   0   1 2 1 2   0                 

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These reduce to 1 1 5 30 181 0 2 10 56 336 , , , ,  0   1   13   93   600   0   1   8   54   343                      1 1 5 30 181 1 0 0 0 3 0 2 10 56 336 0 1 0 0 −1 , row echelon form:  0 1 13 93 600   0 0 1 0 −11  −  0 1 8 54 343   0 0 0 1 8      Thus the minimal polynomial is  

λ4 8λ3 + 11λ2 + λ + 3 − Then you can find the rational canonical form from this. Also, 1 1 1 5 30 − 1 2 1 1 1 1 5 30 0 0 0 3 0 2 10 56 2 3 0− 2 0 2 10 56 1 0 0 −1 =  0 1 13 93   1 3 2 4   0 1 13 93   0 1 0 −11  −  0 1 8 54   1 2 1 2   0 1 8 54   0 0 1 8                  7. Let A : Q3 Q3 be linear. Suppose the minimal polynomial is (λ 2) λ2 + 2λ + 7 . Find the rational→ canonical form. Can you give generalizations of− this rather simple problem to other situations? That second factor is irreducible over the rationals and so you would need the following for the rational canonical form. 2 0 0 0 0 7  0 1 −2  −   8. Find the rational canonical form with respect to the field of scalars equal to Q for the matrix 0 0 1 A = 1 0 1  0 1− 1    Observe that this particular matrix is already a companion matrix of λ3 λ2 + λ 1. Then find the rational canonical form if the field of scalars equals C or Q−+ iQ. − First note that the minimal polynomial is λ3 λ2 + λ 1 and a use of the rational root theorem shows that this factors over the rationals− as− (λ 1) λ2 + 1 . Then you − need to locate cyclic sets starting with a vector in ker A2 + I . Lets find this first. 2 0 0 1 1 0 0 1 1 1 1 0 1 + 0 1 0 = 0 0 0  0 1− 1   0 0 1   1 1 1        Clearly the kernel is the vectors of the form

s t − s−  t   

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Starting with one of these vectors, lets ﬁnd cyclic sets.

1 2 1 0 0 1 1 0 0 1 1 −1 , 1 0 1 −1 , 1 0 1 −1  0   0 1− 1   0   0 1− 1   0            1 0 1 −1 , 1 , 1  0   −1   −0  These of course will be dependent  since otherwise  we don’t really have the minimal polynomial. Thus the cycle can consist of only the first two. Thus we will have these two as the first two columns and then we will have the eigenvector corresponding to ker (A I) − 1 0 1 ker −1 1 1  0− 1− 0    1 This eigenvector is just 0 . Then the rational canonical form of this matrix is  1    1 1 0 1 − 0 0 1 1 0 1 −1 1 0 1 0 1 −1 1 0  0− 1 1   0 1− 1   0− 1 1        1 1 1 2 2 2 0 0 1 1 0 1 − 1 1 1 − = 2 2 2 1 0 1 1 1 0  −1 −1 1   −   −  2 2 2 0 1 1 0 1 1       0 1 0 1− 0 0  0 0 1  If the field of scalars is Q + iQ, then the minimal polynomial factors as (λ 1) (λ + i)(λ i) and so the rational canonical form is just the diagonal matrix, −1 0 0 − 0 i 0  0 0 i  −   9. Let q (λ) be a polynomial and C its companion matrix. Show the characteristic and minimal polynomial of C are the same and both equal q (λ). n 1 n Let q (λ) = a0 + a1λ + + an 1λ − + λ . Then its companion matrix is ··· − 0 0 a ··· − 0 1 0 a1 C =  . · · · −.  .. .    0 1 an 1   − −    Then you can show by induction that the characteristic polynomial of C is q (λ). See Problem 43 on Page 66 It was also shown that q (C) = 0. (Cayley Hamilton theorem for companion matrices.) Let p (λ) be the minimal polynomial. Thus it has degree no more than the degree of q (λ) and thus must divide q (λ). Could it have smaller degree

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than q (λ)? No, it couldn’t. To see this, consider, for example a companion matrix for a monic polynomial of degree 4 denoted as A. The following gives the sequence A0, A, A2,A3. Note that these are linearly independent. Watch the ones to see this. 1 0 0 0 0 0 0 a 0 0 a ad 0 1 0 0 1 0 0 b 0 0 b a + bd  0 0 1 0   0 1 0 c   1 0 c b + cd   0 0 0 1   0 0 1 d   0 1 d d2 + c         0 a ad  a d2 + c    0 b a + bd b d2 + c + ad   0 c b + cd a + c d2 + c + bd  1 d d2 + c b + d d2 + c + cd      This pattern will occur for any sized companion matrix. You start with ones down the main diagonal and then the string of ones moves down one diagonal and then another and so forth. After n 1 multiplications, you get a 1 in the lower left corner. Each multiplication moves the− ones into slots which had formerly been occupied with zeros. That is why these are independent. Therefore, the degree of the minimal polynomial is at least n and so the minimal and characteristic polynomials must coincide for any companion matrix. 10. Use the existence of the rational canonical form to give a proof of the Cayley Hamilton theorem valid for any field, even fields like the integers mod p for p a prime. The earlier proof was fine for fields like Q or R where you could let λ but it is not clear the same result holds in general. → ∞ The linear transformation has a matrix of the following form,

M1 . M =  ..  M  q    mk where Mk comes from ker(φk (λ) ) , φk (λ) being irreducible over the ﬁeld of scalars. Say the minimal polynomial is

q mk p (λ) = φk (λ) kY=1 Furthermore, Mk is of the form

k C1 .  ..  Ck  lk    k k k rj where Cj is the companion matrix of ηj (λ) where ηj (λ) = φk (λ) with rj mk. By k rj ≤ the above problem, the minimum polynomial for Cj is just φk (λ) , rj mk which is also the characteristic polynomial of this matrix. The characteristic polynomial≤ is then q q lk q (λ) = det (λI M ) = det λI Ck − k − j k=1 i=1 j=1 Y Y Y q lk q lk = ηk (λ) = φ (λ)rj , r m j k j ≤ k j=1 j=1 kY=1 Y kY=1 Y

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It follows that q li rj q (A) = φk (A) i=1 j=1 Y Y k rj We know that φk Cj = 0.Also when you have a block diagonal matrix, B1 . B =  ..  B  q    and φ (λ) is a polynomial,

φ (B1) . φ (B) =  ..  φ (B )  q    Therefore, the above product for q (A) is the product of block diagonal matrices such that every position on the diagonal has a zero matrix in at least one of the block matrices in the product. This equals the zero matrix. Like this:

0 0 0 D 0 0 F 0 0 0 B 0 0 0 0 0 H 0  0 0 C   0 0 E   0 0 0        When you multiply the block matrices, it involves simply multiplying together the blocks in the corresponding positions. Since at least one is zero, it follows the product is the zero matrix. 11. Suppose you have two n n matrices A, B whose entries are in a field F and suppose G is an extension of F. For× example, you could have F = Q and G = C. Suppose A and B are similar with respect to the field G. Can it be concluded that they are similar with respect to the field F? Hint: Use the uniqueness of the rational canonical form.

Do A, B have the same minimal polynomial over F? Denote these by pA (λ) and pB (λ) You have pA (A) = 0

and so pA (B) = 0 also because A and B are similar with respect to G. It follows that pA (λ) must divide pB (λ) . Similar reasoning implies that pB (λ) must divide pA (λ). Therefore, these two minimal polynomials having coeﬃcients in F coincide. Otherwise, you could write pB (λ) = pA (λ) + r (λ)

where the degree of r (λ) is smaller than the common degree of pA (λ) and pB (λ). Then plugging in λ = B, you get a contradiction to the deﬁnition of pB (λ) as having the smallest degree possible. Say the common minimal polynomial over F is

q mi φi (λ) i=1 Y

where φi (λ) is irreducible over F. Thus, each of these φi is a polynomial with coeﬃ- cients in F and hence in G, and so

φi (A) , φi (B)

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G are similar over . Letting βvi and βwj denote the A cyclic sets associated with

mi mi ker (φi (A)) , ker (φi (B))

respectively, it follows from Lemma 10.8.3 that the lengths of these βvi and βwi com- prise exactly the same set of positive numbers. These lengths are identiﬁed with the size of the Jordan blocks of the two similar matrices, φi (A) , φi (B). However, the proof of this lemma exhibits a basis which yields the Jordan form, which happens to involve only the ﬁeld F. The companion matrices corresponding to two of these which have the same length are therefore exactly the same, both being companion matrices k of φi (λ) for a suitable k. It follows that the rational canonical form for A, B over F is exactly the same. Hence, by uniqueness, the two matrices are similar over F.

F.29 Exercises

11.4 1. Suppose the migration matrix for three locations is .5 0 .3 .3 .8 0 .  .2 .2 .7  Find a comparison for the populations in the three locations after a long time. .5 0 .3 1 0 0 0.5 0 0.3 .3 .8 0 0 1 0 = −0.3 0.2 0  .2 .2 .7  −  0 0 1   0.2− 0.2 0.3  −  1 3    3  2 0 10 0 1 0 5 0 −3 1 0 0 , row echelon form: 0 1 − 9 0  10 5   10  1 −1 3 0 0 0− 0 0 5 5 − 10 It follows that the populations are   .6 t .9  1  where t is chosen such that the sum of the entries corresponds to the total amount at the beginning.

2. Show that if i aij = 1, then if A = (aij) , then the sum of the entries of Av equals the sum of the entries of v. Thus it does not matter whether aij 0 for this to be so. P ≥ i j aijvj = j i aijvj = j vj n 3.P If APsatisﬁes theP conditionsP of theP above problem, can it be concluded that limn A exists? →∞ Not necessarily. Consider the following matrix. 4 9 3 8 − − 4 4 9 311 936 = etc. The eigenvalues of this matrix are 3 8 −312− 937 − − 1, 5. Thus there is no way that the limit can exist. Just do the powers of the matrix to− the eigenvector which corresponds to 5. −

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4. Give an example of a non regular Markov matrix which has an eigenvalue equal to 1. − 1 0 0 0 0 1 , eigenvalues: 1, 1  0 1 0  −   5. Show that when a Markov matrix is non defective, all of the above theory can be proved n very easily. In particular, prove the theorem about the existence of limn A if the eigenvalues are either 1 or have absolute value less than 1. →∞ n n It suﬃces to show that limn A v exists for each v R . This is because you could T→∞n ∈ th n then assert that limn ei A ej exists. But this is just the ij entry of A . Let the eigen pairs be →∞ (1, v ) , (λ , v ) , i = 2, , n 1 i i ··· and by assumption, these vi form a basis. Therefore, if v is any vector, there exist unique scalars ci such that v = civi i X It follows that n Amv = c v + c λmv c v 1 1 i i → 1 1 j=2 X Thus this is really easy. 6. Find a formula for An where

5 1 2 2 0 1 5− 0 0 −4 A =  7 1 1 −5  2 − 2 2 − 2  7 1 0 2   2 − 2 −    n Does limn A exist? Note that all the rows sum to 1. Hint: This matrix is similar to a diagonal→∞ matrix. The eigenvalues are 1, 1, 1 , 1 . − 2 2 5 1 2 2 0 1 5− 0 0 −4  7 1 1 −5  2 − 2 2 − 2  7 1 0 2   2 − 2 −   1 1 1 0  1 0 0 0 1 0 0 1 3 2 1 0 −0 1 0 0 −1 1 0 2 = 2  2 5 1 1   0 0 1 0   1 1 0− 1  4 −  2 1 1 0   0 0 0 1   1 1 1 1     2   − 4 − 4 − 2  Now it follows that     1 1 1 0 ( 1)n 0 0 0 1 0 0 1 − 1 − n 3 2 1 0 0 2n 0 0 1 1 0 2 A =  5     −  2 4 1 1 0 0 1 0 1 1 0 1 1 1 −1 1  2 1 1 0   0 0 0 n   1     2   − 4 − 4 − 2  1 n  1 n 2   2n ( 1) + 1 2n 1 0 ( 1) 2n + 1 2 − − n 2 − − n− 4 2n 3 ( 1) + 1 2n 1 0 3 ( 1) 2n + 1 =  1 − − n 1 − 1 − n − 3  2n 2 ( 1) + 1 2n 1 2n 2 ( 1) 2n + 1 1 − − n 1 − − n − 2  n 2 ( 1) + 1 n 1 0 2 ( 1) n + 1   2 − − 2 − − − 2   

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7. Find a formula for An where

1 1 2 2 2 1 4− 0 1 −4 A =  5 1 1 −2  2 − 2 −  3 1 1 2   − 2 2 −    Note that the rows sum to 1 in this matrix also. Hint: This matrix is not similar to a diagonal matrix but you can ﬁnd the Jordan form and consider this in order to obtain a formula for this product. The eigenvalues are 1, 1, 1 , 1 . − 2 2 1 1 2 2 2 1 4− 0 1 −4  5 1 1 −2  2 − 2 −  3 1 1 2   − 2 2 −   7 1 7 9 1 6 9 1 0 0 0 1 0 0 1 −23 1 − 1 14 −0 1 0 0 1 1 0− 1 = 9 3 9  − 13 1 − 1 4   0 0 1 1   0 −6 14 20  − 9 − 6 9 2 − −  16 1 1 7   0 0 0 1   1 0 3 2   − 9 − 6 9   2   −  Now consider the Jordan block on the bottom.   It when raising this to the nth power, you get

2 1 k n 2n−k 0 0 1 1 k 0 n−k 0 0 k=0 2 X 1 1 2n 0 2n−1 0 0 1 = 1 + n 1 0 0 −1 0 0 2n 2n 1 21 nn = 2n − 0 1 2n It follows that An is given by the formula 7 1 7 n 9 1 6 9 ( 1) 0 0 0 1 0 0 1 −23 − 1 14 − − 9 1 3 9 0 1 0 0 1 1 0 1  − 13 − 1 4   1 1 n   −  9 1 6 9 0 0 2n 2 − n 0 6 14 20 − 16 − 1 7 1 − −  1   0 0 0 n   1 0 3 2   − 9 − 6 9   2   −   7 1 1 n 7  n 1 1 1n 7 n 16  1 1 n 9 2n 6 2 − n 9 ( 1) + 1 2n 1 2 2 − n 9 ( 1) 9 2n 3 2 − n + 1 14× − 1 1 n − 23 − n 2 − 1 n 23 − n− ×32 − 2 1 n 9 2n 3 2 − n 9 ( 1) + 1 2n 1 2 − n 9 ( 1) 9 2n 3 2 − n + 1  ×4 − 1 1 n − 13 − n 1 − 1 1 n 1 13 − n − ×22 − 1 1 n  9 2n 6 2 − n 9 ( 1) + 1 2n 1 2 2 − n + 2n 9 ( 1) 9 2n 3 2 − n + 1 ×7 − 1 1 n − 16 − n 1 − 1 1 n 16 − n − ×16 − 1 1 n  n 2 − n ( 1) + 1 n 1 2 − n ( 1) n 2 − n + 1   9 2 − 6 − 9 − 2 − 2 9 − − 9 2 − 3   × ×  n 8. Find limn A if it exists for the matrix →∞ 1 1 1 2 2 2 0 1 −1 − 1 2 2 2 0 A =  −1 1 −3  2 2 2 0  3 3 3 1   2 2 2    1 The eigenvalues are 2 , 1, 1, 1. The matrix equals

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1 1 1 0 1 0 0 0 1 0 1 0 − − 1 1 1 0 0 0 2 0 0 1 1 1 0  −2 1 1 0   0 0 1 0   1 1 0 0  − −  4 3 1 1   0 0 0 1   0 2 1 1   −     −  Hence the matrix raised  to the nth power  is  1 1 1 0 1 0 0 0 1 0 1 0 −1 1− 0 0 0 1 0 0 1 1 1 0 2n and so, in the limit,  −2 1 1 0   0 0 1 0   1 1 0 0  − −  4 3 1 1   0 0 0 1   0 2 1 1   −     −  you get      1 1 1 0 1 0 0 0 1 0 1 0 −1 1− 0 0 0 0 0 0 1 1 1 0  −2 1 1 0   0 0 1 0   1 1 0 0  − −  4 3 1 1   0 0 0 1   0 2 1 1   −     −   0 1 1 0     1− 0 −1 0 =  −1 1− 2 0   3 3 3 1      9. Given an example of a matrix A which has eigenvalues which are either equal to 1, 1, − n or have absolute value strictly less than 1 but which has the property that limn A does not exist. →∞ 1 0 0 An easy example is 0 0 1 , eigenvalues: 1, 1  0 1 0  − n 1 0 0   1 0 0 0 0 1 alternates between the identity and 0 0 1 .  0 1 0   0 1 0      10. If A is an n n matrix such that all the eigenvalues have absolute value less than 1, × n show limn A = 0. →∞ The Jordan form is J (λ1) .  ..  J (λ )  r   m  where each λ < 1. Then as shown above, J (λ ) 0. | r| r → n 11. Find an example of a 3 3 matrix A such that limn A does not exist but 5r × →∞ limr A does exist. →∞ An easy example is the following. ei(2π/5) 0 0 A = 0 ei(2π/5) 0   0 0 ei(2π/5)   5 ei(2π/5) 0 0 A5 = 0 ei(2π/5) 0 = I.   0 0 ei(2π/5)  5r  r Therefore, limr A exists. However, limr A cannot exist because it will just yield diagonal matrices→∞ which have various ﬁfth→∞ roots of 1 on the diagonal.

