Linear Algebra and Its Applications, 2ed.Solution of Exercise Problems

Version 1.0.4, last revised on 2014-08-13.

AbstractThis is a solution manual for Linear algebra and its applications, 2nd edition, by Peter Lax [8]. This

version omits the following problems: exercise 2, 9 of Chapter 8; exercise 3, 4 of Chapter 17; exercises ofChapter 18; exercise 3 of Appendix 3; exercises of Appendix 4, 5, 8 and 11.

If you would like to correct any typos/errors, please send email to zypublic@hotmail.com.

Contents1 Fundamentals 3

2 Duality 6

3 Linear Mappings 9

4 Matrices 13

5 Determinant and Trace 15

6 Spectral Theory 19

7 Euclidean Structure 24

8 Spectral Theory of Self-Adjoint Mappings of a Euclidean Space into Itself 29

9 Calculus of Vector- and Matrix- Valued Functions 33

10 Matrix Inequalities 36

11 Kinematics and Dynamics 39

12 Convexity 41

13 The Duality Theorem 45

14 Normed Linear Spaces 46

15 Linear Mappings Between Normed Linear Spaces 48

16 Positive Matrices 50

17 How to Solve Systems of Linear Equations 51

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18 How to Calculate the Eigenvalues of Self-Adjoint Matrices 52

A Appendix 52A.1 Special Determinants . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52A.2 The Pfaan . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53A.3 Symplectic Matrices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54A.4 Tensor Product . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56A.5 Lattices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56A.6 Fast Matrix Multiplication . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56A.7 Gershgorins Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57A.8 The Multiplicity of Eigenvalues . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57A.9 The Fast Fourier Transform . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57A.10 The Spectral Radius . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57A.11 The Lorentz Group . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58A.12 Compactness of the Unit Ball . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58A.13 A Characterization of Commutators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58A.14 Liapunovs Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59A.15 The Jordan Canonical Form . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59A.16 Numerical Range . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59

2

1 FundamentalsThe books own solution gives answers to Ex 1, 3, 7, 10, 13, 14, 16, 19, 20, 21.

I 1. (page 2) Show that the zero of vector addition is unique.Proof. Suppose 0 and 00 are two zeros of vector addition, then by the denition of zero and commutativity,we have 00 = 00 + 0 = 0 + 00 = 0.I 2. (page 3) Show that the vector with all components zero serves as the zero element of classical vectoraddition.Proof. For any x = (x1; ; xn) 2 Kn, we have

x+ 0 = (x1; ; xn) + (0; ; 0) = (x1 + 0; ; xn + 0) = (x1; ; xn) = x:So 0 = (0; ; 0) is the zero element of classical vector addition.I 3. (page 3) Show that (i) and (iv) are isomorphic.Proof. The isomorphism T can be dened as T ((a1; ; an)) = a1 + a2x+ + anxn1.I 4. (page 3) Show that if S has n elements, (i) and (iii) are isomorphic.Proof. Suppose S = fs1; ; sng. The isomorphism T can be dened as T (f) = (f(s1); ; f(sn)) ,8f 2KS .I 5. (page 4) Show that when K = R, (iv) is isomorphic with (iii) when S consists of n distinct points of R.Proof. For any p(x) = a1 + a2x+ + anxn1, we dene

T (p) = p(x);

where p on the left side of the equation is regarded as a polynomial over R while p(x) on the right side ofthe equation is regarded as a function dened on S = fs1; ; sng. To prove T is an isomorphism, it sucesto prove T is one-to-one. This is seen through the observation that2664

1 s1 s21 sn11

1 s2 s22 sn12

1 sn s

2n sn1n

377526664a1a2...an

37775 =2664p(s1)p(s2) p(sn)

3775and the Vandermonde matrix 2664

1 s1 s21 sn11

1 s2 s22 sn12

1 sn s

2n sn1n

3775is invertible for distinct s1, s2, , sn.I 6. (page 4) Prove that Y + Z is a linear subspace of X if Y and Z are.Proof. For any y; y0 2 Y , z; z0 2 Z and k 2 K, we have (by commutativity and associative law)(y + z) + (y0 + z0) = (z + y) + (y0 + z0) = z + (y + (y0 + z0)) = z + ((y + y0) + z0) = z + (z0 + (y + y0))

= (z + z0) + (y + y0) = (y + y0) + (z + z0) 2 Y + Z;and

k(y + z) = ky + kz 2 Y + Z:So Y + Z is a linear subspace of X if Y and Z are.

3

I 7. (page 4) Prove that if Y and Z are linear subspaces of X, so is Y \ Z.Proof. For any x1; x2 2 Y \ Z, since Y and Z are linear subspaces of X, x1 + x2 2 Y and x1 + x2 2 Z.Therefore, x1+x2 2 Y \Z. For any k 2 K and x 2 Y \Z, since Y and Z are linear subspaces of X, kx 2 Yand kx 2 Z. Therefore, kx 2 Y \ Z. Combined, we conclude Y \ Z is a linear subspace of X.I 8. (page 4) Show that the set f0g consisting of the zero element of a linear space X is a subspace of X.It is called the trivial subspace.Proof. By denition of zero vector, 0+0 = 0 2 f0g. For any k 2 K, k0 = k(0+0) = k0+k0. So k0 = 0 2 f0g.Combined, we can conclude f0g is a linear subspace of X.I 9. (page 4) Show that the set of all linear combinations of x1, , xj is a subspace of X, and that it isthe smallest subspace of X containing x1, , xj . This is called the subspace spanned by x1, , xj .Proof. Dene Y = fk1x1 + + kjxj : k1; ; kj 2 Kg. Then clearly x1 = 1x1 + 0x2 + + 0xj 2 Y .Similarly, we can show x2; ; xj 2 Y . Since for any k1; ; kj ; k01; ; k0j 2 K,

(k1x1 + + kjxj) + (k01x1 + + k0jxj) = (k1 + k01)x1 + + (kj + k0j)xj 2 Y

and for any k1; ; kj ; k 2 K,

k(k1x1 + + kjxj) = (kk1)x1 + + (kkj)xj 2 Y;

we can conclude Y is a linear subspace of X containing x1; ; xj . Finally, if Z is any linear subspace ofX containing x1; ; xj , it is clear that Y Z as Z must be closed under scalar multiplication and vectoraddition. Combined, we have proven Y is the smallest linear subspace of X containing x1; ; xj .I 10. (page 5) Show that if the vectors x1, , xj are linearly independent, then none of the xi is the zerovector.Proof. We prove by contradiction. Without loss of generality, assume x1 = 0. Then 1x1 +0x2 + +0xj =0. This shows x1; ; xj are linearly dependent, a contradiction. So x1 6= 0. We can similarly provex2; ; xj 6= 0.I 11. (page 7) Prove that if X is nite dimensional and the direct sum of Y1, , Ym, then

dimX =X

dimYj :

Proof. Suppose Yi has a basis yi1; ; yini . Then it suces to prove y11 ; ; y1n1 ; ; ym1 ; ; ymnm form a basisofX. By denition of direct sum, these vectors spanX, so we only need to show they are linearly independent.In fact, if not, then 0 has two distinct representations: 0 = 0 + + 0 and 0 = Pmi=1(ai1yi1 + + ainiyini)for some a11; ; a1n1 ; ; am1 ; ; amnm , where not all aij are zero. This is contradictory with the denitionof direct sum. So we must have linear independence, which imply y11 ; ; y1n1 ; ; ym1 ; ; ymnm form a basisof X. Consequently, dimX =P dimYi.I 12. (page 7) Show that every nite-dimensional space X over K is isomorphic to Kn, n = dimX. Showthat this isomorphism is not unique when n is > 1.Proof. Fix a basis x1; ; xn of X, any element x 2 X can be uniquely represented as

Pni=1 i(x)xi for some

i(x) 2 K, i = 1; ; n. We dene the isomorphism as x 7! (1(x); ; n(x)). Clearly this isomorphismdepends on the basis and by varying the choice of basis, we can have dierent isomorphisms.I 13. (page 7) Prove (i)-(iii) above. Show furthermore that if x1 x2, then kx1 kx2 for every scalar k.Proof. For any x1; x2 2 X, if x1 x2, i.e. x1 x2 2 Y , then x2 x1 = (x1 x2) 2 Y , i.e. x2 x1. Thisis symmetry. For any x 2 X, x x = 0 2 Y . So x x. This is reexivity. Finally, if x1 x2, x2 x3, thenx1 x3 = (x1 x2) + (x2 x3) 2 Y , i.e. x1 x3. This is transitivity.