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12. If A is a Markov matrix and B is similar to A, does it follow that B is also a Markov matrix? No. Start with a Markov matrix 1/2 1/3 1/2 2/3 Now take a similarity transformation of it. 1 1 2 − 1/2 1/3 1 2 1 1 = 2 3 4 1/2 2/3 3 4 −1 −5 3 This is certainly not a Markov matrix. 13. In Theorem 11.1.3 suppose everything is unchanged except that you assume either a 1 or a 1. Would the same conclusion be valid? What if you don’t j ij ≤ i ij ≤ insist that each aij 0? Would the conclusion hold in this case? P P ≥ You can’t conclude that the limit exists if the sign of aij is not restricted. For example,

4 9 3 8 − − has an eigenvalue of 5 and so the limit of powers of this matrix cannot exist. What − about the case where i aij 1 instead of equal to 1? In this case, you do still get the conclusion of this theorem.≤ The condition was only used to get the entries of An bounded independentP of n and this can be accomplished just as well with the inequality.

14. Let V be an n dimensional vector space and let x V and x = 0. Consider βx m 1 ∈ 6 ≡ x,Ax, ,A − x where ··· m m 1 A x span x,Ax, ,A − x ∈ ··· and m is the smallest such that the above inclusion in the span takes place. Show m 1 that x,Ax, ,A − x must be linearly independent. Next suppose v , , v ··· { 1 ··· n} mi v is a basis for V . Consider βvi as just discussed, having length mi. Thus A i is a m 1 linearly combination of v ,Av , ,A − v for m as small as possible. Let pv (λ) be i i ··· i i the monic polynomial which expresses this linear combination. Thus pvi (A) vi = 0

and the degree of pvi (λ) is as small as possible for this to take place. Show that the minimal polynomial for A must be the monic polynomial which is the least common

multiple of these polynomials pvi (λ).

Let p (λ) be the minimal polynomial. Then there exists ki (λ) , a monic polynomial such that

p (λ) = ki (λ) pvi (λ) + ri (λ)

where ri (λ) must be zero since otherwise, you would have ri (A) vi = 0 and this would

contradict the minimality of the degree of pvi (λ). Therefore, each of these pvi (λ) divides p (λ) . If q (λ) is their least common multiple, then it follows that q (A) = 0 because every vector is a linear combination of these vectors vi. Therefore, p (λ) divides q (λ). Thus q (λ) = p (λ) k (λ)

Now each pvi (λ) divides p (λ) so it is a common multiple. It follows that k (λ) = 1 since otherwise, q (λ) wouldn’t really be the least common multiple.

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15. If A is a complex Hermitian n n matrix which has all eigenvalues nonnegative, show that there exists a complex Hermitian× matrix B such that BB = A.

You have A = U ∗DU for D a diagonal matrix having all nonnegative entries. Then 1/2 1/2 just note that B = U ∗D UU ∗D U,U is unitary. Clearly you should take B = 1/2 U ∗D U. 16. Suppose A, B are n n real Hermitian matrices and they both have all nonnegative eigenvalues. Show that× det (A + B) det (A) + det (B). ≥ Let P 2 = A, Q2 = B where P,Q are Hermitian and nonnegative. Then P A + B = PQ Q Now apply the Cauchy Binet formula to both sides. 2 2 2 det (A + B) = det (Ci) det (P ) + det (Q) i ≥ X = det P 2 + det Q2 = det (A) + det (B)

where the Ci are n n submatrices of PQ . Note that this does not depend on the matrices being real.× You could write P A + B = P Q ∗ ∗ Q Then

det (A + B) = det (Ci∗) det (Ci) = det (Ci∗Ci) i i X X det P 2 + det Q2 = det (A) + det (B) ≥ α c 17. Suppose B = ∗ is an (n + 1) (n + 1) Hermitian nonnegative matrix where b A × α is a scalar and A is n n. Show that α must be real, c = b, and A = A∗,A is nonnegative, and that if α×= 0, then b = 0.

α b∗ B∗ = = B and so you must have α is real and b = c while A = A∗. c A∗ Suppose α = 0 and b = 0. I need to show that B is not nonnegative. 6 B 0 b α α x ∗ ∗ b A x z }| { b x = α x ∗ = αb x+αx b + x Ax ∗ αb + Ax ∗ ∗ ∗ If for some x, x∗Ax < 0, then let α = 0 and you see that B is not nonnegative. Therefore, A must be nonnegative. But now you could let x = b and then pick α be large and negative and again see that B is not nonnegative. Hence if α = 0, then b = 0. Now consider b = 0 and determine the sign of α. Consider 6 B α b β β β 0 ∗ = βα βb ∗ b A 0 ∗ 0 z }| { = β 2α and so for this to be nonnegative, you have to have α > 0.

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18. If A is an n n complex Hermitian and nonnegative matrix, show that there exists ↑ × an upper triangular matrix B such that B∗B = A. Hint: Prove this by induction. It is obviously true if n = 1. Now if you have an (n + 1) (n + 1) Hermitian nonnegative × α2 αb matrix, then from the above problem, it is of the form ∗ , α real. αb A From the above problem, a generic (n + 1) (n + 1) Hermitian nonnegative matrix is of the form just claimed where A is nonnegative× and Hermitian and α is real. In case α = 0, there is nothing to prove. You would just use

0 0∗ 0 0∗ 0 B∗ 0 B where B∗B = A. Otherwise, you consider

α 0∗ α b∗ b B∗ 0 B 19. Suppose A is a nonnegative Hermitian matrix which is partitioned as ↑ A A A = 11 12 A A 21 22

where A11,A22 are square matrices. Show that det(A) det (A11) det (A22). Hint: Use the above problem to factor A getting ≤

B 0 B B A = 11∗ ∗ 11 12 B∗ B∗ 0 B 12 22 22

Next argue that A11 = B11∗ B11,A22 = B12∗ B12 + B22∗ B22. Use the Cauchy Binet theorem to argue that det(A22) = det (B12∗ B12 + B22∗ B22) det (B22∗ B22) . Then explain why ≥

det (A) = det (B11∗ ) det (B22∗ ) det (B11) det (B22)

= det (B11∗ B11) det (B22∗ B22)

The Hint tells pretty much how to do it. A11 = B11∗ B11,A22 = B12∗ B12+B22∗ B22 follows from block multiplication. Why is the determinant of a block triangular matrix equal B B to the product of the determinants of the blocks? Consider 11 12 . If the 0 B 22 rank of B22 is not maximal, then its determinant is zero and also the determinant of the matrix is zero. Assume then that it has full rank. Then suitable row operations can be applied to obtain that the determinant of this matrix equals the determinant of B11 0 0 B 22 This determinant is clearly equal to the product of the determinants of B11 and B22. The inequality det (B12∗ B12 + B22∗ B22) det (B22∗ B22) follows from the Cauchy Binet theorem since there are more nonnegative≥ terms in the sum for the determinant on the left than for the term on the right. Note that these determinants are all real because the matrices are Hermitian. It follows that

det (A) = det (B11∗ B11) det (B22∗ B22) det (A ) det (B∗ B + B∗ B ) = det (A ) det (A ) ≤ 11 12 12 22 22 11 22

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20. Prove the inequality of Hadamard. If A is a Hermitian matrix which is nonnegative, ↑then det (A) A ≤ ii i Y This follows from induction. It is clearly true if n = 1. Now let A be n n, nonnegative and Hermitian and assume the theorem is true for all nonnegative matrices× which are of smaller size. Then B a A = a∗ α It follows that B and α are both nonnegative. Then from the above,

det (A) det (B) det (α) ≤ By induction, this implies det (A) A ≤ ii i Y F.30 Exercises

12.7

1. Find the best solution to the system

x + 2y = 6 2x y = 5 3x +− 2y = 0

T 1 2 1 2 14 6 2 1 2 1 =  −   −  6 9 3 2 3 2     T 1 2 6 14 6 x Solve = 2 1 5 6 9 y  −    3 2 0 14 6 x 16    17 = , Solution is: 15 6 9 y 7 1 45 3 2. Find an orthonormal basis for R , w1, w2, w3 given that w1 is a multiple of the vector (1, 1, 2). { } 1 0 1 A basis consists of 1 , 1 , 0        2 0 0        1 0 1   1 1 0  2 0 0   1 √  1 √ √ 2 √ √ 1 √ 1 √ 6 6 30 5 6 5 5 6 6 6 6 6 1 √ −1 √ √ 1 √ √ 1 √ √ = 6 6 6 5 6 0 0 6 5 6 30 5 6  1 √ 1 √ √ 1 √   − 2 √  3 6 15 5 6 5 5 0 0 5 5  − −   

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Then one example of an orthonormal basis extending the given vector is 1 √6 1 √5√6 2 √5 6 − 30 5 1 √6 , 1 √5√6 , 0  6   6    1 √6 1 √5√6 1 √5 3 15 − 5    −    3. Suppose A = AT is a symmetric real n n matrix which has all positive eigenvalues. Deﬁne × (x, y) (Ax, y) . ≡ Show this is an inner product on Rn. What does the Cauchy Schwarz inequality say in this case? It satisﬁes all the axioms of an inner product obviously. The only ones which are not obvious are whether (Ax, x) 0 and so forth. But A = U ∗DU where D is diagonal and U is orthogonal. Therefore,≥ 2 2 (Ax, x) = (U ∗DUx, x) = (DUx,Ux) µ Ux = µ x ≥ | | | | where µ is the smallest eigenvalue which is assumed positive. Therefore, also (Ax, x) = 0 if and only if x = 0. The Cauchy Schwarz inequality says | | (Ax, y) (Ax, x)1/2 (Ay, y)1/2 | | ≤ 4. Let x max xj : j = 1, 2, , n . || ||∞ ≡ {| | ··· } T Show this is a norm on Cn. Here x = x x . Show 1 ··· n x x (x, x)1/2 || ||∞ ≤ | | ≡ where the above is the usual inner product on Cn. The axioms of a norm are all obvious except for the triangle inequality. However,

x + y = max xi + yi , i n x + y k k∞ {| | ≤ } ≤ k k∞ k k∞ so even this one is all right. Also for some i 1/2 1/2 2 2 x = xi = xi xi = x . k k∞ | | | | ≤ | | | | ! i " X 5. Let n x x . || ||1 ≡ | j | j=1 X T Show this is a norm on Cn. Here x = x x . Show 1 ··· n x x (x, x)1/2 || ||1 ≥ | | ≡ where the above is the usual inner product on Cn. It is obvious that is a norm. What of the inequality? Is k·k1 1/2 x x 2 ? | i| ≥ | i| i ! i " X X Of course. Just square both sides and the left has mixed terms which are not present on the right.

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6. Show that if is any norm on any vector space, then ||·|| x y x y . ||| || − || ||| ≤ || − || x = x y + y x y + y and so k k k − k ≤ k − k k k x y x y k k − k k ≤ k − k Now repeat the argument with the x, y switched. Finally, x y equals either x y or y x either way, you get the inequality. |k k − k k| k k − k k k k − k k 7. Relax the assumptions in the axioms for the inner product. Change the axiom about (x, x) 0 and equals 0 if and only if x = 0 to simply read (x, x) 0. Show the Cauchy Schwarz≥ inequality still holds in the following form. ≥

(x, y) (x, x)1/2 (y, y)1/2 . | | ≤ The proof given included this case.

n 8. Let H be an inner product space and let uk k=1 be an orthonormal basis for H. Show { } n (x, y) = (x, uk) (y, uk). kX=1

You have x = j (x, uj) uj and y = j (y, uj) uj . Therefore, P P (x, y) = (x, u ) u , (y, u ) u  j j k k j j X X  n = (x, uj ) (y, uk)(uj, uk) = (x, uk) (y, uk) Xj,k kX=1 9. Let the vector space V consist of real polynomials of degree no larger than 3. Thus a typical vector is a polynomial of the form

a + bx + cx2 + dx3.

For p, q V deﬁne the inner product, ∈ 1 (p, q) p (x) q (x) dx. ≡ Z0 Show this is indeed an inner product. Then state the Cauchy Schwarz inequality in terms of this inner product. Show 1, x, x2, x3 is a basis for V . Finally, ﬁnd an orthonormal basis for V. This is an example of some orthonormal polynomials. p1 (x) = 1

x (x,p1)p1 x (1/2) p (x) = − = − 1 2 = √3 (2x 1) 2 x (x,p1)p1 1 2 / k − k x 1 − R0 ( − 2 ) 2 2 2 x (x ,√3(2x 1))√3(2x 1) (x ,1)1 x2 x+ 1 p (x) = − − − − = − 6 3 x2 (x2,√3(2x 1))√3(2x 1) (x2,1)1 x2 1 √3√3(2x 1) 1 k − − − − k k − 6 − − 3 k = 6√5 x2 x + 1 − 6

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x3 (x3,6√5(x2 x+ 1 ))6√5(x2 x+ 1 ) (x3,√3(2x 1))√3(2x 1) 1 p (x) = − − 6 − 6 − − − − 4 4 x3 (x3,6√5(x2 x+ 1 ))6√5(x2 x+ 1 ) (x3,√3(2x 1))√3(2x 1) 1 k − − 6 − 6 − − − − 4 k x3 1 √56√5 x2 x+ 1 3 √3√3(2x 1) 1 x3 1 √56√5 x2 x+ 1 3 √3√3(2x 1) 1 − 20 ( − 6 )− 20 − − 4 − 20 ( − 6 )− 20 − − 4 = 3 1 2 1 3 1 = 1 x √56√5(x x+ ) √3√3(2x 1) 2800 √2800 k − 20 − 6 − 20 − − 4 k = 20√7 x3 3 x2 + 3 x 1 − 2 5 − 20 It is clear the vectors given form a basis. The orthonormal basis is 1 3 3 1 1, √3 (2x 1) , 6√5 x2 x + , 20√7 x3 x2 + x − − 6 − 2 5 − 20

10. Let Pn denote the polynomials of degree no larger than n 1 which are defined on an interval [a, b] . Let x , , x be n distinct points in [a, b−] . Now define for p, q P , { 1 ··· n} ∈ n n (p, q) p (x ) q (x ) ≡ j j j=1 X Show this yields an inner product on Pn. Hint: Most of the axioms are obvious. The one which says (p, p) = 0 if and only if p = 0 is the only interesting one. To verify this one, note that a nonzero polynomial of degree no more than n 1 has at most n 1 zeros. − − If (p, p) = 0 then p equals zero at n distinct points and so it must be the zero polynomial because it has degree at most n 1 and so it can have no more than n 1 roots. − − 11. Let C ([0, 1]) denote the vector space of continuous real valued functions defined on [0, 1]. Let the inner product be given as 1 (f, g) f (x) g (x) dx ≡ Z0 Show this is an inner product. Also let V be the subspace described in Problem 9. Using the result of this problem, find the vector in V which is closest to x4. It equals x4, 1 1 + x4, √3 (2x 1) √3 (2x 1) − − 1 1 + x4, 6√5 x2 x + 6√5 x2 x + − 6 − 6 3 3 1 3 3 1 + x4, 20√7 x3 x2 + x 20√7 x3 x2 + x − 2 5 − 20 − 2 5 − 20 Now it is just a matter of working these out. x4, √3 (2x 1) = 1 x4√3 (2x 1) = 2 √3 − 0 − 15 1 x4, 6√5 x2 x+ 1R = x46√5 x2 x + 1 = 2 √5 − 6 0 − 6 35 1 x4, 20√7 x3 3 x2 + 3 xR 1 = x420√7 x3 3 x2 + 3 x 1 = 1 √7 − 2 5 − 20 0 − 2 5 − 20 70 Therefore, the closest point is R 1 2 2 1 + √3√3 (2x 1) + √56√5 x2 x + + 5 15 − 35 − 6 1 3 3 1 + √720√7 x3 x2 + x 70 − 2 5 − 20 9 2 1 = 2x3 x2 + x − 7 7 − 70

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12. A regular Sturm Liouville problem involves the diﬀerential equation, for an un- known function of x which is denoted here by y,

(p (x) y′)′ + (λq (x) + r (x)) y = 0, x [a, b] ∈ and it is assumed that p (t) , q (t) > 0 for any t [a, b] and also there are boundary conditions, ∈

C1y (a) + C2y′ (a) = 0

C3y (b) + C4y′ (b) = 0

where 2 2 2 2 C1 + C2 > 0, and C3 + C4 > 0. There is an immense theory connected to these important problems. The constant, λ is called an eigenvalue. Show that if y is a solution to the above problem corresponding to λ = λ and if z is a solution corresponding to λ = λ = λ , then 1 2 6 1 b q (x) y (x) z (x) dx = 0. (6.37) Za and this deﬁnes an inner product. Hint: Do something like this:

(p (x) y′)′ z + (λ1q (x) + r (x)) yz = 0,

(p (x) z′)′ y + (λ2q (x) + r (x)) zy = 0.

Now subtract and either use integration by parts or show

(p (x) y′)′ z (p (x) z′)′ y = ((p (x) y′) z (p (x) z′) y)′ − −

and then integrate. Use the boundary conditions to show that y′ (a) z (a) z′ (a) y (a) = − 0 and y′ (b) z (b) z′ (b) y (b) = 0. The formula, 6.37 is called an orthogonality relation. It turns out there− are typically inﬁnitely many eigenvalues and it is interesting to write given functions as an inﬁnite series of these “eigenfunctions”. Let y go with λ and z go with µ.

z (p (x) y′)′ + (λq (x) + r (x)) yz = 0

y (p (x) z′)′ + (µq (x) + r (x)) zy = 0

Subtract. z (p (x) y′)′ y (p (x) z′)′ + (λ µ) q (x) yz = 0 − − Now integrate from a to b. First note that d z (p (x) y′)′ y (p (x) z′)′ = (p (x) y′z p (x) z′y) − dx − and so what you get is

p (b) y′ (b) z (b) p (b) z′ (b) y (b) (p (a) y′ (a) z (a) p (a) z′ (a) y (a)) − − − b + (λ µ) q (x) y (x) z (x) dx = 0 − Za

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Look at the stuﬀ on the top line. From the assumptions on the boundary conditions,

C1y (a) + C2y′ (a) = 0

C1z (a) + C2z′ (a) = 0

and so y (a) z′ (a) y′ (a) z (a) = 0 − Similarly, y (b) z′ (b) y′ (b) z (b) = 0 − Hence, that stuﬀ on the top line equals zero and so the orthogonality condition holds. 13. Consider the continuous functions deﬁned on [0, π] ,C ([0, π]) . Show

π (f, g) fgdx ≡ Z0 is an inner product on this vector space. Show the functions 2 sin (nx) ∞ are π n=1 an orthonormal set. What does this mean about the dimensionnq of the vectoro space C ([0, π])? Now let V = span 2 sin (x) , , 2 sin (Nx) . For f C ([0, π]) ﬁnd N π ··· π ∈ a formula for the vector in VN whichq is closest to fqwith respect to the norm determined from the above inner product. This is called the N th partial sum of the Fourier series of f. An important problem is to determine whether and in what way this Fourier series converges to the function f. The norm which comes from this inner product is sometimes called the mean square norm. Note that sin (nx) is a solution to the boundary value problem

2 y′′ + n y = 0, y (0) = y (π) = 0

It is obvious that sin (nx) solves this boundary value problem. Those boundary conditions in the above problem are implied by these. In fact,

1 sin 0 + (0) cos (0) = 0 1 sin π + (0) cos π = 0

Then it follows that π sin (nx) sin (mx) dx = 0 unless n = m. Z0 π 2 1 √2 Now also 0 sin (nx) dx = 2 π and so √π sin (nx) are orthonormal. n o 14. Consider theR subspace V ker (A) where ≡ 1 4 1 1 2 1− 2− 3 A =  4 9 0 1   5 6 3 4      Find an orthonormal basis for V. Hint: You might ﬁrst ﬁnd a basis and then use the Gram Schmidt procedure.

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9 1 4 1 1 x 0 tˆ3 − − − 7 2 1 2 3 y 0 4 tˆ     =  , Solution is:  7 3  4 9 0 1 z 0 tˆ3  5 6 3 4   w   0   0          This is pretty easy. The space has dimension 1. Therefore, an orthonormal basis is just 9 9 √146 − − 146 1 4 2 √146 = 73  7   7 √  √16 + 49 + 81 146 146  0   0          15. The Gram Schmidt process starts with a basis for a subspace v1, , vn and produces an orthonormal basis for the same subspace u , , u {such··· that } { 1 ··· n} span (v , , v ) = span (u , , u ) 1 ··· k 1 ··· k for each k. Show that in the case of Rm the QR factorization does the same thing. More speciﬁcally, if A = v v 1 ··· n and if A = QR q1 qn R ≡ ··· then the vectors q1, , qn is an orthonormal set of vectors and for each k, { ··· } span (q , , q ) = span (v , , v ) 1 ··· k 1 ··· k This follows from the way we multiply matrices and the deﬁnition of an orthogonal matrix. 16. Verify the parallelogram identify for any inner product space,

x + y 2 + x y 2 = 2 x 2 + 2 y 2 . | | | − | | | | | Why is it called the parallelogram identity? x + y 2 + x y 2 = (x + y, x + y) + (x y, x y) | | | − | − − = x 2 + y 2 + 2 (x, y) + x 2 + y 2 2 (x, y) . | | | | | | | | − 17. Let H be an inner product space and let K H be a nonempty convex subset. This means↑ that if k , k K, then the line segment⊆ consisting of points of the form 1 2 ∈ tk + (1 t) k for t [0, 1] 1 − 2 ∈ is also contained in K. Suppose for each x H, there exists P x deﬁned to be a point of K closest to x. Show that P x is unique so∈ that P actually is a map. Hint: Suppose z1 and z2 both work as closest points. Consider the midpoint, (z1 + z2) /2 and use the parallelogram identity of Problem 16 in an auspicious manner.

k1+k2 Suppose there are two closest points to x say k1, k2 both work. Then 2 is also in K and k + k 2 x k x k 2 x 1 2 = 1 + 2 − 2 2 − 2 2 − 2

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k k 2 x k 2 x k 2 = 2 − 1 + 2 1 + 2 2 − 2 2 − 2 2 − 2 2 k2 k1 = − + d2 − 2

where d is the distance between x and ki. 18. In the situation of Problem 17 suppose K is a closed convex subset and that H is↑ complete. This means every Cauchy sequence converges. Recall from calculus a sequence k is a Cauchy sequence if for every ε > 0 there exists N such that { n} ε whenever m, n > Nε, it follows km kn < ε. Let kn be a sequence of points of K such that | − | { } lim x kn = inf x k : k K n →∞ | − | {| − | ∈ } This is called a minimizing sequence. Show there exists a unique k K such that ∈ limn kn k and that k = P x. That is, there exists a well deﬁned projection map onto→∞ the| convex− | subset of H. Hint: Use the parallelogram identity in an auspicious manner to show kn is a Cauchy sequence which must therefore converge. Since K is closed it follows{ this} will converge to something in K which is the desired vector. Let x be given and let k be a minimizing squence. { n} x k λ inf x y : y K | − n| → ≡ {| − | ∈ } Then k k 2 x k x k 2 x k 2 x k 2 n − m + − n + − m = 2 − n + 2 − m 2 2 2 2 2

Now it follows that

k k 2 1 1 k + k 2 n − m = x k 2 + k x 2 x n m 2 2 | − n| 2 | m − | − − 2

1 1 x k 2 + k x 2 λ 2 ≤ 2 | − n| 2 | m − | −

The right side converges to 0 as n, m and so kn is a Cauchy sequence and since the space is given to be complete,→ it follows ∞ that{ this} converges to some k. Since K is closed, it follows that k K. Now ∈

x k = lim x kn = λ n | − | →∞ | − | and so there exists a closest point. 19. Let H be an inner product space which is also complete and let P denote the projection↑ map onto a convex closed subset, K. Show this projection map is characterized by the inequality Re (k P x, x P x) 0 − − ≤ for all k K. That is, a point z K equals P x if and only if the above variational inequality∈ holds. This is what that∈ inequality is called. This is because k is allowed to vary and the inequality continues to hold for all k K. ∈ Consider for t [0, 1] the following. ∈ y (x + t (w x)) 2 | − − |

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where w K and x K. It equals ∈ ∈ f (t) = y x 2 + t2 w x 2 2t Re y x, w x | − | | − | − h − − i

Suppose x is the point of K which is closest to y. Then f ′ (0) 0. However, f ′ (0) = 2 Re y x, w x . Therefore, if x is closest to y, ≥ − h − − i Re y x, w x 0. h − − i ≤ Next suppose this condition holds. Then you have

y (x + t (w x)) 2 y x 2 + t2 w x 2 y x 2 | − − | ≥ | − | | − | ≥ | − | By convexity of K, a generic point of K is of the form x+ t (w x) for w K. Hence x is the closest point. − ∈ 20. Using Problem 19 and Problems 17 - 18 show the projection map, P onto a closed convex↑ subset is Lipschitz continuous with Lipschitz constant 1. That is

P x P y x y | − | ≤ | − |

P x P y,y P y 0 h − − i ≤ P y P x,x P x 0 h − − i ≤ Thus P x P y,x P x 0 h − − i ≥ Hence P x P y,x P x P x P y,y P y 0 h − − i − h − − i ≥ and so P x P y,x y (P x P y) 0 h − − − − i ≥ x y P x P y P x P y,P x P y = P x P y 2 | − | | − | ≥ h − − i | − | 21. Give an example of two vectors in R4 x, y and a subspace V such that x y = 0 but P x P y = 0 where P denotes the projection map which sends x to its closest· point on V . · 6

Try this. V is the span of e1 and e2 and x = e3 + e1, y = e4 + e1.

P x = (e3 + e1, e1) e1 + (e3 + e1, e2) e2 = e1

P y = (e4 + e1, e1) e1 + (e4 + e1, e2) e2 = e1 P x P y = 1 · 22. Suppose you are given the data, (1, 2) , (2, 4) , (3, 8) , (0, 0) . Find the linear regression line using the formulas derived above. Then graph the given data along with your regression line. You draw the graphs. You want to solve

a + b = 2 2a + b = 4 3a + b = 8 b = 0

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Of course there is no solution. The least squares solution is to solve T T 1 1 1 1 1 1 2 2 1 2 1 a 2 1 4 =  3 1   3 1  b  3 1   8   0 1   0 1   0 1   0           14 6  a  34   13  = , Solution is: 5 Then the best line is 6 4 b 14 2 − 5 13 2 y = x 5 − 5

23. Generalize the least squares procedure to the situation in which data is given and you desire to fit it with an expression of the form y = af (x)+bg (x) +c where the problem would be to find a, b and c in order to minimize the error. Could this be generalized to higher dimensions? How about more functions? Say you had ordered pairs given which you wanted the curve to go through. Say these are (x , y ) for i n. Then you would be finding least squares solutions for a, b, c to i i ≤ y = af (x ) + bg (x ) + c, i = 1, , n i i i ··· You would simply obtain a least squares solutions to this. The principles are the same for higher dimensions. 24. Let A (X,Y ) where X and Y are finite dimensional vector spaces with the dimension of∈X Lequal to n. Define rank (A) dim (A (X)) and nullity(A) dim (ker (A)) . ≡ r ≡ Show that nullity(A) + rank (A) = dim (X) . Hint: Let xi i=1 be a basis for ker (A) r n r { } n r and let xi i=1 yi i=1− be a basis for X. Then show that Ayi i=1− is linearly independent{ } and∪ spans { } AX. { } Let Ay m be a basis for AX and let x r be a basis for ker (A). { i}i=1 { i}i=1 Then y1, , ym, x1, , xr is linearly independent. In fact it spans X because if z X,{Then··· ··· } ∈ m Az = ciAyi i=1 X m r and so z i=1 ciyi ker (X) and so it is in the span of the xi i=1 . Therefore, z is in the− span of the∈ given vectors and so this list of vectors{ is a} basis. Hence the dimension ofPX is equal to m + r = rank (A) + dim (ker (A)).

25. Let A be an m n matrix. Show the column rank of A equals the column rank of A∗A. × Next verify column rank of A∗A is no larger than column rank of A∗. Next justify the following inequality to conclude the column rank of A equals the column rank of A∗.

rank (A) = rank (A∗A) rank (A∗) ≤ ≤ = rank (AA∗) rank (A) . ≤ r n r Hint: Start with an orthonormal basis, Axj j=1 of A (F ) and verify A∗Axj j=1 n { } { } is a basis for A∗A (F ) . r Say you have i=1 ciA∗Axi = 0. Then for any appropriate y, P r r 0 = ciA∗Axi, y = ciAxi,Ay i=1 ! i=1 ! X X

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In particular, you could have y = i cixi. Then you would get P r ciAxi = 0 i=1 X and so each c = 0. Now if you have x Fn, Then Ax is a linear combination of the i ∈ Axj and so A∗Ax is a linear combination of the A∗Axi which shows that these vectors n are a basis for A∗AF . It follows that the rank of A and the rank of A∗A are the same. Similarly, the rank of A∗ and AA∗ are the same. Hence the above inequalities follow right away. 26. Let A be a real m n matrix and let A = QR be the QR factorization with Q orthogonal and R upper× triangular. Show that there exists a solution x to the equation

RT Rx = RT QT b

and that this solution is also a least squares solution deﬁned above such that AT Ax = AT b. There exists a soluton to this last equation. Hence you have a solution to

RT QT QRx = RT QT b

But this is just the equations RT Rx = RT QT b Of course this kind of thing is very easy for a computer to solve.

F.31 Exercises

12.9

1. Here are three vectors in R4 : (1, 2, 0, 3)T , (2, 1, 3, 2)T , (0, 0, 1, 2)T . Find the three dimensional volume of the parallelepiped determined− by these three vectors. (1, 2, 0, 3) (1, 2, 0, 3) = 14 · (1, 2, 0, 3) (2, 1, 3, 2) = 10 · − (1, 2, 0, 3) (0, 0, 1, 2) = 6 · (2, 1, 3, 2) (2, 1, 3, 2) = 18 − · − (2, 1, 3, 2) (0, 0, 1, 2) = 1 − · (0, 0, 1, 2) (0, 0, 1, 2) = 5 · 14 10 6 det 10 18 1 = 218  6 1 5    volume is √218 2. Here are two vectors in R4 : (1, 2, 0, 3)T , (2, 1, 3, 2)T . Find the volume of the parallelepiped determined by these two vectors. − (1, 2, 0, 3) (1, 2, 0, 3) = 14 · (1, 2, 0, 3) (2, 1, 3, 2) = 10 · −

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(2, 1, 3, 2) (2, 1, 3, 2) = 18 − · − 14 10 det = 152 10 18 volume is √152 3. Here are three vectors in R2 : (1, 2)T , (2, 1)T , (0, 1)T . Find the three dimensional volume of the parallelepiped determined by these three vectors. Recall that from the above theorem, this should equal 0. It does equal 0. 4. Find the equation of the plane through the three points (1, 2, 3) , (2, 3, 1) , (1, 1, 7) . − 5. Let T map a vector space V to itself. Explain why T is one to one if and only if T is onto. It is in the text, but do it again in your own words. This is because if T is one to one, it takes a basis to a basis. If T is onto, then it takes a basis to a spanning set which must be independent, since otherwise, there would be a linearly independent set of vectors which would also span and yet have fewer vectors than a basis. 6. Let all matrices be complex with complex ﬁeld of scalars and let A be an n n matrix ↑and B a m m matrix while X will be an n m matrix. The problem is to× consider solutions to× Sylvester’s equation. Solve the following× equation for X

AX XB = C − where C is an arbitrary n m matrix. If q (λ) is a polynomial, show first that if AX XB = 0, then q (A) X× Xq (B) = 0. Next define the linear map T which maps the n− m matrices to the n − m matrices as follows. × × TX AX XB ≡ − Show that the only solution to TX = 0 is X = 0 if and only if σ (A) σ (B) = . Conclude that there exists a unique solution for each C to Sylvester’s equation∩ if and∅ only if σ (A) σ (B) = . ∩ ∅ q (A) X Xq (B) = q (A) X = 0 − 1 Now explain why q (A)− exists if and only if σ (A) σ (B) = . ∩ ∅ n 1 n 1 n 1 Consider the first part. Multiply by A − . Then A − AX A − XB = 0. Since AX = XB, you can make multiple switches and write this as −

AnX XBn = 0 − Now it follows that the desired result holds for a polynomial. Consider T which maps the vector space of n m matrices to itself. Let q (λ) be the characteristic polynomial of B as suggested and× suppose TX = 0. Then

q (A) X Xq (B) = 0 = q (A) X. − m Let the characteristic polynomial for B be i=1 (λ λi) . In case there are no shared eigenvalues, it follows that − m Q q (A) = (A λ I) − i i=1 Y

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is invertible. Hence X = 0. Therefore, T is one to one and it follows that it is onto. If there is a shared eigenvalue, then the matrix q (A) will no longer be invertible. Hence there will exist nonzero solutions X to q (A) X = 0. It follows that in this case T is not one to one and so it is not onto either. 7. Compare Deﬁnition 12.8.2 with the Binet Cauchy theorem, Theorem 3.3.14. What is the geometric meaning of the Binet Cauchy theorem in this context? th Letting U be the matrix having ui as the i column the above deﬁnition states that 1/2 the volume is just det U T U and by the Cauchy Binet theorem, in the interesting case that U has more rows than columns, this equals 1/2 2 det (UI ) ! " XI where I is a choice of m rows which are selected from U so that UI is the square matrix which results from these rows. The geometric meaning of the above expression is this. det (UI ) gives the m dimensional volume of the parallelepiped which is obtained from a projection onto the m dimensional subspace of Rn determined by setting the other n m variables equal to 0. Thus the volume is the square root of the sum of the squares− of the areas of these projections.