4

I 14. (page 7) Show that two congruence classes are either identical or disjoint.Proof. For any x1; x2 2 X, we can nd y 2 fx1g \ fx2g if and only if x1 y 2 Y and x2 y 2 Y . Then

x1 x2 = (x1 y) (x2 y) 2 Y:

So fx1g \ fx2g 6= ; if and only if fx1g = fx2g.I 15. (page 8) Show that the above denition of addition and multiplication by scalar is independent of thechoice of representatives in the congruence class.Proof. If fxg = fx0g and fyg = fy0g, then xx0; y y0 2 Y . So (x+ y) (x0+ y0) = (xx0)+ (y y0) 2 Y .This shows fx+ yg = fx0 + y0g. Also, for any k 2 K, kx kx0 = k(x x0) 2 Y . So kfxg = fkxg = fkx0g =kfx0g.I 16. (page 9) Denote by X the linear space of all polynomials p(t) of degree < n, and denote by Y the setof polynomials that are zero at t1, , tj , j < n.

(i) Show that Y is a subspace of X.(ii) Determine dimY .(iii) Determine dimX/Y .

Proof. By theory of polynomials, we have

Y =

(q(t)

jYi=1

(t ti) : q(t) is a polynomial of degree < n j):

Then its easy to see dimY = n j and dimX/Y = dimX dimY = j.I 17. (page 10) Prove Corollary 60.Proof. By Theorem 6, dimX/Y = dimX dimY = 0, which implies X/Y = ff0gg. So X = Y .I 18. (page 11) Show that

dimX1 X2 = dimX1 + dimX2:Proof. Dene Y1 = f(x; 0) : x 2 X1; 0 2 X2g and Y2 = f(0; x) : 0 2 X1; x 2 X2g. Then Y1 and Y2 are linearsubspaces of X1X2. It is easy to see Y1 is isomorphic to X1, Y2 is isomorphic to X2, and Y1\Y2 = f(0; 0)g.So by Theorem 7, dimX1X2 = dimY1+dimY2dim(Y1\Y2) = dimX1+dimX20 = dimX1+dimX2.I 19. (page 11) X is a linear space, Y is a subspace. Show that Y X/Y is isomorphic to X.Proof. By Exercise 18 and Theorem 6, dim(Y X/Y ) = dimY + dim(X/Y ) = dimY + dimX dimY =dimX. Since linear spaces of same nite dimension are isomorphic (by one-to-one mapping between theirbases), Y X/Y is isomorphic to X.I 20. (page 12) Which of the following sets of vectors x = (x1; ; xn) in Rn are a subspace of Rn? Explainyour answer.

(a) All x such that x1 0.(b) All x such that x1 + x2 = 0.(c) All x such that x1 + x2 + 1 = 0.(d) All x such that x1 = 0.(e) All x such that x1 is an integer.

Proof. (a) is not since fx : x1 0g is not closed under the scalar multiplication by 1. (b) is. (c) is notsince x1 + x2 + 1 = 0 and x01 + x02 + 1 = 0 imply (x1 + x01) + (x2 + x02) + 1 = 1. (d) is. (e) is not since x1being an integer does not guarantee rx1 is an integer for any r 2 R.

5

I 21. (page 12) Let U , V , and W be subspaces of some nite-dimensional vectors space X. Is the statement

dim(U + V +W ) = dimU + dimV + dimW dim(U \ V ) dim(U \W )dim(V \W ) + dim(U \ V \W ):

true or false? If true, prove it. If false, provide a counterexample.Proof. (From the textbooks solutions, page 279.) The statement is false; here is an example to the contrary:

X = R2 = (x; y) spaceU = fy = 0g; V = fx = 0g;W = fx = yg:U + V +W = R2; U \ V = f0g; U \W = f0gV \W = f0g; U \ V \W = 0:

2 DualityThe books own solution gives answers to Ex 4, 5, 6, 7.

I 1. (page 15) Given a nonzero vector x1 in X, show that there is a linear function l such that

l(x1) 6= 0:Proof. We let Y = fkx1 : k 2 Kg. Then Y is a 1-dimensional linear subspace of X. By Theorem 2 andTheorem 4,

dimY ? = dimX dimY < dimX = dimX 0

So there must exist some l 2 X 0 n Y ? such that l(x1) 6= 0.Remark 1. When K is R or C, the proof can be constructive. Indeed, assume e1, , en is a basis for Xand x1 =

Pni=1 aiei. In the case of K = R, dene l by setting l(ei) = ai, i = 1; ; n; in the case of K = C,

dene l by setting l(ei) = ai (the conjugate of ai), i = 1; ; n. Then in both cases, l(x1) =Pn

i=1 jjaijj2 > 0.I 2. (page 15) Verify that Y ? is a subspace of X 0.Proof. For any l1 and l2 2 Y ?, we have (l1+ l2)(y) = l1(y)+ l2(y) = 0+0 = 0 for any y 2 Y . So l1+ l2 2 Y ?.For any k 2 K, (kl)(y) = k(l(y)) = k0 = 0 for any y 2 Y . So kl 2 Y ?. Combined, we conclude Y ? is asubspace of X 0.I 3. (page 17) Prove Theorem 6.Proof. Since S Y , Y ? S?. For , let x1; ; xm be a maximal linearly independent subset of S.Then S = span(x1; ; xm) and Y = f

Pmi=1 ixi : 1; ; m 2 Kg by Exercise 9 of Chapter 1. By the

denition of annihilator, for any l 2 S? and y =Pmi=1 ixi 2 Y , we havel(y) =

mXi=1

il(xi) = 0:

So l 2 Y ?. By the arbitrariness of l, S? Y ?. Combined, we have S? = Y ?.I 4. (page 18) In Theorem 7 take the interval I to be [1; 1], and take n to be 3. Choose the three pointsto be t1 = a, t2 = 0, and t3 = a.

(i) Determine the weights m1, m2, m3 so that (9) holds for all polynomials of degree < 3.(ii) Show that for a >

p1/3, all three weights are positive.

(iii) Show that for a =p

3/5, (9) holds for all polynomials of degree < 6.

6

Proof. Suppose three linearly independent polynomials p1, p2 and p3 are applied to formula (9). Then m1,m2 and m3 must satisfy the linear equations24p1(t1) p1(t2) p1(t3)p2(t1) p2(t2) p2(t3)

p3(t1) p3(t2) p3(t3)

3524m1m2m3

35 =264R 11 p1(t)dtR 11 p2(t)dtR 11 p3(t)dt

375We take p1(t) = 1, p2(t) = t and p3(t) = t2. The above equation becomes24 1 1 1a 0 a

a2 0 a2

3524m1m2m3

35 =242023

35So 24m1m2

m3

35 =24 1 1 1a 0 aa2 0 a2

351 242023

35 =240 12a 12a21 0 1a20 12a

12a2

351 242023

35 =24 13a22 23a2

13a2

35Then its easy to see that for a >

p1/3, all three weights are positive.

To show formula (9) holds for all polynomials of degree < 6 when a =p

3/5, we note for any odd n 2 N,Z 11

xndx = 0; m1p(a) +m3p(a) = 0 since m1 = m2 and p(x) = p(x); and m2p(0) = 0:

So (9) holds for any xn of odd degree n. In particular, for p(x) = x3 and p(x) = x5. For p(x) = x4, we haveZ 11

x4dx =2

5; m1p(t1) +m2p(t2) +m3p(t3) = 2m1a

4 =2

3a2:

So formula (9) holds for p(x) = x4 when a =p

3/5. Combined, we conclude for a =p

3/5, (9) holds for allpolynomials of degree < 6.Remark 2. In this exercise problem and Exercise 5 below, Theorem 6 is corrected to Theorem 7.I 5. (page 18) In Theorem 7 take the interval I to be [1; 1], and take n = 4. Choose the four points to bea, b, b, a.

(i) Determine the weights m1, m2, m3, and m4 so that (9) holds for all polynomials of degree < 4.(ii) For what values of a and b are the weights positive?