F.32 Exercises

13.12

1. Show (A∗)∗ = A and (AB)∗ = B∗A∗.

(A∗)∗ x, y = (x, A∗y) = (Ax, y) . Since y is arbitrary, (A∗)∗ x = Ax.

(AB)∗ x, y = (x, ABy) = (A∗x, By) = (B∗A∗x, y) Since y is arbitrary, this shows (AB)∗ x = B∗A∗x and since x is arbitrary, this shows the formula. 2. Prove Corollary 13.3.9. This is just like the proof of the theorem.

+ 1 3. Show that if A is an n n matrix which has an inverse then A = A− . × 1 1 1 1 1 You have AA− A = A, A− AA− = A− , and AA− = I which is Hermitian and 1 A− A = I which is also Hermitian. 4. Using the singular value decomposition, show that for any square matrix A, it follows that A∗A is unitarily similar to AA∗.

You have A = U ∗ΣV where Σ is the singular value matrix and U,V are unitary of the 2 2 right size. Therefore, A∗A = V ∗ΣUU ∗ΣV = V ∗Σ V . Similarly, AA∗ = U ∗Σ U. Then 2 Σ = UAA∗U ∗ and so A∗A = V ∗UAA∗U ∗V

Since these matrices are all square, this does it. U ∗V is unitary. 5. Let A, B be a m n matrices. Deﬁne an inner product on the set of m n matrices by × × (A, B) trace (AB∗) . F ≡

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Show this is an inner product satisfying all the inner product axioms. Recall for M an n n n matrix, trace (M) i=1 Mii. The resulting norm, F is called the Frobenius norm× and it can be used≡ to measure the distance between||·|| two matrices. P This was done earlier. 6. Let A be an m n matrix. Show × A 2 (A, A) = σ2 || ||F ≡ F j j X

where the σj are the singular values of A. 2 A F = trace (AA∗) = trace (U ∗ΣVV ∗Σ∗U) = trace (ΣΣ∗) = sum of the squares of thek k singular values. 7. If A is a general n n matrix having possibly repeated eigenvalues, show there is a × sequence Ak of n n matrices having distinct eigenvalues which has the property {th } × th that the ij entry of Ak converges to the ij entry of A for all ij. Hint: Use Schur’s theorem.

A = U ∗TU where T is upper triangular and U is unitary. Change the diagonal entries of T slightly so that the resulting upper triangular matrix Tk has all distinct diagonal entries and T T in the sense that the ijth entry of T converges to the ijth entry k → k of T . Then let Ak = U ∗TkU. It follows that Ak A in the sense that corresponding entries converge. → 8. Prove the Cayley Hamilton theorem as follows. First suppose A has a basis of eigenvectors v n ,Av = λ v . Let p (λ) be the characteristic polynomial. Show { k}k=1 k k k p (A) vk = p (λk) vk = 0. Then since vk is a basis, it follows p (A) x = 0 for all x and so p (A) = 0. Next in the general case,{ } use Problem 7 to obtain a sequence A { k} of matrices whose entries converge to the entries of A such that Ak has n distinct eigenvalues and therefore by Theorem 7.1.7 Ak has a basis of eigenvectors. There- fore, from the ﬁrst part and for pk (λ) the characteristic polynomial for Ak, it follows pk (Ak) = 0. Now explain why and the sense in which

lim pk (Ak) = p (A) . k →∞

k k First say A has a basis of eigenvectors v1, , vn .A vj = λj vj . Then it follows { ··· } n that p (A) vk = p (λk) vk. Hence if x is any vector, let x = k=1 xkvk and it follows that P n n p (A) x = p (A) xkvk = xkp (A) vk k=1 ! k=1 n X n X = xkp (λk) vk = xk0vk = 0 Xk=1 Xk=1 Hence p (A) = 0. Now drop the assumption that A is nondefective. From the above, there exists a sequence Ak which is non defective which converges to A and also pk (λ) p (λ) uniformly on compact sets because these characteristic polynomials are deﬁned→ in terms of determinants of the corresponding matrix. See the above construction of the Ak. It is probably easiest to use the Frobinius norm for the last part. p (A ) p (A) p (A ) p (A ) + p (A ) p (A) k k k − kF ≤ k k k − k kF k k − kF

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The first term converges to 0 because the convergence of Ak to A implies all entries of Ak lie in a compact set. The second term converges to 0 also because the entries of Ak converge to the corresponding entries of A. 9. Prove that Theorem 13.4.6 and Corollary 13.4.7 can be strengthened so that the condition on the Ak is necessary as well as sufficient. Hint: Consider vectors of the x form where x Fk. 0 ∈ Say the matrix is positive definite. Then you could just look at

x xT 0 M (A) = xT M (A) x 0 k and this needs to be positive. Hence those principle minors have all positive determinants.

10. Show directly that if A is an n n matrix and A = A∗ (A is Hermitian) then all the eigenvalues and eigenvectors are× real and that eigenvectors associated with distinct eigenvalues are orthogonal, (their inner product is zero). λ (x, x) = (Ax, x) = (x,Ax) = λ¯ (x, x) so λ = λ¯. Now if x is an eigenvector, A¯x =Ax = λ¯x and so x + ¯x is also an eigenvector. Hence, you can assume that the eigenvectors are real. If ¯x is pure imaginary, you could simply multiply by i and get an eigenvector which is real. 11. Let v , , v be an orthonormal basis for Fn. Let Q be a matrix whose ith column 1 ··· n is vi. Show Q∗Q = QQ∗ = I.

This follows from how we multiply matrices. 12. Show that an n n matrix Q is unitary if and only if it preserves distances. This means Qv = v×. This was done in the text but you should try to do it for yourself. | | | | n 13. Suppose v1, , vn and w1, , wn are two orthonormal bases for F and suppose Q is{ an n··· n matrix} { satisfying··· Qv} = w . Then show Q is unitary. If v = 1, × i i | | show there is a unitary transformation which maps v to e1. This is easy because you show it preserves distances. 14. Finish the proof of Theorem 13.6.5.

15. Let A be a Hermitian matrix so A = A∗ and suppose all eigenvalues of A are larger than δ2. Show (Av, v) δ2 v 2 ≥ | | Where here, the inner product is

n (v, u) v u . ≡ j j j=1 X

Since it is Hermitian, there exists unitary U such that U ∗AU = D. Then

2 2 2 2 (Ax, x) = (UDU ∗x, x) = (DU ∗x,U ∗x) δ U ∗x = δ x ≥ | | | |

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16. Suppose A + A∗ has all negative eigenvalues. Then show that the eigenvalues of A have all negative real parts. You have

0 > ((A + A∗) x, x) = (Ax, x) + (A∗x, x) = (Ax, x) + (Ax, x) Now let Ax = λx. Then you get 0 > λ x 2 + λ¯ x 2 = Re (λ) x 2 | | | | | | 17. The discrete Fourier transform maps Cn Cn as follows. → n 1 1 − i 2π jk F (x) = z where z = e− n x . k √n j j=0 X 1 Show that F − exists and is given by the formula

n 1 − 2 1 1 i π jk F − (z) = x where x = e n z j √n k j=0 X Here is one way to approach this problem. Note z = Ux where

i 2π 0 0 i 2π 1 0 i 2π 2 0 i 2π (n 1) 0 e− n · e− n · e− n · e− n − · i 2π 0 1 i 2π 1 1 i 2π 2 1 ··· i 2π (n 1) 1  e− n · e− n · e− n · e− n − ·  1 i 2π 0 2 i 2π 1 2 i 2π 2 2 ··· i 2π (n 1) 2 U = e− n · e− n · e− n · e− n − · √n  ···   . . . .   . . . .   i 2π 0 (n 1) i 2π 1 (n 1) i 2π 2 (n 1) i 2π (n 1) (n 1)   e− n · − e− n · − e− n · − e− n − · −   ···    Now argue U is unitary and use this to establish the result. To show this verify each row has length 1 and the inner product of two diﬀerent rows gives 0. Now 2π 2π i n jk i n jk Ukj = e− and so (U ∗)kj = e . Take the inner product of two rows.

n 1 n 1 n 1 − 2 − 2 2 − 2 i π jk i 2π jl i π jk i π jl i π j(l k) e− n e− n = e− n e n = e− n − j=0 j=0 j=0 X X X i 2π n(l k) 1 e− n − = − = 0 if l = k i 2π (l k) 1 e n 6 − − − i 2π n(l k) n 1 1 1 n − because e− − = 1. What if l = k? then the sum reduces to j=0 √n √n = 1 and so this matrix is unitary. Therefore, its inverse is just the transpose conjugate P which yields the other formula. 18. Let f be a periodic function having period 2π. The Fourier series of f is an expression of the form n ∞ ikx ikx cke lim cke ≡ n k= →∞ k= n X−∞ X− and the idea is to ﬁnd ck such that the above sequence converges in some way to f. If

∞ ikx f (x) = cke k= X−∞

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imx and you formally multiply both sides by e− and then integrate from 0 to 2π, interchanging the integral with the sum without any concern for whether this makes sense, show it is reasonable from this to expect

2π 1 imx c = f (x) e− dx. m 2π Z0 Now suppose you only know f (x) at equally spaced points 2πj/n for j = 0, 1, , n. Consider the Riemann sum for this integral obtained from using the left endpoint··· of 2π n the subintervals determined from the partition n j j=0. How does this compare with the discrete Fourier transform? What happens as n to this approximation? → ∞ imx Multiply by e− and formally integrate both sides. This leads to the formula for cm. Now use a left sum. This would be

n 1 1 − 2π im 2π j 2π 2π f j e− n (j + 1) j 2π n n − n j=0 X n 1 1 − 2π im 2π j = f j e− n n n j=0 X 1 f Thus an approximation to cm comes from multiplying √n by the unitary matrix U th 2π above where f is the vector whose j component is f n j /√n. 19. Suppose A is a real 3 3 orthogonal matrix (Recall this means AAT = AT A = I.) having determinant 1.× Show it must have an eigenvalue equal to 1. Note this shows there exists a vector x = 0 such that Ax = x. Hint: Show ﬁrst or recall that any orthogonal matrix must6 preserve lengths. That is, Ax = x . | | | | If Ax = λx, then you can take the norm of both sides and conclude that λ = 1. It iθ iθ | | follows that the eigenvalues of A are e , e− and another one which has magnitude 1 and is real. This can only be 1 or 1. Since the determinant is given to be 1, it follows that it is 1. Therefore, there exists− an eigenvector for the eigenvalue 1. 20. Let A be a complex m n matrix. Using the description of the Moore Penrose inverse in terms of the singular× value decomposition, show that

1 + lim (A∗A + δI)− A∗ = A δ 0+ → where the convergence happens in the Frobenius norm. Also verify, using the singular value decomposition, that the inverse exists in the above formula. Recall that the Moore Penrose inverse is σ 1 0 V − U 0 0 ∗ σ 0 where A = U V . The matrices U,V are unitary and of the right size. First 0 0 ∗ of all, observe that there is no problem about the existence of the inverse mentioned in the formula. This is because

2 ((A∗A + δI) x, x) δ x ≥ | | and so it is a one to one matrix. Now consider the limit assertion. The left side equals

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1 σ 0 σ 0 − σ 0 lim V UU V ∗ + δI V U ∗ δ 0+ 0 0 0 0 0 0 → 1 σ2 0 − σ 0 = lim V V ∗ + δI V U ∗ δ 0 0 0 0 0 → Now 1 σ2 0 − σ 0 V V + δI V U 0 0 ∗ 0 0 ∗

1 σ2 0 − σ 0 = V V ∗ + δV V ∗ V U ∗ 0 0 0 0 1 σ2 0 − σ 0 = V + δI U 0 0 0 0 ∗

1 σ2 + δI 0 − σ 0 = V U 0 δI 0 0 ∗ 1 σ2 + δI − 0 σ 0 = V 1 U ∗ 0 δ− I 0 0 ! " 1 σ2 + δI − σ 0 = V U ∗ 0 0 ! 1 Of course σ and σ2 and so forth are diagonal matrices. Thus σ2 + δI − is also a 1 2 − diagonal matrix having on the diagonal an expression of the form! σi +δ σi and 1 this clearly converges as δ 0 to σ− . Therefore, the above converges to → i ! 1 σ− 0 + V U ∗ = A 0 0

+ + + 21. Show that A = (A∗A) A∗. Hint: You might use the description of A in terms of the singular value decomposition. σ 1 0 σ 0 Recall that A+ = V − U where A = U V . The σ was the 0 0 ∗ 0 0 ∗ diagonal matrix which consisting of square roots of eigenvalues of A∗A, arranged in decreasing order from top left toward lower right.

σ2 0 A∗A = V V ∗ 0 0 2 2 + where the σ are the square roots of the eigenvalues of (A∗A) . Thus (A∗A) A∗ =

2 1 σ− 0 σ 0 σ− 0 + V V ∗V U ∗ = V U ∗ = A 0 0 0 0 0 0

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14.7

1. Solve the system 4 1 1 x 1 1 5 2 y = 2  0 2 6   z   3  using the Gauss Seidel method and the  Jacobi method.  Check your answer by also solving it using row operations. 9 4 1 1 x 1 100 21 1 5 2 y = 2 , Solution is: 100        43  0 2 6 z 3 100         2. Solve the system 4 1 1 x 1 1 7 2 y = 2  0 2 4   z   3  using the Gauss Seidel method and the  Jacobi method.  Check your answer by also solving it using row operations. 5 4 1 1 x 1 94 7 1 7 2 y = 2 , Solution is: 94        67  0 2 4 z 3 94         3. Solve the system 5 1 1 x 1 1 7 2 y = 2  0 2 4   z   3  using the Gauss Seidel method and the  Jacobi method.  Check your answer by also solving it using row operations. 5 5 1 1 x 1 118 9 1 7 2 y = 2 , Solution is: 118        42  0 2 4 z 3 59         1 4. If you are considering a system of the form Ax = b and A− does not exist, will either the Gauss Seidel or Jacobi methods work? Explain. What does this indicate about ﬁnding eigenvectors for a given eigenvalue? No. These methods only work when there is a unique solution.

5. For x max xj : j = 1, 2, , n , the parallelogram identity does not hold. Explain.|| ||∞ ≡ {| | ··· } Let x = (1, 0) , y = (0, 1) . Then

x + y 2 + x y 2 = 1 + 1 = 2 k k k − k 2 x 2 + 2 y 2 = 4 k k k k 6. A norm is said to be strictly convex if whenever x = y , x = y, it follows ||·|| || || || || 6 x + y < x = y . 2 || || || ||

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Show the norm which comes from an inner product is strictly convex. |·| Let x, y be as described.

x + y 2 x y 2 1 1 + − = x 2 + y 2 = x 2 2 2 2 k k 2 k k k k

If x = y, then 6 2 x + y < x 2 2 k k

7. A norm is said to be uniformly convex if whenever x , y are equal to 1 for ||·|| || n|| || n|| all n N and limn xn + yn = 2, it follows limn xn yn = 0. Show the norm ∈ coming from→∞ an|| inner product|| is always uniformly→∞ || convex.− || Also show that uniform|·| convexity implies strict convexity which is deﬁned in Problem 6. The usual norm satisﬁes the parallelogram law. Thus

x y 2 = 2 x 2 + 2 y 2 x + y 2 | n − n| | n| | n| − | n n| = 4 x + y 2 − | n n| and the right side converges to 0. Thus x y 0. | n − n| → 8. Suppose A : Cn Cn is a one to one and onto matrix. Define → x Ax . || || ≡ | | Show this is a norm. If x = 0, then Ax = 0 ans since A is one to one, it follows that x = 0. Clearly cxk k= c x . What about the triangle inequality? This is also easy. k k | | k k x + y A (x + y) Ax + Ay = x + y k k ≡ | | ≤ | | | | k k k k 9. If X is a finite dimensional normed vector space and A, B (X,X) such that 1 ∈ L B < A , can it be concluded that A− B < 1? || || || || 1 0 1/2 0 1 0 Maybe not. Say A = ,B = . Then A 1 = and 0 1/2 0 3/4 − 0 2 1 1 2 0 A− B = so it has norm 3/2 > 1. Also B < A . 0 3 k k k k 2 10. Let X be a vector space with a norm and let V = span (v1, , vm) be a finite dimensional subspace of X such that v||·||, , v is a basis for V.···Show V is a closed { 1 ··· m} subspace of X. This means that if wn w and each wn V, then so is w. Next show that if w / V, → ∈ ∈ dist (w, V ) inf w v : v V > 0 ≡ {|| − || ∈ } is a continuous function of w and

dist (w, V ) dist (w ,V ) w w | − 1 | ≤ k 1 − k Next show that if w / V, there exists z such that z = 1 and dist (z, V ) > 1/2. For those who know some∈ advanced calculus, show that|| if|| X is an inﬁnite dimensional vector space having norm , then the closed unit ball in X cannot be compact. Thus closed and bounded is never||·|| compact in an inﬁnite dimensional normed vector space.

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1/2 m n m 2 Say wn = k=1 ck vk. Deﬁne a norm on V by w k=1 ck where w = m k k ≡ | | k=1 ckvk.P This gives a norm on V which is equivalentP to the given norm on V . n m m Thus if wn w, it follows that c c in C and so w = ckvk and so P → → k=1 V is closed. Let w, w′ be two points and suppose without loss of generality that P dist (w, V ) dist (w′,V ) 0. Let v V be such that w′ v < dist (w′,V ) + ε. Then − ≥ ∈ k − k

dist (w, V ) dist (w′,V ) = dist (w, V ) dist (w′,V ) | − | − w v w′ v + ε ≤ k − k − k − k w w′ + w′ v w′ v + ε ≤ k − k k − k − k − k = w w′ + ε k − k

Then since ε is arbitrary, this shows that dist (w, V ) dist (w′,V ) w w′ and so this is continuous as claimed. Now let w| / V. Thus− dist (w, V ) >|0 ≤ because k − ifk this ∈ is not so, there wouild exist vn w and by the ﬁrst part, w V . Let v V be such → w ∈v ∈ that 2 dist (w, V ) > w v . Then you can consider z = w−v . If v1 V, k − k k − k ∈ w v v w v z v = − 1 k − k k − 1k w v − w v k − k k − k 1 w v w v v ≥ 2 dist (w, V ) k − − k − k 1k dist (w, V ) 1 = ≥ 2 dist (w, V ) 2

1 Since v1 is arbitrary, it follows that dist (z, V ) 2 . If you have an inﬁnite dimensional normed linear space, the above argument shows≥ that you can obtain an inﬁnite sequence of vectors xn , each xn = 1 and such that dist (xn+1, span (x1, , xn)) 1/2 for every n. Therefore,{ } therek isk no convergent subsequence because no subsequence··· ≥ is a Cauchy sequence. Therefore, the unit ball is not sequentially compact. 11. Suppose ρ (A) < 1 for A (V,V ) where V is a p dimensional vector space having a norm . You can use∈R Lp or Cp if you like. Show there exists a new norm such that||·|| with respect to this new norm, A < 1 where A denotes the operator|||·||| norm of A taken with respect to this new||| norm||| on V , ||| |||

A sup Ax : x 1 ||| ||| ≡ {||| ||| ||| ||| ≤ } Hint: You know from Gelfand’s theorem that

An 1/n < r < 1 || || provided n is large enough, this operator norm taken with respect to . Show there exists 0 < λ < 1 such that ||·|| A ρ < 1. λ You can do this by arguing the eigenvalues of A/λ are the scalars µ/λ where µ σ (A). ∈ Now let Z+ denote the nonnegative integers.

An x sup n x ||| ||| ≡ n Z+ λ ∈

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First show this is actually a norm. Next explain why

An+1 Ax λ sup x λ x . ||| ||| ≡ Z n+1 ≤ ||| ||| n + λ ∈

First, it is obvious that there exists λ < 1 such that

A ρ < 1 λ Just pick λ larger than the absolute value of all the eigenvalues which are each less than 1. Now use the norm suggested. First of all, why is this a norm? If x = 0, then x = 0 because ||| ||| k k x x k k ≤ ||| ||| It is also clear that cx = c x . What about the triangle inequality? ||| ||| | | ||| ||| An An An x + y sup n (x + y) sup n x + n y ||| ||| ≡ n Z+ λ ≤ n Z+ λ λ ∈ ∈

An An sup n x + sup n y x + y ≤ n Z+ λ n Z+ λ ≡ ||| ||| ||| ||| ∈ ∈

Now why is the operator norm with respect to this new norm less than 1? An An+1 Ax sup n Ax = λ sup n+1 x λ x ||| ||| ≡ n Z+ λ n Z+ λ ≤ ||| ||| ∈ ∈

12. Establish a similar result to Problem 11 without using Gelfand’s theorem. Use an argument which depends directly on the Jordan form or a modification of it. 1 ρ (A) < 1 and A = S− JεS where Jε is in modified Jordan form having ε or 0 on the super diagonal the diagonal entries of every block being less than 1 in absolute value. Let λ < 1 be larger than the absolute values of all eigenvalues. Now what if you define x Sx ? Then ||| ||| ≡ k k∞

Ax SAx = JεSx ||| ||| ≡ k k∞ k k∞ max µi (Sx) + ε (Sx) ≤ i i i+1 max λ (Sx) + ε max ( S x) ≤ i { | i|} i {| i|} (λ + ε) Sx < Sx x ≤ k k∞ k k∞ ≡ ||| ||| provided ε is chosen small enough. 13. Using Problem 11 give an easier proof of Theorem 14.6.6 without having to use Corol- lary 14.6.5. It would suffice to use a different norm of this problem and the contraction mapping principle of Lemma 14.6.4. 1 The suggestion gives it away. You were given ρ B− C < 1. Now with the above 1 problem, there is another norm such that with respect to this other norm, B− C is a contraction mapping. Therefore, it has a fixed point by the easier result on contraction maps.

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14. A matrix A is diagonally dominant if aii > j=i aij . Show that the Gauss Seidel method converges if A is diagonally dominant.| | 6 | | P 1 It is the eigenvalues of B− C which are important. Say

1 B− Cx = λx

Then you must have det (λB C) = 0 − Illustrating what happens in the Gauss Seidel method in three dimensions, this requires

λa11 a12 a13 det λa −λa −a = 0  21 22 − 23  λa31 λa32 λa33   In other words the above matrix must have a zero determinant. However, since the original matrix was given to be diagonally dominant, this matrix cannot have a zero determinant unless λ < 1. | | n 15. Suppose f (λ) = k∞=0 anλ converges if λ < R. Show that if ρ (A) < R where A is an n n matrix, then | | × P ∞ f (A) a An ≡ n k 0 X− converges in (Fn, Fn) . Hint: Use Gelfand’s theorem and the root test. L Since the given series converges if λ < R, it follows that | | n 1/n lim sup anλ < 1 n | | →∞ and so 1/n lim sup an λ < 1 if λ < R n | | | | | | →∞ Thus you have 1/n 1 lim sup an n | | ≤ R →∞ You need to look at a An 1/n . k n k n 1/n 1/n n 1/n lim sup anA = lim sup an A n k k n | | k k →∞ →∞ 1/n n 1/n lim sup an lim sup A ≤ n | | n k k →∞ →∞ 1 < R = 1 R and so the series converges. 16. Referring to Corollary 14.4.3, for λ = a + ib show

exp (λt) = eat (cos (bt) + i sin (bt)) .

at Hint: Let y (t) = exp (λt) and let z (t) = e− y (t) . Show

2 z′′ + b z = 0, z (0) = 1, z′ (0) = ib.

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Now letting z = u + iv where u, v are real valued, show

2 u′′ + b u = 0, u (0) = 1, u′ (0) = 0 2 v′′ + b v = 0, v (0) = 0, v′ (0) = b.

Next show u (t) = cos (bt) and v (t) = sin (bt) work in the above and that there is at most one solution to

2 w′′ + b w = 0 w (0) = α, w′ (0) = β.

Thus z (t) = cos (bt) + i sin (bt) and so y (t) = eat (cos (bt) + i sin (bt)). To show there is at most one solution to the above problem, suppose you have two, w1, w2. Subtract them. Let f = w1 w2. Thus − 2 f ′′ + b f = 0

and f is real valued. Multiply both sides by f ′ and conclude

d (f )2 f 2 ′ + b2 = 0 dt 2 2 !

Thus the expression in parenthesis is constant. Explain why this constant must equal 0. The first claim is a routine computation. The next claim is also routine computations. What about the uniqueness assertion? It suffices to show that if α, β = 0 then the only solution is 0. Multiply both sides of the differential equation by w′. Then you get

2 1 d 2 b d 2 (w′) + w = 0 2 dt 2 dt " It follows that 2 2 (w′) + w is a constant. From the initial conditions, this constant can only be 0. The rest follows. 17. Let A (Rn, Rn) . Show the following power series converges in (Rn, Rn). ∈ L L ∞ tkAk k! Xk=0 You might want to use Lemma 14.4.2. This is how you can deﬁne exp (tA). Next show using arguments like those of Corollary 14.4.3 d exp (tA) = A exp (tA) dt so that this is a matrix valued solution to the diﬀerential equation and initial condition

Ψ′ (t) = AΨ(t) , Ψ (0) = I.

This Ψ (t) is called a fundamental matrix for the diﬀerential equation y′ = Ay. Show t Ψ(t) y gives a solution to the initial value problem → 0

y′ = Ay, y (0) = y0.

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The series converges absolutely by the ratio test and the observation that Ak k ≤ A . Now we need to show that you can diﬀerentiate it. Writing the diﬀerence quotientk k gives

k k k k 1 k 1 1 ∞ (t + h) t A ∞ ks (k, h) − Ak ∞ s (k, h) − Ak − = = h k! k! (k 1)! Xk=0 kX=1 Xk=1 −

where s (k, h) is between t and t + h. Thus for each k, limh 0 s (k, h) = t. I want to −1 → tk Ak ∞ show that the limit of the series is k=1 (k 1)! . − P k 1 k 1 − k 1 k ∞ s (k, h) − Ak ∞ tk 1Ak ∞ s (k, h) t − A − = − (k 1)! − (k 1)! (k 1)! Xk=1 − kX=1 − kX=1 − k 2 k k 2 k ∞ (k 1) p (k, h) − (s (k, h) t) A ∞ p (k, h) − (s (k, h) t) A = − − = − (k 1)! (k 2)! Xk=1 − Xk=2 − where p (k, h) is also between t and t + h. Now the norm of this is dominated by

k 2 k ∞ p (k, h) − A h | | k k | | (k 2)! Xk=2 − The series converges by the ratio test and so

k 1 ∞ s (k, h) − Ak ∞ tk 1Ak − 0 (k 1)! − (k 1)! → Xk=1 − kX=1 − as h 0. Hence you can differentiate the series and it gives you → ∞ tk 1Ak ∞ tk 1Ak 1 − = A − − = A exp (At) (k 1)! (k 1)! Xk=1 − kX=1 − When you plug in t = 0, you get I. Therefore, this is the solution to the given differential equation. 18. In Problem 17 Ψ (t) is defined by the given series. Denote by exp (tσ (A)) the numbers exp (tλ) where λ σ (A) . Show exp (tσ (A)) = σ (Ψ (t)) . This is like Lemma 14.4.7. Letting J be the Jordan∈ canonical form for A, explain why

∞ k k ∞ k k t A t J 1 Ψ(t) = S S− ≡ k! k! Xk=0 kX=0 and you note that in J k, the diagonal entries are of the form λk for λ an eigenvalue of A. Also J = D + N where N is nilpotent and commutes with D. Argue then that

∞ tkJ k k! Xk=0 is an upper triangular matrix which has on the diagonal the expressions eλt where λ σ (A) . Thus conclude ∈ σ (Ψ (t)) exp (tσ (A)) ⊆

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Next take etλ exp (tσ (A)) and argue it must be in σ (Ψ (t)) . You can do this as follows: ∈

k ∞ tkAk ∞ tkλ Ψ(t) etλI = I − k! − k! Xk=0 Xk=0 ∞ tk = Ak λkI k! − Xk=0 k k 1 ∞ t − k j j = A − λ (A λI) (6.38)  k!  − j=1 Xk=0 X   Now you need to argue k k 1 ∞ t − k j j A − λ k! j=1 Xk=0 X converges to something in (Rn, Rn). To do this, use the ratio test and Lemma 14.4.2 after ﬁrst using the triangleL inequality. Since λ σ (A) , Ψ(t) etλI is not one to one and so this establishes the other inclusion. You∈ ﬁll in the details.− This theorem is a special case of theorems which go by the name “spectral mapping theorem”. (tA)kx ∞ λt If λ σ (A) , then Ax = λx and so k=0 k! = e x and so exp (tσ (A)) ∈ 1 ⊆ σ (exp (tA)) . Now consider exp (tA) . You have A = S− JS where J is Jordan form. Thus it is simple to write that P

k ∞ 1 (tJ) exp (tA) = S− S k! kX=0 Now J is block diagonal and the blocks have a constant down the main diagonal. Say

α 1 . .  .. ..  α 1    α      When you raise such a block to the exponent k, you get an upper triangular matrix which has αk down the diagonal. Therefore, exp (tA) is of the form

1 S− BS

where B is an upper triangular matrix which has for its diagonal entries eλt for λ σ (A). These diagonal entries are the eigenvalues of exp (tA). It follows that ∈

σ (exp (tA)) exp (tσ (A)) . ⊆ 19. Suppose Ψ(t) (V,W ) where V,W are finite dimensional inner product spaces and t Ψ(t) is continuous∈ L for t [a, b]: For every ε > 0 there there exists δ > 0 such that if→s t < δ then Ψ(t) Ψ(∈ s) < ε. Show t (Ψ (t) v, w) is continuous. Here it is the| inner− | product|| in W. −Also define|| what it means→ for t Ψ(t) v to be continuous and show this is continuous. Do it all for differentiable in→ place of continuous. Next show t Ψ(t) is continuous. → || ||

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(Ψ (t) v, w) (Ψ (s) v, w) Ψ(t) v Ψ(s) v w | − | ≤ | − | | | Ψ(t) Ψ(s) v w ≤ k − k | | | | and so this is continuous.

Ψ(t) v Ψ(s) v Ψ(t) Ψ(s) v k − k ≤ k − k | | so t Ψ(t) v is continuous. Diﬀerentiable works out similarly. If t Ψ(t) is diﬀerentiable,→ then those other things are too. For example, →

Ψ(t + h) v Ψ(t) v Ψ′ (t) hv k − − k Ψ(t + h) Ψ(t) Ψ′ (t) h v ≤ k − − k | | = o (h) v = o (h) | | As to continuity of t Ψ(t) , → k k Ψ(t) Ψ(s) Ψ(t) Ψ(s) |k k − k k| ≤ k − k from the triangle inequality. 20. If z (t) W, a finite dimensional inner product space, what does it mean for t z (t) to be continuous∈ or differentiable? If z is continuous, define →

b z (t) dt W ∈ Za as follows. b b w, z (t) dt (w, z (t)) dt. ≡ Za ! Za Show that this deﬁnition is well deﬁned and furthermore the triangle inequality,

b b z (t) dt z (t) dt, a ≤ a | | Z Z

and fundamental theorem of calculus, d t z (s) ds = z (t) dt Za hold along with any other interesting properties of integrals which are true. Diﬀerentiability is the same as usual.

z (t + h) z (t) z′ (t) h = o (h) − − If t z (t) is continuous, then from the above problem, so is t (z (t) , w) and so b → → a (z (t) , w) dt makes sense. Does there exist an element I of the inner product space b Rsuch that (w, I) = a (w, z (t)) dt? This is so in a ﬁnite dimensional inner product space because R b w (w, z (t)) dt → Za

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is a linear map and so it can be represented by a unique I W so that ∈ b (w, I) = (w, z (t)) dt Za for all w. Now what about the fundamental theorem of calculus? t+h z (s) ds 1 t+h w, t (w, z (s)) ds (w, z (t)) h ≡ h → R ! Zt Since the space is ﬁnite dimensional, this is the same as saying that

t+h z (s) ds lim t = z (t) . h 0 h → R You can just apply the above to the ﬁnitely many Fourier coeﬃcients. Therefore, the usual Fundamental theorem of calculus holds. What about the triangle inequality?

b b z (t) dt = sup w, z (t) dt a w =1 a ! Z | | Z

b b = sup (w, z (t)) dt sup (w, z (t)) dt w =1 a ≤ w =1 a | | | | Z | | Z b

z (t) dt. ≤ | | Za 21. In the situation of Problem 19 deﬁne b Ψ(t) dt (V,W ) ∈ L Za as follows. b b w, Ψ(t) dt (v) (w, Ψ(t) v) dt. ≡ Za ! Za b Show this is well deﬁned and does indeed give a Ψ(t) dt (V,W ) . Also show the triangle inequality ∈ L b bR Ψ(t) dt Ψ(t) dt a ≤ a || || Z Z

where is the operator norm and verify the fundamental theorem of calculus holds. ||·|| t ′ Ψ(s) ds = Ψ (t) . Za Also verify the usual properties of integrals continue to hold such as the fact the integral is linear and

b c c Ψ(t) dt + Ψ(t) dt = Ψ(t) dt Za Zb Za and similar things. Hint: On showing the triangle inequality, it will help if you use the fact that

w W = sup (w, v) . | | v 1 | | | |≤

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You should show this also. That last assertion follows by the Cauchy Schwarz inequality which shows the right side is no larger than w and then by taking v = w/ w . That the integral is well | | b | | defined follows because w a (w, Ψ(t) v) dt is a linear mapping and so, by the Riesz representation theorem,→ there exists a unique I (v) W such that (w, I (v)) = b R ∈ a (w, Ψ(t) v) dt. Now also, I is a linear function of v because of the fact that each b RΨ(t) is linear. Therefore, you can denote by a Ψ(t) dt (V,W ) this I. Thus the definition of the integral is well defined. The standard properties∈ L of the integral are now fairly obvious. As for the triangle inequality,R it goes the same way as in the above problem.