Proof. We take p1(t) = 1, p2(t) = t, p3(t) = t2, and p4(t) = t3. Then m1, m2, m3, and m4 solve the followingequation: 2664

1 1 1 1a b b aa2 b2 b2 a2

a3 b3 b3 a3

37752664m1m2m3m4

3775 =2664

20

2/30

3775

7

Then 2664m1m2m3m4

3775 =2664

1 1 1 1a b b aa2 b2 b2 a2

a3 b3 b3 a3

37751 2664

20

2/30

3775

=

26664b2

2a2+2b2b2

2a32ab21

2a22b21

2a3+2ab2a2

2a22b2a2

2a2b+2b31

2a2+2b21

2a2b2b3a2

2a22b2a2

2a2b2b31

2a2+2b21

2a2b+2b3b2

2a2+2b2b2

2a3+2ab21

2a22b21

2a32ab2

377752664

20

2/30

3775

=

2666643b2+13(a2b2)3a21

3(a2b2)3a21

3(a2b2)3b2+13(a2b2)

377775So the weights are positive if and only if one of the following two mutually exclusive cases hold1) b2 > 13 , a2 < b2, a2 > 13 ;2) b2 < 13 , a2 > b2, a2 < 13 .

I 6. (page 18) Let P2 be the linear space of all polynomials

p(x) = a0 + a1x+ a2x2

with real coecients and degree 2. Let 1, 2, 3 be three distinct real numbers, and then dene

lj = p(j) for j = 1; 2; 3.

(a) Show that l1, l2, l3 are linearly independent linear functions on P2.(b) Show that l1, l2, l3 is a basis for the dual space P 02.(c) (1) Suppose fe1; ; eng is a basis for the vector space V . Show there exist linear functions fl1; ; lng

in the dual space V 0 dened by

li(ej) =

(1 if i = j0 if i 6= j

Show that fl1; ; lng is a basis of V 0, called the dual basis.(2) Find the polynomials p1(x), p2(x), p3(x) in P2 for which l1, l2, l3 is the dual basis in P 02.

(a)Proof. (From the textbooks solutions, page 280) Suppose there is a linear relation

al1(p) + bl2(p) + cl3(p) = 0:

Set p = p(x) = (x 2)(x 3). Then p(2) = p(3) = 0, p1(1) 6= 0; so we get from the above relation thata = 0. Similarly b = 0, c = 0.

(b)Proof. Since dimP2 = 3, dimP 02 = 3. Since l1, l2, l3 are linearly independent, they span P 02.

(c1)

8

Proof. We dene l1 by setting

l1(ej) =

(1; if j = 10; if j 6= 1

and extending l1 to V by linear combination, i.e. l1(Pn

j=1 jej) :=Pn

j=1 j l1(ej) = 1. l2, , ln can beconstructed similarly. If there exist a1, , an such that a1l1 + + anln = 0, we have

0 = a1l1(ej) + anln(ej) = aj ; j = 1; ; n:So l1, , ln are linearly independent. Since dimV 0 = dimV = n, fl1; ; lng is a basis of V 0.

(c2)Proof. We dene

p1(x) =(x x2)(x x3)(x1 x2)(x1 x3) ; p2(x) =

(x x1)(x x3)(x2 x1)(x2 x3) ; p3(x) =

(x x1)(x x2)(x3 x1)(x3 x2) :

I 7. (page 18) Let W be the subspace of R4 spanned by (1; 0;1; 2) and (2; 3; 1; 1). Which linear functionsl(x) = c1x1 + c2x2 + c3x3 + c4x4 are in the annihilator of W?Proof. (From the textbooks solutions, page 280) l(x) has to be zero for x = (1; 0;1; 2) and x = (2; 3; 1; 1).These yield two equations for c1, , c4:

c1 c3 + 2c4 = 0; 2c1 + 3c2 + c3 + c4 = 0:We express c1 and c2 in terms of c3 and c4. From the rst equation, c1 = c3 2c4. Setting this into thesecond equation gives c2 = c3 + c4.

3 Linear MappingsThe books own solution gives answers to Ex 1, 2, 4, 5, 6, 7, 8, 10, 11, 13.

F Comments: To memorize Theorem 5 (R?T = NT 0), recall for a given l 2 U 0, (l; Tx) = 0 for any x 2 Xif and only if T 0l = 0.

I 1. (page 20) Prove Theorem 1.(a)

Proof. For any y; y0 2 T (X), there exist x; x0 2 X such that T (x) = y and T (x0) = y0. So y + y0 =T (x) + T (x0) = T (x + x0) 2 T (X). For any k 2 K, ky = kT (x) = T (kx) 2 T (X). Combined, we concludeT (X) is a linear subspace of U .

(b)Proof. Suppose V is a linear subspace of U . For any x; x0 2 T1(V ), there exist y; y0 2 V such that T (x) = yand T (x0) = y0. Since T (x + x0) = T (x) + T (x0) = y + y0 2 V , x + x0 2 T1(V ). For any k 2 K, sinceT (kx) = kT (x) = ky 2 V , kx 2 T1(V ). Combined, we conclude T1(V ) is a linear subspace of X.I 2. (page 24) Let

nX1

tijxj = ui; i = 1; ;m

be an overdetermined system of linear equationsthat is, the number m of equations is greater than thenumber n of unknowns x1, , xn. Take the case that in spite of the overdeterminacy, this system ofequations has a solution, and assume that this solution is unique. Show that it is possible to select a subsetof n of these equations which uniquely determine the solution.

9

Proof. (From the textbooks solution, page 280) Suppose we drop the ith equation; if the remaining equationsdo not determine x uniquely, there is an x 6= 0 that is mapped into a vector whose components except theith are zero. If this were true for all i = 1; ;m, the range of the mapping x! u would be m-dimensional;but according to Theorem 2, the dimension of the range is n < m. Therefore one of the equations may bedropped without losing uniqueness; by induction m n of the equations may be omitted.

Alternative solution: Uniqueness of the solution x implies the column vectors of the matrix T = (tij) arelinearly independent. Since the column rank of a matrix equals its row rank (see Chapter 3, Theorem 6 andChapter 4, Theorem 2), it is possible to select a subset of n of these equations which uniquely determine thesolution.Remark 3. The textbooks solution is a proof that the column rank of a matrix equals its row rank.I 3. (page 25) Prove Theorem 3.

(i)Proof. S T (ax+ by) = S(T (ax+ by)) = S(aT (x)+ bT (y)) = aS(T (x))+ bS(T (y)) = aS T (x)+ bS T (y).So S T is also a linear mapping.

(ii)Proof. (R + S) T (x) = (R + S)(T (x)) = R(T (x)) + S(T (x)) = (R T + S T )(x) and S (T + P )(x) =S((T + P )(x)) = S(T (x) + P (x)) = S(T (x)) + S(P (x)) = (S T + S P )(x).I 4. (page 25) Show that S and T in Examples 8 and 9 are linear and that ST 6= TS.Proof. For Example 8, the linearity of S and T is easy to see. To see the non-commutativity, consider thepolynomial p(s) = s. We have TS(s) = T (s2) = 2s 6= s = S(1) = ST (s). So ST 6= TS.

For Example 9, 8x = (x1; x2; x3) 2 X, S(x) = (x1; x3;x2) and T (x) = (x3; x2;x1). So its easy tosee S and T are linear. To see the non-commutativity, note ST (x) = S(x3; x2;x1) = (x3;x1;x2) andTS(x) = T (x1; x3;x2) = (x2; x3;x1). So ST 6= TS in general.Remark 4. Note the problem does not specify the direction of the rotation, so it is also possible thatS(x) = (x1;x3; x2) and T (x) = (x3; x2; x1). There are total of four choices of (S; T ), and each of thecorresponding proofs is similar to the one presented above.I 5. (page 25) Show that if T is invertible, TT1 is the identity.Proof. TT1(x) = T (T1(x)) = x by denition. So TT1 = id.I 6. (page 25) Prove Theorem 4.

(i)Proof. Suppose T : X ! U is invertible. Then for any y; y0 2 U , there exist a unique x 2 X and a uniquex0 2 X such that T (x) = y and T (x0) = y0. So T (x+ x0) = T (x) + T (x0) = y + y0 and by the injectivity ofT , T1(y + y0) = x + x0 = T1(y) + T1(y0). For any k 2 K, since T (kx) = kT (x) = ky, injectivity of Timplies T1(ky) = kx = kT1(y). Combined, we conclude T1 is linear.

(ii)Proof. Suppose T : X ! U and S : U ! V . First, by the denition of multiplication, ST is a linear map.Second, if x 2 X is such that ST (x) = 0 2 V , the injectivity of S implies T (x) = 0 2 U and the injectivityof T further implies x = 0 2 X. So, ST is one-to-one. For any z 2 V , there exists y 2 U such that S(y) = z.Also, we can nd x 2 X such that T (x) = y. So ST (x) = S(y) = z. This shows ST is onto. Combined, weconclude ST is invertible.

By associativity, we have (ST )(T1S1) = ((ST )T1)S1 = (S(TT1))S1 = SS1 = idV . ReplaceS with T1 and T with S1, we also have (T1S1)(ST ) = idX . Therefore, we can conclude (ST )1 =T1S1.

10

I 7. (page 26) Show that whenever meaningful,

(ST )0 = T 0S0; (T +R)0 = T 0 +R0; and (T1)0 = (T 0)1:

(i)Proof. Suppose T : X ! U and S : U ! V are linear maps. Then for any given l 2 V 0, ((ST )0l; x) =(l; STx) = (S0l; Tx) = (T 0S0l; x), 8x 2 X. Therefore, (ST )0l = T 0S0l. Let l run through every element ofV 0, we conclude (ST )0 = T 0S0.

(ii)Proof. Suppose T and R are both linear maps from X to U . For any given l 2 U 0, we have ((T +R)0l; x) =(l; (T + R)x) = (l; Tx + Rx) = (l; Tx) + (l; Rx) = (T 0l; x) + (R0l; x) = ((T 0 + R0)l; x), 8x 2 X. Therefore(T +R)0l = (T 0 +R0)l. Let l run through every element of V 0, we conclude (T +R)0 = T 0 +R0.

(iii)Proof. Suppose T is an isomorphism from X to U , then T1 is a well-dened linear map. We rst showT 0 is an isomorphism from U 0 to X 0. Indeed, if l 2 U 0 is such that T 0l = 0, then for any x 2 X, 0 =(T 0l; x) = (l; Tx). As x varies and goes through every element of X, Tx goes through every element ofU . By considering the identication of U with U 00, we conclude l = 0. So T 0 is one-to-one. For any givenm 2 X 0, dene l = mT1, then l 2 U 0. For any x 2 X, we have (m;x) = (m;T1(Tx)) = (l; Tx) = (T 0l; x).Since x is arbitrary, m = T 0l and T 0 is therefore onto. Combined, we conclude T 0 is an isomorphism fromU 0 to X 0 and (T 0)1 is hence well-dened.

By part (i), (T1)0T 0 = (TT1)0 = (idU )0 = idU 0 and T 0(T1)0 = (T1T )0 = (idX)0 = idX0 . This shows(T1)0 = (T 0)1.I 8. (page 26) Show that if X 00 is identied with X and U 00 with U via (5) in Chapter 2, then

T 00 = T:

Proof. Suppose : X ! X 00 and : U ! U 00 are the isomorphisms dened in Chapter 2, formula (5), whichidentify X with X 00 and U with U 00, respectively. Then for any x 2 X and l 2 U 0, we have

(T 00x; l) = (x; T 0l) = (T 0l; x) = (l; Tx) = (Tx; l):

Since l is arbitrary, we must have T 00x = Tx, 8x 2 X. Hence, T 00 = T , which is the preciseinterpretation of T 00 = T .I 9. page 28) Show that if A in L(X;X) is a left inverse of B in L(X;X), that is AB = I, then it is alsoa right inverse: BA = I.Proof. If Bx = 0, by applying A to both sides of the equation and AB = I, we conclude x = 0. So B isinjective. By Corollary B of Theorem 2, B is surjective. Therefore the inverse of B, denoted by B1, alwaysexists, and A = A(BB1) = (AB)B1 = IB1 = B1, which implies BA = I.Remark 5. For a general algebraic structure, e.g. a ring with unit, its not always the case that an elementsright inverse equals to its left inverse. In the proof above, we used the fact that for nite dimensional linearvector space, a linear mapping is injective if and only if its surjective.I 10. (page 30) Show that if M is invertible, and similar to K, then K also is invertible, and K1 is similarto M1.Proof. Suppose K = MS . Then K(M1)S = SMS1SM1S1 = I. By Exercise 9, K is also invertibleand K1 = (M1)S .I 11. (page 30) Prove Theorem 9.

11

Proof. Suppose A is invertible, we have AB = AB(AA1) = A(BA)A1. So AB and BA are similar. Thecase of B being invertible can be proved similarly.I 12. (page 31) Show that P dened above is a linear map, and that it is a projection.Proof. For any ; 2 K and x = (x1; ; xn), y = (y1; ; yn), we have

P (x+ y) = P ((x1 + y1; ; xn + yn))= (0; 0; x3 + y3; ; xn + yn)= (0; 0; x3; ; xn) + (0; 0; y3; ; yn)= (0; 0; x3; ; xn) + (0; 0; y3; ; yn)= P (x) + P (y):

This shows P is a linear map. Furthermore, we have

P 2(x) = P ((0; 0; x3; ; xn)) = (0; 0; x3; ; xn) = P (x):

So P is a projection.I 13. (page 31) Prove that P dened above is linear, and that it is a projection.Proof. For any ; 2 K and f; g 2 C[1; 1], we have

P (f + g)(x) =1

2[(f + g)(x) + (f + g)(x)]

=

2[f(x) + f(x)] +

2[g(x) + g(x)]

= P (f)(x) + P (g)(x):

This shows P is a linear map. Furthermore, we have

(P 2f)(x) = (P (Pf))(x) = P

f() + f()

2

(x) =

1

2

f(x) + f(x)

2+f(x) + f(x)

2

=

1

2(f(x) + f(x)) = (Pf)(x):

So P is a projection.I 14. (page 31) Suppose T is a linear map of rank 1 of a nite dimensional vector space into itself.

(a) Show there exists a unique number c such that T 2 = cT .(b) Show that if c 6= 1 then I T has an inverse. (As usual I denotes the identity map Ix = x.)(a)

Proof. Since dimRT = 1, it suces to prove the following claim: if T is a linear map on a 1-dimensionallinear vector space X, there exists a unique number c such that T (x) = cx, 8x 2 X. We assume theunderlying led K is either R or C. We further assume S : X ! K is an isomorphism. Then S T S1 isa linear map on K. Dene c = S T S1(1), we have

S T S1(k) = S T S1(k 1) = k c; 8k 2 K:

So T S1(k) = S1(c k) = cS1(k), 8k 2 K. This shows T is a scalar multiplication.(b)

Proof. If c 6= 1, its easy to verify I + 11cT is the inverse of I T .

12

I 15. (page 31) Suppose T and S are linear maps of a nite dimensional vector space into itself. Show thatthe rank of ST is less than or equal the rank of S. Show that the dimension of the nullspace of ST is lessthan or equal the sum of the dimensions of the nullspaces of S and of T .Proof. Because RST RS , rank(ST ) = dim(RST ) dimRS = rank(S). Moreover, since the columnrank of a matrix equals its row rank (see Chapter 3, Theorem 6 and Chapter 4, Theorem 2), we haverank(ST ) = rank(T 0S0) rank(T 0) = rank(T ). Combined, we conclude rank(ST ) minfrank(S); rank(T )g.

Also, we note NST /NT is isomorphic to NS \RT , with the isomorphism dened by (fxg) = Tx, wherefxg := x + NT . Its easy to see is well-dened, is linear, and is both injective and surjective. So byTheorem 6 of Chapter 1,

dimNST = dimNT + dimNST /NT = dimNT + dim(NS \RT ) dimNT + dimNS :

Remark 6. The result rank(ST ) minfrank(S); rank(T )g is used in econometrics. Cf. Greene [4, page 985]Appendix A.

4 MatricesThe books own solution gives answers to Ex 1, 2, 4.

I 1. (page 35) Let A be an arbitrary m n matrix, and let D be an m n diagonal matrix,

Dij =

(di if i = j0 if i 6= j:

Show that the ith row of DA equals di times the ith row of A, and show that the jth column of AD equalsdj times the jth column of A.Proof. It looks the phrasement of the exercise is problematic: when m 6= n, AD or DA may not be well-

dened. So we will assume m = n in the below. We can write A in the row form

2664r1r2 rm

3775. Then DA can bewritten as

DA =

2664d1 0 00 d2 0 0 0 dn

37752664r1r2 rn

3775 =2664d1r1d2r2 dnrn

3775We can also write A in the column form [c1; c2; ; cn], then AD can be written as

AD = [c1; c2; ; cn]

2664d1 0 00 d2 0 0 0 dn

3775 = [d1c1; d2c2; ; dncn]

I 2. (page 37) Look up in any text the proof that the row rank of a matrix equals its column rank, andcompare it to the proof given in the present text.Proof. Proofs in most textbooks are lengthy and complicated. For a clear, although still lengthy, proof, see [12, page 112], Theorem 3.5.3.