b b b Ψ(t) dt sup Ψ(t) dt (v) = sup sup w, Ψ(t) dt (v) a ≡ v 1 a v 1 w 1 a ! Z | |≤ Z | |≤ | |≤ Z

b b = sup sup (w, Ψ( t)(v)) dt Ψ(t) dt. v 1 w 1 a ≤ a k k | |≤ | |≤ Z Z

Now consider the fundamental theorem of calculus. For simplicity, let h 0 + . → 1 t+h 1 t+h Ψ(s) ds Ψ(t) = Ψ(s) Ψ(t) ds h t − h t − Z Z t+h 1 Ψ(s) Ψ(t) ds ≤ h k − k Zt which converges to 0 by continuity of Ψ. 22. Prove Gronwall’s inequality. Suppose u (t) 0, continuous, and for all t [0,T ] , ≥ ∈ t u (t) u + Ku (s) ds. ≤ 0 Z0 where K is some nonnegative constant. Then

u (t) u eKt. ≤ 0 t Hint: w (t) = 0 u (s) ds. Then using the fundamental theorem of calculus, w (t) satisﬁes the following. R

u (t) Kw (t) = w′ (t) Kw (t) u , w (0) = 0. − − ≤ 0 Now use the usual techniques you saw in an introductory diﬀerential equations class. Kt Multiply both sides of the above inequality by e− and note the resulting left side is now a total derivative. Integrate both sides from 0 to t and see what you have got. If you have problems, look ahead in the book. This inequality is proved later in Theorem C.4.3. Following the hint, Kt Kt we− ′ u e− ≤ 0 and therefore, " t Kt Kt Ks 1 e− w (t) e− u e− ds = u ≤ 0 0 K − K Z0

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Kt Therefore, w (t) u e 1 . Also, you get ≤ 0 K − K u (t) u + Kw (t) ≤ 0 If K > 0, this implies

eKt 1 u (t) u + K u = u eKt ≤ 0 0 K − K 0 If K = 0, then the result requires no proof. 23. With Gronwall’s inequality and the integral deﬁned in Problem 21 with its properties listed there, prove there is at most one solution to the initial value problem

y′ = Ay, y (0) = y0.

Hint: If there are two solutions, subtract them and call the result z. Then

z′ = Az, z (0) = 0.

It follows t z (t) = 0+ Az (s) ds Z0 and so t z (t) A z (s) ds || || ≤ k k || || Z0 Now consider Gronwall’s inequality of Problem 22. Use Problem 20 as needed. From Gronwall’s inequality,

A t z (t) 0e−k k = 0. k k ≤ 24. Suppose A is a matrix which has the property that whenever µ σ (A) , Re µ < 0. Consider the initial value problem ∈

y′ = Ay, y (0) = y0.

The existence and uniqueness of a solution to this equation has been established above in preceding problems, Problem 17 to 23. Show that in this case where the real parts of the eigenvalues are all negative, the solution to the initial value problem satisﬁes

lim y (t) = 0. t →∞ Hint: A nice way to approach this problem is to show you can reduce it to the consideration of the initial value problem

z′ = Jεz, z (0) = z0

where Jε is the modiﬁed Jordan canonical form where instead of ones down the main 1 diagonal, there are ε down the main diagonal. You have y′ = S− JεSy and so you just let z = Sy. Then z 0 if and only if y 0. → → Then z′ = Dz + Nεz

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where D is the diagonal matrix obtained from the eigenvalues of A and Nε is a nilpotent matrix commuting with D which is very small provided ε is chosen very small. Now let Ψ (t) be the solution of Ψ′ = DΨ, Ψ (0) = I − described earlier as k ∞ ( 1) tkDk − . k! Xk=0 Thus Ψ (t) commutes with D and Nε. Tell why. The reason these commute is that D is block diagonal, each block having constant entries down the diagonal while the same is true of Nε. The blocks associated with Nε commute with the blocks associated with D and so D and Nε commute. Thus Ψ (t) commutes with D and Nε. Next argue (Ψ (t) z)′ = Ψ (t) Nεz (t) and integrate from 0 to t.

Just do the diﬀerentiation. (Ψ (t) z)′ =

Ψ′ (t) z (t) + Ψ (t) z′ (t) = DΨ(t) z (t) + Ψ (t)(D + N ) z (t) − ε = Ψ (t) Nεz (t)

Then integrating, t Ψ(t) z (t) z = Ψ(s) N z (s) ds. − 0 ε Z0 It follows from the triangle inequality discussed above that

t Ψ(t) z (t) z + N Ψ(s) z (s) ds. || || ≤ || 0|| || ε|| || || Z0 It follows from Gronwall’s inequality

N t Ψ(t) z (t) z e|| ε|| || || ≤ || 0|| Now look closely at the form of Ψ (t) to get an estimate which is interesting. Explain why eµ1t 0 . Ψ(t) =  ..  0 eµnt     and now observe that if ε is chosen small enough, Nε is so small that each component of z (t) converges to 0. || || It follows right away from the deﬁnition of Ψ (t) that Ψ (t) is given by the above diagonal matrix where here the µi are negatives of the eigenvalues of A. Thus their real parts are all bounded below by some δ > 0. It follows that

(Ψ (t) x) eδt x | i| ≥ | i| and so Ψ(t) z (t) eδt z (t) || || ≥ k k

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Therefore, from the above inequality,

t δt δt N s z (t) z e− + e− z e|| ε|| ds k k ≤ k 0k || 0|| Z0 Of course the right side converges to 0 if you choose ε small enough that N < δ. k εk 25. Using Problem 24 show that if A is a matrix having the real parts of all eigenvalues less than 0 then if Ψ′ (t) = AΨ(t) , Ψ (0) = I it follows lim Ψ(t) = 0. t →∞ Hint: Consider the columns of Ψ (t)? From the above, problem, this happens. If Ψ is just deﬁned, then the solution to xi′ = Axi (t) , xi (0) = ei is Ψ (t) ei. This follows from the fact that Ψ (t) ei satisﬁes the equation and uniqueness. Thus the ith column of Ψ (t) converges to 0. In particular the entries of Ψ (t) converge to 0. Thus Ψ(t) 0 also. k k → 26. Let Ψ (t) be a fundamental matrix satisfying

Ψ′ (t) = AΨ(t) , Ψ (0) = I.

n Show Ψ (t) = Ψ (nt) . Hint: Subtract and show the diﬀerence satisﬁes Φ′ = AΦ, Φ (0) = 0. Use uniqueness. 27. If the real parts of the eigenvalues of A are all negative, show that for every positive t, lim Ψ(nt) = 0. n →∞ Hint: Pick Re (σ (A)) < λ < 0 and use Problem 18 about the spectrum of Ψ(t) and Gelfand’s theorem for− the spectral radius along with Problem 26 to argue that λnt Ψ(nt) /e− < 1 for all n large enough.

n iH (iH) 28. Let H be a Hermitian matrix. (H = H∗) . Show that e ∞ is unitary. ≡ n=0 n! You have H = U ∗DU where U is unitary and D is a real diagonalP matrix. Then you have iλ1 n e ∞ iH (iD) . e = U ∗ U = U ∗ . U n!  .  n=0 eiλn X   and this is clearly unitary because each matrix in the product is. 29. Show the converse of the above exercise. If V is unitary, then V = eiH for some H Hermitian.

To do this, note ﬁrst that V is normal because V ∗V = VV ∗ = I. Therefore, there exists a unitary U such that V = U ∗DU where D is the diagonal matrix consisting of the eigenvalues of V down the main diagonal. Since V is unitary, it preserves all lengths and so each diagonal entry of D is of magnitude 1. Thus you have

iλ1 e k . ∞ (iD) V = U ∗ . U = U ∗ U  .  k! eiλn k=0   X  

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λ1 . where D =  ..  , each λk real. Then the above equals λ  n    k ∞ (iU DU) ∗ = eiH k! kX=0 where H = U ∗DU is obviously Hermitian. 1 30. If U is unitary and does not have 1 as an eigenvalue so that (I + U)− exists, show that − 1 H = i (I U)(I + U)− − is Hermitian. Then, verify that 1 U = (I + iH)(I iH)− . − 1 To see this, note ﬁrst that (I U) , (I + U)− commute. Taking the adjoint of the above operator, it is easy to see− that this equals 1 1 i (I + U ∗)− (I U ∗) = i (U ∗U + U ∗)− (U ∗U U ∗) − − − 1 − = i (U + I)− UU ∗ (U I) − 1 − 1 = i (I + U)− (I U) = i (I U)(I + U)− . − − Next, using the fact that everything commutes, 1 1 1 − I + i i (I U)(I + U)− I i i (I U)(I + U)− − − − 1 1 1 − = I (I U)(I + U)− I + (I U)(I + U)− − − − 1 1 1 − = (I + U (I U)) (I + U)− (I + U + I U)(I + U)− − − − 1 1 1 − 1 1 = 2U (I + U)− 2I (I + U)− = 2U (I + U)− (I + U) = U 2 31. Suppose that A (V,V ) where V is a normed linear space. Also suppose that A < 1 where this∈ Lrefers to the operator norm on A. Verify that k k ∞ 1 i (I A)− = A − i=0 X This is called the Neumann series. Suppose now that you only know the algebraic 1 condition ρ (A) < 1. Is it still the case that the Neumann series converges to (I A)− ? − Consider partial sums of the series.

q q q Ai Ai A i ≤ ≤ k k i=p i=p i=p X X X

i 1 which converges to 0 as p, q because ∞ A = < . Therefore, by → ∞ i=0 k k 1 A ∞ completeness of (V,V ) , this series converges. Why is is the−k desiredk inverse? This is P obvious when youL multiply both sides by (I A) . Thus − n ∞ (I A) Ai = lim (I A) Ai = lim I An+1 = I − n − n − i=0 →∞ i=0 →∞ X X

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because An+1 A n+1 and this converges to 0 as n because A < 1. Yes, it the series still converges.≤ k k Let ρ (A) < r < R < 1. Then→ for ∞ all n largek enough,k you

have An 1/n < R k k and so An < Rn. It follows that the partial sums of the series form a Cauchy sequencek andk so the series converges.

F.34 Exercises

15.3

1. In Example 15.1.10 an eigenvalue was found correct to several decimal places along with an eigenvector. Find the other eigenvalues along with their eigenvectors. 1 2 3 The matrix was 2 2 1  3 1 4  20   1 2 3 8. 485 7 1015 6. 190 4 1015 1. 188 8 1016 2 2 1 = 6. 190 4 × 1015 4. 515 9 × 1015 8. 672 7 × 1015  3 1 4   1. 188 8 × 1016 8. 672 7 × 1015 1. 665 6 × 1016  × × ×  0.534 92 0.470 81 0.701 57  − − = 0.390 23 0.598 82 0.699 38  0.749 39− 0.647 89 0.136 60  1. 586 4 1016 1. 157 3 1016 2. 222 5 1016 0× 2. 212 9 × 1011 8. 403 1 × 1011  0 0× 4. 160 5 × 1011  ×   T 0.534 92 0.470 81 0.701 57 1 2 3 0.390 23 −0.598 82− 0.699 38 2 2 1  0.749 39− 0.647 89 0.136 60   3 1 4   0.534 92 0.470 81 0.701 57    0.390 23 −0.598 82− 0.699 38  0.749 39− 0.647 89 0.136 60   4 5 6. 662 1 1. 218 8 10− 7. 374 9 10− 4 × × = 1. 218 8 10− 1. 139 5 1. 156 9  × 5 −  7. 374 9 10− 1. 156 9 0.801 49 × − −   0.534 92 The largest eigenvalue is close to 6. 662 1 and an eigenvector is 0.390 23 . The  0.749 39  others can be found by looking at the lower right block. You could do the same as the above or you could just graph the characteristic function and zoom in on the roots. They are approximately 1.7 and 1.3. Next use shifted inverse power method to ﬁnd them more exactly and to also ﬁnd− the eigenvectors. 1 1 2 3 1 0 0 − 0.977 92 5. 047 3 3. 47 − − 2 2 1 1.7 0 1 0 = 5. 047 3 33. 47 21. 136  3 1 4  −  0 0 1   − 3. 47− 21. 136 13. 281  −      

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Rather than use this method as described, I will just use the variation of it based on the QR algorithm which involves raising the matrix to a power which was just used to find the first eigenvalue. 20 0.977 92 5. 047 3 3. 47 − 5. 047 3 − 33. 47 21. 136 =  − 3. 47− 21. 136 13. 281  −  6. 044 8 1031 3. 893 4 1032 2. 458 8 1032 × × − × 3. 893 4 1032 2. 507 7 1033 1. 583 7 1033 :  2. 458 8× 1032 1. 583 7× 1033 −1. 000 2 ×1033  − × − × ×  2  0.130 15 3. 322 8 10− 0.990 94 0.838 32 − 0.529× 98 0.127 88  − − 2  0.529 42 0.847 36 4. 112 3 10− − − ×  4. 644 3 1032 2. 991 4 1033 1. 889 2 1033 0× 9. 738 8 × 1027 −3. 859 8 × 1028  0 0× − 3. 342 9 ×1027  ×  2 T  0.130 15 3. 322 8 10− 0.990 94 0.838 32 − 0.529× 98 0.127 88  − − 2  0.529 42 0.847 36 4. 112 3 10− − − ×  0.977 92 5. 047 3 3. 47  − 5. 047 3 − 33. 47 21. 136  − 3. 47− 21. 136 13. 281  −  2  0.130 15 3. 322 8 10− 0.990 94 − × 0.838 32 0.529 98 0.127 88 =  − − 2  0.529 42 0.847 36 4. 112 3 10− − − ×  5  5 47. 602 6. 755 1 10− 2. 994 7 10− − 5 × 2 − × 6. 755 1 10− 6. 320 5 10− 0.232 90  × 5 × −  2. 994 7 10− 0.232 90 0.190 38 − × − − Thus you find the eigenvector by solving 1 = 47. 602, Solution is: 1. 679 0 and a λ 1.7 − − 0.130 15 suitable eigenvector is just the first column of the orthogonal matrix 0.838 32 .  0.529 42  − How well does it work?   1 2 3 0.130 15 0.218 53 2 2 1 0.838 32 = 1. 407 5  3 1 4   0.529 42   0.888 91  − −  0.130 15    0.218 52  0.838 32 (1. 679 0) = 1. 407 5  0.529 42   0.888 90  − − This worked well.   Now find the pair which goes with 1.3. − 1 1 2 3 1 0 0 − 2 2 1 + 1.3 0 1 0  3 1 4   0 0 1   16. 948 7. 810 9 8. 119 2 = −7. 810 9 3. 278 5 3. 802 7  8. 119 2 −3. 802 7 −3. 689 6  − −  

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20 16. 948 7. 810 9 8. 119 2 − 7. 810 9 3. 278 5 3. 802 7 =  8. 119 2 −3. 802 7 −3. 689 6  − −  3. 827 4 1027 1. 745 5 1027 1. 823 0 1027 1. 745 5× 1027 −7. 960 3 ×1026 −8. 313 7 ×1026 :  −1. 823 0 × 1027 8. 313 7 × 1026 8. 682 7 × 1026  − × × ×  2  0.834 83 0.544 13 8. 355 9 10− 0.380 73 −0.461 03 0.801× 56  −0.397 63 −0.700 98 0.592 05  − − −  4. 584 7 1027 2. 090 8 1027 2. 183 7 1027 0× − 1. 658 9 ×1022 −2. 792 9 ×1022  0 0× 4. 620 1 × 1021  ×  2 T  0.834 83 0.544 13 8. 355 9 10− 0.380 73 −0.461 03 0.801× 56  −0.397 63 −0.700 98 0.592 05  − − −  16. 948 7. 810 9 8. 119 2  −7. 810 9 3. 278 5 3. 802 7  8. 119 2 −3. 802 7 −3. 689 6  − −   2 0.834 83 0.544 13 8. 355 9 10− 0.380 73 −0.461 03 0.801× 56  −0.397 63 −0.700 98 0.592 05  − − −  5  5 24. 377 9. 597 9 10− 5. 267 10− − 5 × × 2 = 9. 597 9 10− 0.127 02 1. 802 1 10−  × 5 2 − ×  5. 267 10− 1. 802 1 10− 0.334 17 × − × 1  λ+1.3 = 24. 377, Solution is: 1. 341. The approximate eigenvector is the ﬁrst column. How− well does it work? − 1 2 3 0.834 83 1. 119 5 − 2 2 1 0.380 73 = 0.510 57  3 1 4   −0.397 63   0.533 24  −  0.834 83    1. 119 5  0.380 73 ( 1. 341) = −0.510 56  −0.397 63  −  0.533 22  − It works well. You could do more iterations if desired. 3 2 1 2. Find the eigenvalues and eigenvectors of the matrix A = 2 1 3 numerically.  1 3 2  In this case the exact eigenvalues are √3, 6. Compare with the exact answers. 13 ± 3 2 1 4. 353 6 109 4. 353 6 109 4. 353 6 109 × × × 2 1 3 = 4. 353 6 109 4. 353 6 109 4. 353 6 109 =  1 3 2   4. 353 6 × 109 4. 353 6 × 109 4. 353 6 × 109  × × ×    29  0.577 35 0.816 50 1. 437 8 10− 0.577 35− 0.408 25 − 0.707× 11  0.577 35 0.408 25− 0.707 11   7. 540 7 109 7. 540 7 109 7. 540 7 109 × × 19 × 19 0 1. 533 3 10− 1. 533 3 10−  × × 48  0 0 2. 736 9 10− ×  