13

I 3. (page 38) Show that the product of two matrices in 2 2 block form can be evaluated asA11 A12A21 A22

B11 B12B21 B22

=

A11B11 +A12B21 A11B12 +A12B22A21B11 +A22B21 A21B12 +A22B22

Proof. The calculation is a bit messy. We refer the reader to [12, page 190], Theorem 4.6.1.I 4. (page 38) Construct two 2 2 matrices A and B such that AB = 0 but BA 6= 0.

Proof. Let A =1 10 0

and B =

1 21 2

. Then AB = 0 yet BA =

1 11 1

6= 0.

I 5. (page 40) Show that x1, x2, x3, and x4 given by (20)j satisfy all four equations (20).Proof. 2664

1 2 3 12 5 4 32 3 4 11 4 2 2

37752664

1221

3775 =26641 1 + 2 2 + 3 (2) + (1) 12 1 + 5 2 + 4 (2) + (3) 12 1 + 3 2 + 4 (2) + 1 1

1 1 + 4 2 + 2 (2) + (2) 1

3775 =26642113

3775

I 6. (page 41) Choose values of u1, u2, u3, u4 so that condition (23) is satised, and determine all solutionsof equations (22).Proof. We choose u1 = u2 = u3 = 1 and u4 = 2. Then x3 = 5x4 u3 u2 + 3u1 = 5x4 + 1,x2 = 7x4 + u4 3u1 = 7x4 1, and x1 = u1 x2 2x3 3x4 = 1 (7x4 1) 2(5x4 + 1) 3x4 = 0.I 7. (page 41) Verify that l = (1;2;1; 1) is a left nullvector of M :

lM = 0:

Proof.

[1;2;1; 1]

26641 1 2 31 2 3 12 1 2 33 4 6 2

3775= [1 1 2 1 1 2 + 1 3; 1 1 2 2 1 1 + 1 4; 1 2 2 3 1 2 + 1 6; 1 3 2 1 1 3 + 1 2]= 0:

I 8. (page 42) Show by Gaussian elimination that the only left nullvectors ofM are multiples of l in Exercise7, and then use Theorem 5 of Chapter 3 to show that condition (23) is sucient for the solvability of thesystem (22).Proof. Suppose a row vector x = (x1; x2; x3; x4) satises xM = 0. Then we can proceed according toGaussian elimination8>>>>>:

x1 + x2 + 2x3 + 3x4 = 0

x2 + 2x2 + x3 + 4x4 = 0

2x1 + 3x2 + 2x3 + 6x4 = 0

3x1 + x2 + 3x3 + 2x4 = 0

)

8>:x2 x3 + x4 = 0x2 2x3 = 02x2 3x3 7x4 = 0

)(x3 x4 = 05x3 5x4 = 0:

14

So we have x1 = x4, x2 = 2x4, and x3 = x4, i.e. x = x4(1;2;1; 1), a multiple of l in Exercise 7.

Equation (22) has a solution if and only if u =

2664u1u2u3u4

3775 is in RM . By Theorem 5 of Chapter 3, this is equivalentto yu = 0, 8y 2 NM 0 (elements of NM 0 are seen as row vectors). We have proved y is a multiple of l. Hencecondition (23), which is just lu = 0, is sucient for the solvability of the system (22).

5 Determinant and TraceThe books own solution gives answers to Ex 1, 2, 3, 4, 5.

F Comments:1) For a more intuitive proof of Theorem 2 (det(BA) = detAdetB), see Munkres [10, page 18], Theorem

2.10.2) The following proposition is one version of Cramers rule and will be used in the proof of Lemma 6,

Chapter 6 (formula (21) on page 68).Proposition 5.1. Let A be an n n matrix and B dened as the matrix of cofactors of A; that is,

Bij = (1)i+j detAji;

where Aji is the (ji)th minor of A. Then AB = BA = detA Inn.Proof. Suppose A has the column form A = (a1; ; an). By replacing the jth column with the ith columnin A, we obtain

M = (a1; ; aj1; ai; aj ; ; an):On one hand, Property (i) of a determinant gives detM = ij detA; on the other hand, Laplace expansionof a determinant gives

detM =nX

k=1

(1)k+jaki detAkj =nX

k=1

akiBjk = (Bj1; ; Bjn)

0B@ a1i...ani

1CA :Combined, we can conclude detA Inn = BA. By replacing the ith column with the jth column in A, wecan get similar result for AB.

I 1. (page 47) Prove properties (7).(a)

Proof. By formula (5), we have jP (x1; ; xn)j = jQi

Proof. To see (c) is true, we suppose t interchange i0 and j0. Without loss of generality, we assume i0 < j0.Then

P (t(x1; ; xn)) = P (x1; ; xj0 ; ; xi0 ; ; xn)= (xj0 xi0)

Yi

I 5. (page 49) Show that Property (iv) implies Property (i), unless the eld K has characteristic two, thatis, 1 + 1 = 0.Proof. By property (iv), D(a1; ; ai; ; ai; ; an) = D(a1; ; ai; ; ai; ; an): So add to bothsides of the equations D(a1; ; ai; ; ai; ; an) , we have 2D(a1; ; ai; ; ai; ; an) = 0. If thecharacter of the eld K is not two, we can conclude D(a1; ; ai; ; ai; ; an) = 0.Remark 7. This exercise and Exercise 5.4 together show formula (16) is equivalent to Properties (i)-(iii),provided the character of K is not two. Therefore, for K = R or C, we can either use (16) or properties(i)-(iii) as the denition of determinant.I 6. (page 52) Verify that C(A11) has properties (i)-(iii).

Proof. If two column vectors ai and aj (i 6= j) of A11 are equal, we have0ai

=

0aj

. So C(A11) = 0 and

property (i) is satised. Since any linear operation on a column vector ai of A11 can be naturally extendedto0ai

, property (ii) is also satised. Finally, we note when A11 = I(n1)(n1),

1 00 A11

= Inn. So

property (iii) is satised.I 7. (page 52) Deduce Corollary 5 from Lemma 4.Proof. We rst move the j-th column to the position of the rst column. This can be done by interchangingneighboring columns (j 1) times. The determinant of the resulted matrix A1 is (1)j1 detA. Then wemove the i-th row to the position of the rst row. This can be done by interchanging neighboring rows(i 1) times. The resulted matrix A2 has a determinant equal to (1)i1 detA1 = (1)i+j detA. On theother hand, A2 has the form of

1 0 Aij

. By Lemma 4, we have detAij = detA2 = (1)i+j detA. So

detA = (1)i+j detAij .Remark 8. Rigorously speaking, we only proved that swapping two neighboring columns will give a minussign to the determinant (Property (iv)), but we havent proved this property for neighboring rows. This canbe made rigorous by using detA = detAT (Exercise 8 of this chapter).I 8. (page 54) Show that for any square matrix

detAT = detA; AT = transpose of A

[Hint: Use formula (16) and show that for any permutation (p) = (p1).]Proof. We rst show for any permutation p, (p) = (p1). Indeed, by formula (7)(b), we have 1 = (id) =(p p1) = (p)(p1). By formula (7)(a), we conclude (p) = (p1). Second, we denote by bij the(i; j)-th entry of AT . Then bij = aji. By formula (16) and the fact that p 7! p1 is a one-to-one and ontomap from the set of all permutations to itself, we have

detAT =X

(p)bp11 bpnn=

X(p)a1p1 anpn

=X

(p1)a(p1p)1p1 a(p1p)npn=

X(p1)ap1(p1)p1 ap1(pn)pn

=X

(p1)ap1(1)1 ap1(n)n= detA:

17

I 9. (page 54) Given a permuation p of n objects, we dene an associated so-called permutation matrix Pas follows:

Pij =

(1; if j = p(i),0; otherwise.

Show that the action of P on any vector x performs the permutation p on the components of x. Show that ifp, q are two permutations and P , Q are the associated permutation matrices, then the permutation matrixassociated with p q is the product PQ.Proof. By Exercise 2, it suces to prove the property for transpositions. Suppose p interchanges i1; i2 andq interchanges j1; j2. Denote by P and Q the corresponding permutation matrices, respectively. Then forany x = (x1; ; xn)T 2 Rn, we have (ij is the Kronecker sign)

(Px)i =X

Pijxj =X

p(i)jxj =

8>:xi2 if i = i1xi1 if i = i2xi otherwise.

This shows the action of P on any column vector x performs the permutation p on the components of x.Similarly, we have

(Qx)i =

8>:xj2 if i = j1xj1 if i = j2xi otherwise.

Since (PQ)(x) = P (Q(x)), the action of matrix PQ on x performs rst the permutation q and then thepermutation p on the components of x. Therefore, the permutation matrix associated with p q is theproduct of P and Q.I 10. (page 56) Let A be an m n matrix, B an nm matrix. Show that

trAB = trBA

Proof.

tr(AB) =mXi=1

(AB)ii =mXi=1

nXj=1

aijbji =mXj=1

nXi=1

ajibij =nXi=1

mXj=1

bijaji =nXi=1

(BA)ii = tr(BA);

where the third equality is obtained by interchanging the names of the indices i; j.I 11. (page 56) Let A be an n n matrix, AT its transpose. Show that

trAAT =X

a2ij :

The square root of the double sum on the right is called the Euclidean, or Hilbert-Schmidt, norm of thematrix A.Proof.

tr(AAT ) =Xi

(AAT )ii =Xi

Xj

AijATji =

Xi

Xj

aijaij =Xij

a2ij :

I 12. (page 56) Show that the determinant of the 2 2 matrixa bc d

is D = ad bc.

18

Proof. Apply Laplace expansion of a determinant according to its columns (Theorem 6).I 13. (page 56) Show that the determinant of an upper triangular matrix, one whose elements are zerobelow the main diagonal, equals the product of its elements along the diagonal.Proof. Apply Laplace expansion of a determinant according to its columns (Theorem 6) and work by induc-tion.I 14. (page 57) How many multiplications does it take to evaluate detA by using Gaussian elimination tobring it into upper triangular form?Proof. Denote by M(n) the number of multiplications needed to evaluate detA of an n n matrix A byusing Gaussian elimination to bring it into upper triangular form. To use the rst row to eliminate a21,a31, , an1, we need n(n 1) multiplications. So M(n) = n(n 1) + M(n 1) with M(1) = 0. SoM(n) =

Pnk=1 k(k 1) = n(n+1)(2n+1)6 n(n+1)2 = (n1)n(n+1)3 .

I 15. (page 57) How many multiplications does it take to evaluate detA by formula (16)?Proof. Denote byM(n) the number of multiplications needed to evaluate the determinant of an nn matrixby formula (16). Then M(n) = nM(n 1). So M(n) = n!.I 16. (page 57) Show that the determinant of a (3 3) matrix

A =

0@ a b cd e fg h i

1Acan be calculated as follows. Copy the rst two columns of A as a fourth and fth column:0@ a b c a bd e f d e

g h i g h

1AdetA = aei+ bfg + cdh gec hfa idb:

Show that the sum of the products of the three entries along the dexter diagonals, minus the sum of theproducts of the three entries along the sinister diagonals is equal to the determinant of A.Proof. We apply Laplace expansion of a determinant according to its columns (Theorem 6):

det24a b cd e fg h i

35 = a det e fh i

d det

b ch i

+ g det

b ce f

= a(ie fh) d(ib ch) + g(bf ce)= aei+ bfg + cdh gec afh idb:

6 Spectral TheoryThe books own solution gives answers to Ex 2, 5, 7, 8, 12.

F Comments:1) 2 C is an eigenvalue of a square matrix A if and only if it is a root of the characteristic polynomial

det(aI A) = pA(a) (Corollary 3 of Chapter 5). The spectral mapping theorem (Theorem 4) extends thisresult further to polynomials of A.

19

2) The proof of Lemma 6 in this chapter (formula (21) on page 68) used Proposition 5.1 (see the Commentsof Chapter 5).

3) On p.72, the fact that that from a certain index on, Nds become equal can be seen from the followingline of reasoning. Assume Nd1 6= Nd while Nd = Nd+1. For any x 2 Nd+2, we have (AaI)x 2 Nd+1 = Nd.So x 2 Nd+1 = Nd. Then we work by induction.

4) Theorem 12 can be enhanced to a statement on necessary and sucient conditions, which leads to theJordan canonical form (see Appendix A.15 for details).

F Supplementary notes:1) Minimal polynomial is dened from the algebraic point of view as the generator of the polynomial

ring fp : p(A) = 0g. So the powers of its linear factors are given algebraically. Meanwhile, the index of aneigenvalue is dened from the geometric point of view. Theorem 11 says they are equal.

2) As a corollary of Theorem 11, we claim an n n matrix A can be diagonalized over the eld F if andonly if its minimal polynomial can be decomposed into the product of distinct linear factors (polynomials ofdegree 1 over the eld F). Indeed, by the uniqueness of minimal polynomial, we have

mA is the product of distinct linear factors

() Fn =kMj=1

N1(aj)

() Fn has a basis fxigni=1 consisting of eigenvectors of A() A can be diagonalized by the matrix U = (x1; ; xn), such that U1AU = diagf1; ; ng:The above sequence of equivalence also gives the steps to diagonalize a matrix Ai) Compute the characteristic polynomial pA(s) = det(sI A).ii) Solve the equation pA(s) = 0 in F to obtain the eigenvalues of A: a1, , ak.iii) For each aj (j = 1; ; k), solve the homogenous equation (ajI A)x = 0 to obtain the eigenvectors

pertaining to aj : xj1, , xjmj , where mj = dimN1(aj).iv) If Pkj=1mj < n, A cannot be diagonalized in F. If Pkj=1mj = n, A can be diagonalized by the

matrixU = (x11; ; x1m1 ; x21; ; x2m2 ; ; xk1; ; xkmk);

such that U1AU = diagfa1; ; a1| {z }dimN1(a1)

; ; ak; ; ak| {z }dimN1(ak)

g.

3) Suppose X is a linear subspace of Fn that is invariant under the n n matrix A. If A can bediagonalized, the matrix corresponding to AjX can also be diagonalized. This is due to the observation thatX =

Lkj=1(N1(aj) \X).

4) We summarize several relationships among index, algebraic multiplicity, geometric multiplicity, andthe dimension of the space of generalized eigenvectors pertaining to a given eigenvalue. The rst result is anelementary proof of Lemma 10, Chapter 9, page 132.Proposition 6.1 (Geometric and algebraic multiplicities). Let A be an n n matrix over a eld Fand an eigenvalue of A. If m() is the multiplicity of as a root of the characteristic polynomial pA ofA, then dimN1() m().

m() is called the algebraic multiplicity of ; dimN1() is called the geometric multiplicity of and is the dimension of the linear space spanned by the eigenvectors pertaining to . So this result saysgeometric multiplicity dimN1() algebraic multiplicity m().Proof. Let v1; ; vs be a basis of N1() and extend it to a basis of Fn: v1, , vs, u1, , ur. DeneU = (v1; ; vs; u1; ; ur). Then

U1AU = U1A(v1; ; vs; u1; ; ur)= U1(v1; ; vs; Au1; ; Aur)= (U1v1; ; U1vs; U1Au1; ; U1Aur):

20

Because U1U = I, we must have U1AU =Iss B

0 C

and det(I A) = det(I U1AU) =

det( )Iss B

0 I(ns)(ns) C= ( )s det(I C).1 So s m.

We continue to use the notation from Proposition 6.1, and we dene d() as the index of . Then wehaveProposition 6.2 (Index and algebraic multiplicity). d() m().Proof. Let q(s) = pA(s)/(s )m. Then Cn = NpA = Nq

LNm(). For any v 2 Nm+1(), v can be

uniquely written as v = v0 + v00 with v0 2 Nq and v00 2 Nm(). Then v0 = v v00 2 Nm+1() \ Nq.Similar to the second part of the proof of Lemma 9, we can show v0 = 0. So v = v00 2 Nm(). This showsNm+1() = Nm() and hence d() m().

Using the notations from Propositions 6.1 and 6.2, we haveProposition 6.3 (Algebraic multiplicity and the dimension of the space of generalized eigen-vectors). m() = dimNd()().Proof. See Theorem 11 of Chapter 9, page 133.

In summary, we havedimN1(); d() m() = dimNd()():

In words, it becomes

geometric multiplicity of , index of algebraic multiplicity of as a root of the characteristic polynomial= dim. of the space of generalized eigenvectors pertaining to :

I 1. (page 63) Calculate f32.Proof. f32 = a321 /

p5 = 2178309.

I 2. (page 65) (a) Prove that if A has n distinct eigenvalues aj and all of them are less than one in absolutevalue, then all h in Cn

ANh! 0; as N !1;that is, all components of ANh tend to zero.