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29 T 0.577 35 0.816 50 1. 437 8 10− 3 2 1 0.577 35− 0.408 25 − 0.707× 11 2 1 3  0.577 35 0.408 25− 0.707 11   1 3 2   29    0.577 35 0.816 50 1. 437 8 10− 0.577 35− 0.408 25 − 0.707× 11  0.577 35 0.408 25− 0.707 11   6. 000 0 0 0  29 = 1. 262 2 10− 1. 5 0.866 03  − × 29  2. 524 4 10− 0.866 03 1. 5 − × −  0.577 35 An eigenvalue is 6 and an eigenvector is 0.577 35 . Now lets ﬁnd the others.  0.577 35  1  3 2 1 1 0 0 − 2 1 3 1.5 0 1 0  1 3 2  −  0 0 1   2. 740 7 0.592 59 1. 925 9 − 2 − = 0.592 59 7. 407 4 10− 0.740 74  − 1. 925 9 0.740× 74 1. 407 4  −  20 2. 740 7 0.592 59 1. 925 9 − 2 − 0.592 59 7. 407 4 10− 0.740 74 =  − 1. 925 9 0.740× 74 1. 407 4  −  3. 034 1 1012 8. 129 9 1011 2. 221 1 1012 8. 129 9× 1011 −2. 178 4 ×1011 −5. 951 5 ×1011 :  −2. 221 1 × 1012 5. 951 5 × 1011 1. 626 0 × 1012  − × × ×  0.788 68 0.425 47 0.443 81  0.211 33 0.490 28 0.845 56  −0.577 35− 0.760 66 0.296 76  −  3. 847 1 1012 1. 030 8 1012  2. 816 3 1012 0× − 3. 850 7 ×106 −3. 843 7 ×107  0 0× 1. 927 6 × 107  ×  T  0.788 68 0.425 47 0.443 81 2. 740 7 0.592 59 1. 925 9 − 2 − 0.211 33 0.490 28 0.845 56 0.592 59 7. 407 4 10− 0.740 74  −0.577 35− 0.760 66 0.296 76   − 1. 925 9 0.740× 74 1. 407 4  − −  0.788 68 0.425 47 0.443 81    0.211 33 0.490 28 0.845 56  −0.577 35− 0.760 66 0.296 76  −   6 5 4. 309 4 9. 614 1 10− 1. 276 5 10− 6 − × − × = 9. 614 1 10− 0.223 59 0.195 59  − × 5 −  1. 276 5 10− 0.195 59 0.136 42 − ×  0.788 68 1 λ 1.5 = 4. 309 4, Solution is: 1. 732 1 and an eigenvector is 0.211 33 . How well −  −0.577 35  − does it work?   3 2 1 0.788 68 1. 366 2 1 3 0.211 33 = 0.366 02  1 3 2   −0.577 35   − 1.0  − −      

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0.788 68 1. 366 1 0.211 33 1. 732 1 = 0.366 04  −0.577 35   − 1.0  − − Next you can ﬁnd the other one.  1 3 2 1 1 0 0 − 2 1 3 + 1.5 0 1 0  1 3 2   0 0 1    2   4. 444 4 10− 0.711 11 0.622 22 × − = 0.711 11 2. 622 2 2. 044 4  0.622 22− 2. 044 4 1. 288 9  − −  2 20 4. 444 4 10− 0.711 11 0.622 22 0.711× 11 2. 622 2− 2. 044 4 =  0.622 22− 2. 044 4 1. 288 9  − −  2. 178 4 1011 8. 129 9 1011 5.951 5 1011 × − × × 8. 129 9 1011 3. 034 1 1012 2. 221 1 1012 :  −5. 951 5 ×1011 2. 221 1× 1012 −1. 626 0 ×1012  × − × ×  0.211 32 0.490 28 0.845 56  0.788 68− 0.425 47 0.443 81  −0.577 35 0.760 66 0.296 76   1. 030 8 1012 3. 847 1 1012 2. 816 3 1012 0× − 1. 437 1 ×107 2. 791 7× 107  0 0× 1. 927 6 × 107  ×  T  0.211 32 0.490 28 0.845 56 0.788 68− 0.425 47 0.443 81  −0.577 35 0.760 66 0.296 76   2  4. 444 4 10− 0.711 11 0.622 22 0.211 32 0.490 28 0.845 56 0.711× 11 2. 622 2− 2. 044 4 0.788 68− 0.425 47 0.443 81 :  0.622 22− 2. 044 4 1. 288 9   −0.577 35 0.760 66 0.296 76  − −  6   5  4. 309 4 2. 468 3 10− 1. 167 8 10− − 6 − × − × 2 2. 468 3 10− 0.280 95 6. 479 4 10−  − × 5 2 − ×  1. 167 8 10− 6. 479 4 10− 0.161 74 − × − ×   0.211 32 1 λ+1.5 = 4. 309 4, Solution is: 1. 732 1 and an eigenvector is 0.788 68 . How − −  −0.577 35  well does it work?   3 2 1 1 0 0 0.211 32 2 1 3 + 1. 732 1 0 1 0 0.788 68  1 3 2   0 0 1   −0.577 35    5     2. 262 8 10− − × 5 = 6. 262 8 10− which is pretty close to 0 so this worked well.  − × 6  7. 935 10− ×   3 2 1 3. Find the eigenvalues and eigenvectors of the matrix A = 2 5 3 numerically.  1 3 2  The exact eigenvalues are 2, 4 + √15, 4 √15. Compare your numerical results with the exact values. Is it much fun to compute− the exact eigenvectors?

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20 3 2 1 2 5 3 =  1 3 2   1. 448 7 1017 2. 713 5 1017 1. 632 8 1017 × × × 2. 713 5 1017 5. 082 3 1017 3. 058 1 1017 =  1. 632 8 × 1017 3. 058 1 × 1017 1. 840 1 × 1017  × × ×  0.415 99 0.806 58 0.419 97  2 0.779 18 7. 803 9 10− 0.621 92  0.468 86 − 0.585× 95− 0.660 94  −  3. 482 5 1017 6. 522 6 1017 3. 924 8 1017 0× 1. 539 ×1013 1. 324 1 × 1013  0 0× 1. 399 9 × 1012  ×  T  0.415 99 0.806 58 0.419 97 3 2 1 2 0.779 18 7. 803 9 10− 0.621 92 2 5 3  0.468 86 − 0.585× 95− 0.660 94   1 3 2  −  0.415 99 0.806 58 0.419 97    2 0.779 18 7. 803 9 10− 0.621 92  0.468 86 − 0.585× 95− 0.660 94  −  5  5 7. 873 0 8. 768 9 10− 3. 295 6 10− 5 × × = 8. 768 9 10− 1. 746 2 0.641 05  × 5  3. 295 6 10− 0.641 05 0.380 82 ×   0.415 99 An eigenvalue is 7. 873 0 and the approximate eigenvector is 0.779 18 . How well  0.468 86  does it work?   3 2 1 1 0 0 0.415 99 2 5 3 7. 873 0 0 1 0 0.779 18  1 3 2  −  0 0 1   0.468 86    4     1. 007 3 10− × 5 = 2. 414 10−  − × 5  8. 478 10− − × 4 +√15 = 7. 873 0 which is what was just obtained to four decimal places. It works pretty well. Next we ﬁnd the other eigenvalues and eigenvectors which go with them. 1 3 2 1 1 0 0 − 2 5 3 1.75 0 1 0 =  1 3 2  −  0 0 1  3. 295 6 1. 006 3 1. 106 9 1. 006 3− 0.276 73− 0.704 4  − 2  1. 106 9 0.704 4 2. 515 7 10− − − ×  20 3. 295 6 1. 006 3 1. 106 9 1. 006 3− 0.276 73− 0.704 4  − 2  1. 106 9 0.704 4 2. 515 7 10− − − ×  8. 995 9 1011 2. 998 7 1011 2. 998 6 1011 = 2. 998 7× 1011 −9. 995 6 ×1010 −9. 995 4 ×1010 =  −2. 998 6 × 1011 9. 995 4 × 1010 9. 995 2 × 1010  − × × ×  

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0.904 53 0.404 72 0.134 23 0.301 52 −0.829 69 0.469 80  −0.301 51 −0.384 47− 0.872 51  − −  9. 945 4 1011 3. 315 2 1011 3. 315 1 1011 0× − 2. 995 0 ×106 − 1. 376 ×106  0 0× 5. 368 8× 105  ×  T  0.904 53 0.404 72 0.134 23 3. 295 6 1. 006 3 1. 106 9 0.301 52 −0.829 69 0.469 80 1. 006 3− 0.276 73− 0.704 4  − − −   − 2  0.301 51 0.384 47 0.872 51 1. 106 9 0.704 4 2. 515 7 10− − − − − ×  0.904 53 0.404 72 0.134 23    0.301 52 −0.829 69 0.469 80 =  −0.301 51 −0.384 47− 0.872 51  − −  6  6 4. 000 0 1. 828 8 10− 8. 371 4 10− 6 × − × 2 1. 828 8 10− 0.155 70 7. 669 2 10−  × 6 2 − ×  8. 371 4 10− 7. 669 2 10− 0.608 53 − × − × −  0.904 53 1 λ 1.75 = 4, Solution is: 2.0. An eigenvector is then 0.301 52 . How well does it −  −0.301 51  − work?   3 2 1 1 0 0 0.904 53 0.000 02 2 5 3 2 0 1 0 0.301 52 = −0.000 03  1 3 2  −  0 0 1   −0.301 51   −0.000 03  − −      1    3 2 1 1 0 0 − 2 5 3 0.38 0 1 0 =  1 3 2  −  0 0 1     2  0.493 54 7. 815 4 10− 0.449 38 2 × − 7. 815 4 10− 1. 056 5 1. 908 3  0.449× 38− 1. 908 3 2. 639 1  − −  2 20 0.493 54 7. 815 4 10− 0.449 38 2 × − 7. 815 4 10− 1. 056 5 1. 908 3 =  0.449× 38− 1. 908 3 2. 639 1  − −  7. 593 8 109 4. 459 9 1010 6. 738 1010 4. 459 9× 1010 −2. 619 4 ×1011 3. 957× 3 1011 :  −6. 738 ×1010 3. 957 3× 1011 −5. 978 7 ×1011  × − × ×  2  9. 356 7 10− 0.107 43 0.989 8 × − 2 0.549 53 0.834 58 3. 863 5 10−  −0.830 22 0.540 31 0.137× 12   8. 115 9 1010 4. 766 6 1011 7. 201 3 1011 0× − 4. 418 3 ×106 1. 380 8× 106  0 0× 6. 663 6 × 105  ×  2 T  9. 356 7 10− 0.107 43 0.989 8 × − 2 0.549 53 0.834 58 3. 863 5 10−  −0.830 22 0.540 31 0.137× 12   2  0.493 54 7. 815 4 10− 0.449 38 2 × − 7. 815 4 10− 1. 056 5 1. 908 3  0.449× 38− 1. 908 3 2. 639 1  − −  

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2 9. 356 7 10− 0.107 43 0.989 8 × − 2 0.549 53 0.834 58 3. 863 5 10− =  −0.830 22 0.540 31 0.137× 12   5  5 3. 952 9 2. 044 4 10− 1. 021 4 10− − 5 − × × 2. 044 4 10− 0.182 25 0.145 62  − × 5  1. 021 4 10− 0.145 62 0.568 55 ×  1  λ .38 = 3. 952 9, Solution is: 0.127 02 and an approximate eigenvector is then − − 2 9. 356 7 10− × 0.549 53 . How well does it work?  −0.830 22    2 3 2 1 1 0 0 9. 356 7 10− 2 5 3 0.127 02 0 1 0 0.549× 53  1 3 2  −  0 0 1   −0.830 22    5     2. 388 10− − × 5 = 5. 469 9 10−  − × 5  3. 754 4 10− − × Note that 4 √15 = 0.127 02 which is what was just obtained. − 0 2 1 4. Find the eigenvalues and eigenvectors of the matrix A = 2 5 3 numerically.  1 3 2  I don’t know the exact eigenvalues in this case. Check your answers by multiplying your numerically computed eigenvectors by the matrix. 50 0 2 1 5. 045 4 1042 1. 454 4 1043 8. 826 9 1042 2 5 3 = 1. 454 4 × 1043 4. 192 3 × 1043 2. 544 4 × 1043  1 3 2   8. 826 9 × 1042 2. 544 4 × 1043 1. 544 2 × 1043  × × ×  0.284 32 0.756 78 0.588 6  = 0.819 59 0.510 39 0.260 32  0.497 42− 0.408 40 0.765 36  · − 1. 774 5 1043 5. 115 1 1043 3. 104 5 1043 0× 7. 091 8 × 1038 8. 072 9 × 1037  0 0× 3. 007 5 × 1038  × A matrix similar to the original is  T 0.284 32 0.756 78 0.588 6 0 2 1 0.819 59 0.510 39 0.260 32 2 5 3  0.497 42− 0.408 40 0.765 36   1 3 2  −  0.284 32 0.756 78 0.588 6    0.819 59 0.510 39 0.260 32 =  0.497 42− 0.408 40 0.765 36  −  5  5 7. 514 5 9. 130 9 10− 1. 248 7 10− 5 × × 9. 130 9 10− 0.541 46 0.344 28  × 5 − − 2  1. 248 7 10− 0.344 28 2. 686 9 10− × − ×   0.284 32 Thus an eigenvalue is 7. 514 5 and an eigenvector is approximately 0.819 59 . How  0.497 42  well does it work?  