(b) Prove that if all aj are greater than one in absolute value, then for all h 6= 0,

ANh!1; as N !1;

that is, some components of ANh tend to innity.(a)

Proof. Denote by hj the eigenvector corresponding to the eigenvalue aj . For any h 2 Cn, there exist1; ; n 2 C such that h =

Pj jhj . So ANh =

Pj ja

Nj hj . Dene b = maxfja1j; ; janjg. Then for any

1 k n, j(ANh)kj = jP

j jaNj (hj)kj bN

Pj jj jj(hj)kj ! 0, as N ! 1, since 0 b < 1. This shows

ANh! 0 as N !1.(b)

1For the last equality, see, for example, Munkres [10, page 24], Problem 6, or [6, page 173].

21

Proof. We use the same notation as in part (a). Since h 6= 0, there exists some k0 2 f1; ; ng so thatthe k0th coordinate of h satises hk0 =

Pj j(hj)k0 6= 0. Then j(ANh)k0 j = j

Pj ja

Nj (hj)k0 j. Dene

b1 = max1infjaij : i 6= 0; (hi)k0 6= 0g. Then b1 > 1 and j(ANh)k0 j = jb1jNPni=1 i aNibN1 (hi)k0 ! 1 as

N !1.I 3. (page 66) Verify for the matrices discussed in Examples 1 and 2,

3 21 4

and

0 11 1

;

that the sum of the eigenvalues equals the trace, and their product is the determinant of the matrix.Proof. The verication is straightforward.I 4. (page 69) Verify (25) by induction on N .Proof. Formula (24) gives us Af = af + h, which is formula (25) when N = 1. Suppose (25) holds for anyn N , then AN+1f = A(ANf) = A(aNf +NaN1h) = aNAf +NaN1Ah = aN (af + h) +NaN1ah =aN+1f + (N + 1)aNh. So (25) also holds for N + 1. By induction, (25) holds for any N 2 N.I 5. (page 69) Prove that for any polynomial q,

q(A)f = q(a)f + q0(a)h;

where q0 is the derivative of q and f satises (22).Proof. Suppose q(s) = Pni=0 bisi, then by formula (25), q(A)f = Pni=0 biAif = Pni=0 bi(aif + iai1h) =(Pn

i=0 biai)f + (

Pni=1 ibia

i1)h = q(a)f + q0(a)h.I 6. (page71) Prove (32) by induction on k.Proof. By Lemma 9, Np1pk = Np1 Np2pk = Np1 (Np2 Np3pk) = Np1 Np2 Np3pk = =Np1 Np2 Npk .I 7. (page 73) Show that A maps Nd into itself.Proof. For any x 2 Nd(a), we have (A aI)d(Ax) = (A aI)d+1x+ a(A aI)dx = 0. So Ax 2 Nd(a).I 8. (page 73) Prove Theorem 11.Proof. A number is an eigenvalue of A if and only if its a root of the characteristic polynomial pA. So pA(s)can be written as pA(s) =

Qk1(s ai)mi with each mi a positive integer (i = 1; ; k). We have shown

in the text that pA is a multiple of mA, so we can assume mA(s) =Qki=1(s ai)ri with each ri satisfying

0 ri mi (i = 1; ; k). We argue ri = di for any 1 i k.Indeed, we have

Cn = NpA =kMj=1

Nmj (aj) =kMj=1

Ndj (aj):

where the last equality comes from the observation Nmj (aj) Nmj+dj (aj) = Ndj (aj) by the denition ofdj . This shows the polynomial

Qkj=1(s aj)dj 2 } := fpolynomials p : p(A) = 0g. By the denition of

minimal polynomial, rj dj for j = 1; ; n.Assume for some j, rj < dj , we can then nd x 2 Ndj (aj) n Nrj (aj) with x 6= 0. Dene q(s) =Qk

i=1;i6=j(s ai)ri , then by Corollary 10 x can be uniquely decomposed into x0 + x00 with x0 2 Nq andx00 2 Nrj (aj). We have 0 = (A ajI)djx = (A ajI)djx0 + 0. So x0 2 Nq \ Ndj (aj) = f0g. This impliesx = x00 2 Nrj (aj). Contradiction. Therefore, ri di for any 1 i k.

Combined, we conclude mA(s) =Qki=1(s ai)di .

22

Remark 9. Along the way, we have shown that the index d of an eigenvalue is no greater than the algebraicmultiplicity of the eigenvalue in the characteristic polynomial. Also see Proposition 6.2.I 9. (page 75) Prove Corollary 15.Proof. The extension is straightforward as the key feature of the proof, B maps N (j) into N (j), remainsthe same regardless of the number of linear maps, as far as they commute pairwise.I 10. (page 76) Prove Theorem 18.Proof. For any i 2 f1; ; ng, by Theorem 17, (li; xj) = 0 for any j 6= i. Since x1, , xn span the wholespace and li 6= 0, we must have(li; xi) 6= 0, i = 1; ; n. This proves (a) of Theorem 18. For (b), we note ifx =

Pkj=1 kjxj , then (li; x) = ki(li; xi). So ki = (li; x)/(li; xi).

I 11. (page 76) Take the matrix 0 11 1

from equation (10)0 of Example 2.

(a) Determine the eigenvector of its transpose.(b) Use formulas (44) and (45) to determine the expansion of the vector (0,1) in terms of the eigenvectors

of the original matrix. Show that your answer agrees with the expansion obtained in Example 2.(a)

Proof. The matrix is symmetric, so its equal to its transpose and the eigenvectors are the same: for eigenvaluea1 =

1+p5

2 , the eigenvector is h1 =1a1

; for eigenvalue a2 = 1

p5

2 , the eigenvector is h2 =1a2

.

(b)

Proof. We note (h1; h1) = 1 + a21 = 5+p5

2 and (h2; h2) = 1 + a22 = 5p5

2 . For x =01

, we have (h1; x) = a1

and (h2; x) = a2. So using formula (44) and (45), x = c1h1 + c2h2 with

c1 = a1/5 +

p5

2= 1/

p5; c2 = a2/

5p52

= 1/p5:

This agrees with the expansion obtained in Example 2.I 12. (page 76) In Example 1 we have determined the eigenvalues and corresponding eigenvector of thematrix

3 21 4

as a1 = 2, h1 =

21

, and a2 = 5, h2 =

11

.

Determine eigenvectors l1 and l2 of its transpose and show that

(li; hj) =

(0 for i 6= j6= 0 for i = j

Proof. The transpose of the matrix has the same eigenvalues a1 = 2, a2 = 5. Solving the equation3 12 4

xy

= 2

xy

, we have l1 =

1 1. Solving the equation 3 1

2 4

xy

= 5

xy

, we have l2 =

1 2

.

Then its easy to calculate (l1; h1) = 3, (l1; h2) = 0, (l2; h1) = 0, and (l2; h2) = 3.

23

I 13. (page 76) Show that the matrix

A =

0@ 0 1 11 0 11 1 0

1Ahas -1 as an eigenvalue. What are the other two eigenvalues?Solution.

det(I A) = det24 1 11 11 1

35 = det24 0 1 1 + 20 + 1 1 1 1

35= [(+ 1)2 (2 1)(+ 1)] = (+ 1)2( 2):

So the eigenvalues of A are 1 and 2, and the eigenvalue 2 has a multiplicity of 2.

7 Euclidean StructureThe books own solution gives answers to Ex 1, 2, 3, 5, 6, 7, 8, 14, 17, 19, 20.

F Erratum: In the Note on page 92, the innite-dimensional version of Theorem 15 is Theorem 5 inChapter 15, not Theorem 15.

I 1. (page 80) Prove Theorem 2.Proof. By letting y = xjjxjj , we get jjxjj maxjjyjj=1(x; y). By Schwartz Inequality, maxjjyjj=1(x; y) jjxjj.Combined, we must have jjxjj = maxjjyjj=1(x; y).I 2. (page 87) Prove Theorem 11.Proof. 8x; y and suppose their decomposition are x1 + x2, y1 + y2, respectively. Here x1; y1 2 Y andx2; y2 2 Y ?. Then (P Y y; x) = (y; PY x) = (y1 + y2; x1) = (y1; x1) = (y1; x) = (PY y; x). By the arbitrarinessof x and y, PY = P Y .I 3. (page 89) Construct the matrix representing reection of points in R3 across the plane x3 = 0. Showthat the determinant of this matrix is 1.Proof. Under the reection across the plane f(x1; x2; x3) : x3 = 0g, point (x1; x2; x3) will be mapped to

(x1; x2;x3). So the corresponding matrix is0@1 0 00 1 00 0 1

1A, whose determinant is 1.I 4. (page 89) Let R be reection across any plane in R3.

(i) Show that R is an isometry.(ii) Show that R2 = I.(iii) Show that R = R.