Saylor URL: http://www.saylor.org/courses/ma212/ The Saylor Foundation 176 Exercises

0 2 1 0.284 32 2. 136 6 2 5 3 0.819 59 = 6. 158 9  1 3 2   0.497 42   3. 737 9   0.284 32   2. 136 5  0.819 59 7. 514 5 = 6. 158 8 .  0.497 42   3. 737 9  This has essentially found it. However, we don’t know the others yet. begin with the one which is closest to 0.541 46 − 1 0 2 1 1 0 0 − 2 5 3 + .541 46 0 1 0  1 3 2   0 0 1   5. 323 8 2. 181 4  0.480 23 − − = 2. 181 4 0.393 89 0.393 38  0.480 23 −0.393 38− 1. 046 8  − −  20 5. 323 8 2. 181 4 0.480 23 −2. 181 4 0.393 89 −0.393 38 =  0.480 23 −0.393 38− 1. 046 8  − −  5. 483 4 1015 2. 055 8 1015  2. 530 4 1014 2. 055 8× 1015 −7. 707 7 ×1014 9. 487 0× 1013 :  −2. 530 4 ×1014 9. 487 0× 1013 −1. 167 7 ×1013  × − × ×  2  0.935 48 0.353 06 1. 491 6 10− × 2 0.350 73 0.932 81 8. 286 3 10−  − 2 2 − ×  4. 316 9 10− 7. 228 5 10− 0.996 45 × − × −  5. 861 6 1015 2. 197 6 1015 2. 704 9 1014  0× − 2. 167 1 ×1010 1. 796 2× 109  0 0× − 8. 270 5 ×107  ×  2 T 0.935 48 0.353 06 1. 491 6 10− × 2 0.350 73 0.932 81 8. 286 3 10−  − 2 2 − ×  4. 316 9 10− 7. 228 5 10− 0.996 45 × − × −  5. 323 8 2. 181 4 0.480 23  −2. 181 4 0.393 89 −0.393 38  0.480 23 −0.393 38− 1. 046 8  − −   2 0.935 48 0.353 06 1. 491 6 10− × 2 0.350 73 0.932 81 8. 286 3 10− =  − 2 2 − ×  4. 316 9 10− 7. 228 5 10− 0.996 45 × − × −  5 6  6. 163 8 1. 742 0 10− 1. 159 3 10− − 5 × − × 1. 742 0 10− 0.513 51 0.577 10  × 6  1. 159 3 10− 0.577 10 0.979 41 − ×  1 = 6. 163 8, Solution is: 0.703 70  λ+.541 46 − − 0.935 48 0.350 73 . How well does it work?  − 2  4. 316 9 10− ×  0 2 1  1 0 0 0.935 48 2 5 3 + 0.703 70 0 1 0 0.350 73 :      − 2  1 3 2 0 0 1 4. 316 9 10− ×      

Saylor URL: http://www.saylor.org/courses/ma212/ The Saylor Foundation Exercises 177

6 6. 276 10− × 6 8. 299 10−  × 6  6. 025 3 10− × Next find the eigenvalue closest to 0. 1 0 2 1 − 1.0 1.0 1.0 2 5 3 = −1.0 1.0 −2.0  1 3 2   1.0 2.0− 4.0  − −    20  1.0 1.0 1.0 −1.0 1.0 −2.0  1.0 2.0− 4.0  − −  1. 287 1 1013 2. 778 8 1013 5. 314 3 1013 = 2. 778 8 × 1013 5. 999 6 × 1013 −1. 147 4 × 1014 :  5. 314 3× 1013 1. 147 4× 1014 −2. 194 2 ×1014  − × − × × 0.209 85 0.930 85 0.299 13  0.453 05 −0.178 55 0.873 42  0.866 43 −0.318 81 −0.384 26  − − −  6. 133 5 1013 1. 324 3 1014 2. 532 5 1014 0× 1. 509 3× 109 −1. 706 9 ×109  0 0× 6. 192 3 × 109  ×  T  0.209 85 0.930 85 0.299 13 0.453 05 −0.178 55 0.873 42  0.866 43 −0.318 81 −0.384 26  − − −  1.0 1.0 1.0 0.209 85  0.930 85 0.299 13 − − − 1.0 1.0 2.0 0.453 05 0.178 55 0.873 42 =  1.0 2.0− 4.0   0.866 43 −0.318 81 −0.384 26  − − − − −    5 5  5. 288 0 2. 017 9 10− 2. 856 3 10− 5 × − × 2. 017 9 10− 0.916 86 0.727 58  × 5 −  2. 856 3 10− 0.727 58 0.371 12 − × −   0.209 85 1 λ = 5. 288 0, Solution is: 0.189 11. An approximate eigenvector is 0.453 05 .  0.866 43  − How well does it work?   5 0 2 1 1 0 0 0.209 85 1. 473 4 10− − × 5 2 5 3 0.189 11 0 1 0 0.453 05 = 1. 628 6 10−   −      − × 6  1 3 2 0 0 1 0.866 43 9. 422 7 10− . − − ×         0 2 1 5. Find the eigenvalues and eigenvectors of the matrix A = 2 0 3 numerically.  1 3 2  I don’t know the exact eigenvalues in this case. Check your answers by multiplying your numerically computed eigenvectors by the matrix. I will do this one a little differently. You can easily find that the eigenvalues are approximately 4.9, 2.7,.3. To find the eigen pair which goes with .3 to the following. − 1 0 2 1 1 0 0 − 2 0 3 .3 0 1 0 =  1 3 2  −  0 0 1     

Saylor URL: http://www.saylor.org/courses/ma212/ The Saylor Foundation 178 Exercises

2 1. 138 5 4. 788 7 10− 0.754 22 − 2 − × 4. 788 7 10− 0.180 77 0.347 18  − 0.754× 22− 0.347 18 0.468 10  −  2 20 1. 138 5 4. 788 7 10− 0.754 22 − 2 − × 4. 788 7 10− 0.180 77 0.347 18 =  − 0.754× 22− 0.347 18 0.468 10  −  17805. 3431. 1 12214.  − 3431. 1 661. 21 2353. 7 =  12214. 2353. 7− 8378. 3  − −  0.814 41 0.321 53 0.483 08 0.156 94− 0.923 48 −0.350 07  0.558 67 0.209 28 −0.802 55  − − −  21863. 4213.0 14997. 2 − 2 0 2. 261 7 10− 6. 348 1 10−  0 0× 0.281× 26  A similar matrix is  T 0.814 41 0.321 53 0.483 08 0.156 94− 0.923 48 −0.350 07  0.558 67 0.209 28 −0.802 55  − − −  2 1. 138 5 4. 788 7 10− 0.754 22 − 2 − × 4. 788 7 10− 0.180 77 0.347 18  − 0.754× 22− 0.347 18 0.468 10  −  0.814 41 0.321 53 0.483 08  − − 0.156 94 0.923 48 0.350 07 =  0.558 67 0.209 28 −0.802 55  − − −  6  5 1. 665 1 7. 437 4 10− 1. 163 3 10− − 6 × × 7. 437 4 10− 0.296 62 0.142 05  × 5 − −  1. 163 3 10− 0.142 05 0.174 36 × − Thus an eigenvalue is close to 1.6651 and so the eigenvalue for the original matrix 1 − is the solution of λ .3 = 1.6651, Solution is: 0.300 56. Then what about the eigenvector? We have− − − 1 1 (A .3I)− v = v − λ .3 − It follows that (λ .3) v = (A .3I) v. Thus you have Av = λv. Thus an eigenvector is just the ﬁrst column− of that− orthogonal matrix above. Check it. 0 2 1 0.814 41 0.244 79 2 0 3 0.156 94 = −0.047 19  1 3 2   0.558 67   −0.167 89  −  0.814 41     0.244 78  − 2 0.156 94 ( 0.300 56) = 4. 717 0 10−  0.558 67  −  − 0.167× 91  − This worked very well. Now lets ﬁnd the pair associated with the eigenvalue being close to 2.7. − 1 0 2 1 1 0 0 − 2 0 3 + 2.7 0 1 0  1 3 2   0 0 1     

Saylor URL: http://www.saylor.org/courses/ma212/ The Saylor Foundation Exercises 179

7. 969 8 13. 823 7. 127 4 − = 13. 823 25. 248 13. 175  −7. 127 4 13. 175− 7. 105 8  −  20 7. 969 8 13. 823 7. 127 4 13. 823− 25. 248 13. 175 =  −7. 127 4 13. 175− 7. 105 8  −  1. 896 8 1031 3. 436 7 1031 1. 799 7 1031 × − × × 3. 436 7 1031 6. 226 6 1031 3. 260 7 1031 =  −1. 799 7 ×1031 3. 260 7× 1031 −1. 707 5 ×1031  × − × ×  0.439 25 0.891 55 0.110 45  0.795 85 −0.443 21 0.412 55  − − 2 −  0.416 76 9. 330 8 10− 0.904 21 × −  4. 318 3 1031 7. 823 9 1031 4. 097 2 1031 0× − 7. 478 5 ×1026 3. 772 7× 1026  0 0× − 3. 494 4 ×1026  ×  T  0.439 25 0.891 55 0.110 45 0.795 85 −0.443 21 0.412 55  − − 2 −  0.416 76 9. 330 8 10− 0.904 21 × −  7. 969 8 13. 823 7. 127 4 0.439 25 0.891 55 0.110 45 13. 823− 25. 248 13. 175 0.795 85 −0.443 21 0.412 55 =  − −   − − 2 −  7. 127 4 13. 175 7. 105 8 0.416 76 9. 330 8 10− 0.904 21 − × −   4 5  39. 777 1. 791 2 10− 6. 881 10− 4 − × × 1. 791 2 10− 0.336 06 0.128 95  − × 5 −  6. 881 10− 0.128 95 0.210 75 × −  1  Eigenvalue, λ+2.7 = 39. 777, Solution is: 2. 674 9. The eigenvector would be the ﬁrst column of the above orthogonal matrix on− the right. 0 2 1 0.439 25 1. 174 9 − 2 0 3 0.795 85 = 2. 128 8  1 3 2   −0.416 76   1. 114 8  −  0.439 25    1. 174 9  − 0.795 85 ( 2. 674 9) = 2. 128 8 This worked very well.  −0.416 76  −  1. 114 8  − Now consider the largest eigenvalue.  1 0 2 1 1 0 0 − 2 0 3 4.9 0 1 0  1 3 2  −  0 0 1   1. 753 6 2. 962 0 3. 668 8  = 2. 962 0 4. 446 3 5. 621  3. 668 8 5. 621 6. 735 1   40 1. 753 6 2. 962 0 3. 668 8 2. 962 0 4. 446 3 5. 621 =  3. 668 8 5. 621 6. 735 1   1. 144 5 1044 1. 765 1 1044 2. 164 3 1044 1. 765 1 × 1044 2. 722 1 × 1044 3. 337 8 × 1044  2. 164 3 × 1044 3. 337 8 × 1044 4. 092 8 × 1044  × × ×  

Saylor URL: http://www.saylor.org/courses/ma212/ The Saylor Foundation 180 Exercises

0.379 20 0.739 6 0.556 06 0.584 81 0.657 26 −0.475 40  0.717 08− 0.144 92− 0.681 76   3. 018 2 1044 4. 654 7 1044 5. 707 5 1044 0× 6. 539 3 × 1039 5. 579 8 × 1039  0 0× 4. 797 6 × 1039  ×  T  0.379 20 0.739 6 0.556 06 0.584 81 0.657 26 −0.475 40  0.717 08− 0.144 92− 0.681 76   1. 753 6 2. 962 0 3. 668 8  2. 962 0 4. 446 3 5. 621  3. 668 8 5. 621 6. 735 1   0.379 20 0.739 6 0.556 06 − 0.584 81 0.657 26 0.475 40 =  0.717 08− 0.144 92− 0.681 76   4  5 13. 259 1. 247 5 10− 1. 129 2 10− 4 × × 2 1. 247 5 10− 0.142 61 2. 290 2 10−  × 5 − 2 ×  1. 129 2 10− 2. 290 2 10− 0.181 74 × × −  1  λ 4.9 = 13. 259, Solution is: 4. 975 4. Check this. − 0 2 1 0.379 20 1. 886 7 2 0 3 0.584 81 = 2. 909 6  1 3 2   0.717 08   3. 567 8   0.379 20   1. 886 7  0.584 81 4. 975 4 = 2. 909 7 . It worked well.  0.717 08   3. 567 8      3 2 3 6. Consider the matrix A = 2 1 4 and the vector (1, 1, 1)T . Find the shortest  3 4 0  distance between the Rayleigh quotient determined by this vector and some eigenvalue of A. T 1 3 2 3 1 1 q = 1 2 1 4 1 3 = 7. 333 3  1   3 4 0   1        3 2 3 1 1 2 1 4 1 7. 333 3 1     −   3 4 0 1 1 7. 333 3 λ q       | − | ≤ √3 = 0 .471 41

Thus there is an eigenvalue close to 7.333. 1 2 1 7. Consider the matrix A = 2 1 4 and the vector (1, 1, 1)T . Find the shortest  1 4 5  distance between the Rayleigh quotient determined by this vector and some eigenvalue of A.

Saylor URL: http://www.saylor.org/courses/ma212/ The Saylor Foundation Exercises 181

T 1 1 2 1 1 1 q = 1 2 1 4 1 3 = 7  1   1 4 5   1        7 λ = 2. 449 5 | − q| 3 2 3 8. Consider the matrix A = 2 6 4 and the vector (1, 1, 1)T . Find the shortest  3 4 3  − distance between the Rayleigh quotient determined by this vector and some eigenvalue of A. T 1 3 2 3 1 1 q = 1 2 6 4 1 3 = 8.0  1   3 4 3   1  −       3 2 3 1 1 2 6 4 1 8 1     −   3 4 3 1 1 λ 8 − q       | − | ≤ √3 = 3 . 266 0

9. Using Gerschgorin’s theorem, ﬁnd upper and lower bounds for the eigenvalues of A = 3 2 3 2 6 4 .  3 4 3  − 10 λ 12 − ≤ ≤ 10. Tell how to ﬁnd a matrix whose characteristic polynomial is a given monic polynomial. This is called a companion matrix. Find the roots of the polynomial x3 +7x2 +3x+7. 0 0 7 − The companion matrix is 1 0 3  0 1 −7  − 20   0 0 7 8. 062 3 1014 5. 408 5 1015 3. 628 2 1016 − × − × × 1 0 3 = 2. 253 5 1014 1. 511 7 1015 1. 014 1 1016  0 1 −7   7. 726 4 × 1014 −5. 183 2 × 1015 3. 477 ×1016  − × − × ×  0.707 72 0.258 58 0.657 47  = 0.197 82 0.820 86 0.535 78  0.678 23 0.509 24 −0.529 79  · − − 1. 139 2 1015 7. 642 2 1015 5. 126 6 1016 0× − 4. 570 4 ×1010 2. 079 1 × 1010  0 0× 3. 151 4 × 1011  ×  T  0.707 72 0.258 58 0.657 47 0 0 7 0.197 82 0.820 86 0.535 78 1 0 −3  0.678 23 0.509 24 −0.529 79   0 1 −7  · − − −  0.707 72 0.258 58 0.657 47    0.197 82 0.820 86 0.535 78  0.678 23 0.509 24 −0.529 79  − −  

Saylor URL: http://www.saylor.org/courses/ma212/ The Saylor Foundation 182 Exercises

6. 708 3 5. 850 6 5. 220 9 − 5 = 4. 147 2 10− 0.154 76 1. 187 6  × 5  1. 391 6 10− 0.936 81 0.446 46 − × − − Clearly a real root is close to 6. 708 3. Then the other roots can be obtained from the lower right block. − 0.154 76 1. 187 6 , eigenvalues: 0.145 85 + 1. 011i, 0.145 85 1. 011i 0.936 81 0.446 46 − − − − − How well do these work? Try the last one. Evaluating the polynomial at this value of x gives 4 4 6. 350 5 10− + 2. 065 8 10− i × × 11. Find the roots to x4 + 3x3 + 4x2 + x + 1. It has two complex roots. 20 0 0 0 1 1 0 0 −1 =  0 1 0 −4  −  0 0 1 3   −   36482.  27300.0 49899. 2. 448 9 105 14010.0− 9182.0 −77199. 1. 949 9 × 105 =  1. 318 0 105 95190.0 1.−904 1 105 9. 023 5 × 105  × − − × ×  27300.0 49899. 2. 448 9 105 5. 442 5 105   − × ×   2  0.260 30 6. 965 1 10− 0.941 85 0.200 79 2 − × − 9. 996 0 10− 0.251 72 0.209 32 0.939 59  0.940× 38 0.202 95 0.253 14 0.102 07  − 2 −  0.194 78 0.943 71 7. 090 6 10− 0.257 75   − × −   1. 401 6 105 85984. 2. 474 6 105 1. 037 8 106  0× − 70622. −2. 084 2 × 105 3. 625 1 × 105  0 0− 2. 130× 5 5. 038×  −  0 0 0 2. 683 4     2  T 0.260 30 6. 965 1 10− 0.941 85 0.200 79 2 − × − 9. 996 0 10− 0.251 72 0.209 32 0.939 59  0.940× 38 0.202 95 0.253 14 0.102 07  − 2 −  0.194 78 0.943 71 7. 090 6 10− 0.257 75   − × −   0 0 0 1  1 0 0 −1  0 1 0 −4  −  0 0 1 3   −    2 0.260 30 6. 965 1 10− 0.941 85 0.200 79 9. 996 0 10 2 − 0.251× 72− 0.209 32 0.939 59 − =  0.940× 38 0.202 95 0.253 14 0.102 07  − 2 −  0.194 78 0.943 71 7. 090 6 10− 0.257 75   − × −   0.613 47 4. 251 0 0.485 69 2. 096 8  −0.503 89 −2. 337 6 0.115 42 0.330 94  7 − 6  2. 547 6 10− 8. 418 7 10− 0.157 34 0.304 46 × 6 × 6 −  4. 611 8 10− 7. 585 5 10− 0.974 48 0.108 46   × × −  Now there are two blocks of importance.  0.613 47 4. 251 0 − − , eigenvalues: 1. 475 5 + 1. 182 7i, 1. 475 5 1. 182 7i 0.503 89 2. 337 6 − − − −

Saylor URL: http://www.saylor.org/courses/ma212/ The Saylor Foundation Exercises 183

0.157 34 0.304 46 − , eigenvalues: 0.024 44 + 0.528 23i, 0.024 44 0.528 23i 0.974 48 0.108 46 − − − − How well does one of these work? Try one. ( 0.024 44 + 0.528 23i)4 + 3 ( 0.024 44 + 0.528 23i)3 − − +4 ( 0.024 44 + 0.528 23i)2 + ( 0.024 44 + 0.528 23i) − − 5 6 +1 = 2. 887 9 10− 3. 009 2 10− i × − × 12. Suppose A is a real symmetric matrix and the technique of reducing to an upper Hessenberg matrix is followed. Show the resulting upper Hessenberg matrix is actually equal to 0 on the top as well as the bottom. Let QT AQ = H where H is upper Hessenberg. Then take the transpose of both sides. This will show that H = HT and so H is zero on the top as well.

Saylor URL: http://www.saylor.org/courses/ma212/ The Saylor Foundation