Proof. Suppose the plane L is determined by the equation Ax + By + Cz = D. For any point x =(x1; x2; x3)

0 2 R3, we rst nd y = (y1; y2; y3)0 2 L such that the line segment xy ? L. Then y mustsatisfy the following equations (

Ay1 +By2 + Cy3 = D

(y1 x1; y2 x2; y3 x3) = k(A;B;C)

24

where k is some constant. Solving the equations gives us k = D(Ax1+Bx2+Cx3)A2+B2+C2 and24y1y2y3

35 =24x1x2x3

35+ k24ABC

35 =24x1x2x3

35 1A2 +B2 + C2

24A2 AB ACAB B2 BCCA CB C2

3524x1x2x3

35+ DA2 +B2 + C2

24ABC

35So the symmetric point z = (z1; z2; z3)0 of x with respect to L is given by24z1z2z3

35 = 224y1y2y3

3524x1x2x3

35 =24x1x2x3

35 2A2 +B2 + C2

24A2 AB ACAB B2 BCCA CB C2

3524x1x2x3

35+ 2DA2 +B2 + C2

24ABC

35=

1

A2 +B2 + C2

24A2 +B2 + C2 2AB 2AC2AB A2 B2 + C2 2BC2CA 2CB A2 +B2 C2

3524x1x2x3

35+ 2DA2 +B2 + C2

24ABC

35 :To make the reection R a linear mapping, its necessary and sucient that D = 0. So the problemsstatement should be corrected to let R be reection across any plane in R3 that contains the origin. Then

R =1

A2 +B2 + C2

24A2 +B2 + C2 2AB 2AC2AB A2 B2 + C2 2BC2CA 2CB A2 +B2 C2

35 :R is symmetric, so R = R and by plain calculation, we have RR = R2 = I. By Theorem 12, R is anisometry.I 5. (page 89) Show that a matrix M is orthogonal i its rows are pairwise orthogonal unit vectors.Proof. Suppose M is an n n orthogonal matrix. Let r1; ; rn be its column vectors. Then

I = MMT =

24r1 rn

35 [rT1 ; ; rTn ] =24r1rT1 r1rT2 r1rTn rnr

T1 rnr

T2 rnrTn

35 :So M is orthogonal if and only if rirTj = ij (1 i; j n).I 6. (page 90) Show that jaij j jjAjj.Proof. Note jaij j = sign(aij) eTi Aej , where ek is the column vector that has 1 as the k-th entry and 0elsewhere. Then we apply (ii) of Theorem 13.I 7. (page 94) Show that fAng converges to A i for all x, Anx converges to Ax.Proof. The key is the space X being nite dimensional. See the solution in the textbook.I 8. (page 95) Prove the Schwarz inequality for complex linear spaces with a Euclidean structure.

Proof. For any x; y 2 X and a 2 C, 0 jjx ayjj = jjxjj2 2Re(x; ay) + jaj2jjyjj2. Let a = (x;y)jjyjj2 (assumey 6= 0), then we have

0 jjxjj2 2Re((x; y)

jjyjj2 (x; y))

+j(x; y)j2jjyjj2 ;

which gives after simplication j(x; y)j jjxjjjjyjj.I 9. (page 95) Prove the complex analogues of Theorem 6, 7, and 8.Proof. Proofs are the same as the ones for the real Euclidean space.I 10. (page 95) Prove the complex analogue of Theorem 9.

25

Proof. Proof is the same as the one for real Euclidean space.I 11. (page 96) Show that a unitary map M satises the relations

MM = I

and, conversely, that every map M that satises (45) is unitary.Proof. IfM is a unitary map, then by parallelogram law,M preserves inner product. So 8x; y, (x;MMy) =(Mx;My) = (x; y). Since x is arbitrary, MMy = y, 8y 2 X. So MM = I. Conversely, if MM = I,(x; x) = (x;MMx) = (Mx;Mx). So M is an isometry.I 12. (page 96) Show that if M is unitary, so is M1 and M.Proof. (M1x;M1x) = (M(M1x);M(M1x)) = (x; x). (Mx;Mx) = (x; x) = (MMx;MMx). ByRM = X, (y; y) = (My;My), 8y 2 X. So M1 and M are both unitary.I 13. (page 96) Show that the unitary maps form a group under multiplication.Proof. If M , N are two unitary maps, then (MN)(MN) = NMMN = NN = I. So the set of unitarymaps is closed under multiplication. Exercise 12 shows that each unitary map has a unitary inverse. So theset of unitary maps is a group under multiplication.I 14. (page 96) Show that for a unitary map M , j detM j = 1.

Proof. By Exercise 8 of Chapter 5, detM = detMT = detM . So by MM = I, we have

1 = detM detM = j detM j2;

i.e. jdetM j = 1.I 15. (page 96) Let X be the space of continuous complex-valued functions on [1; 1] and dene the scalarproduct in X by

(f; g) =

Z 11

f(s)g(s)ds:

Let m(s) be a continuous function of absolute value 1: jm(s)j = 1, 1 s 1.Dene M to be multiplication by m:

(Mf)(s) = m(s)f(s):

Show that M is unitary.

Proof. (Mf;Mf) = R 11Mf(s)Mf(s)ds = R 11m(s)f(s) m(s) f(s)ds = R 11 jm(s)j2jf(s)j2ds = (f; f). Thisshows M is unitary.I 16. (page 98) Prove the following analogue of (51) for matrices with complex entries:

jjAjj 0@X

i;j

jaij j21A1/2 :

Proof. The proof is very similar to that of real case, so we omit the details. Note we need the complexversion of Schwartz inequality (Exercise 8).I 17. (page 98) Show that X

i;j

jaij j2 = trAA:

26

Proof. We have

(AA)ij = [ai1; ; ain]24aj1 ajn

35 = nXk=1

aikajk:

So (AA)ii =Pn

k=1 jaikj2 and tr(AA) =P

i;j jaij j2.I 18. (page 99) Show that

trAA = trAA:Proof. This is straightforward from the result of Exercise 17.I 19. (page 99) Find an upper bound and a lower bound for the norm of the 2 2 matrix

A =

1 20 3

The quantityP

i;j jaij j21/2

is called the Hilbert-Schmidt norm of the matrix A.

Solution. Suppose 1 and 2 are two eigenvalues of A. Then by Theorem 3 of Chapter 6, 1+2 = trA = 4and 12 = detA = 3. Solving the equations gives us 1 = 1, 2 = 3. By formula (46), jjAjj 3. Accordingto formula (51), we have jjAjj p12 + 22 + 32 = p14. Combined, we have 3 jjAjj p14 3:7417.I 20. (page 99) (i) w is a bilinear function of x and y. Therefore we write w as a product of x and y, denotedas

w = x y;and called the cross product.

(ii) Show that the cross product is antisymmetric:

y x = x y:

(iii) Show that x y is orthogonal to both x and y.(iv) Let R be a rotation in R3; show that

(Rx) (Ry) = R(x y):

(v) Show thatjjx yjj = jjxjjjjyjj sin ;

where is the angle between x and y.(vi) Show that 0@ 10

0

1A0@ 01

0

1A =0@ 00

1

1A :(vii) Using Exercise 16 in Chapter 5, show that0@ ab

c

1A0@ de

f

1A =0@ bf cecd af

ae bd

1A :(i)

27

Proof. For any 1, 2 2 F, we have

(w(1x1 + 2x2; y); z) = det(1x1 + 2x2; y; z) = 1 det(x1; y; z) + 2 det(x2; y; z)= 1(w(x1; y); z) + 2(w(x2; y); z) = (1w(x1; y) + 2w(x2; y); z):

Since z is arbitrary, we necessarily have w(1x1 + 2x2; y) = 1w(x1; y) + 2w(x2; y). Similarly, we canprove w(x; 1y1 + 2y2) = 1w(x; y1) + 2w(x; y2). Combined, we have proved w is a bilinear function of xand y.

(ii)Proof. We note

(w(x; y); z) = det(x; y; z) = det(y; x; z) = (w(y; x); z) = (w(y; x); z):

By the arbitrariness of z, we conclude w(x; y) = w(y; x), i.e. y x = x y.(iii)

Proof. Since (w(x; y); x) = det(x; y; x) = 0 and (w(x; y); y) = det(x; y; y) = 0, x y is orthogonal to both xand y.

(iv)Proof. We suppose every vector is in column form and R is the matrix that represents a rotation. Then

(RxRy; z) = det(Rx;Ry; z) = (detR) det(x; y;R1z)

and(R(x y); z) = (R(x y))T z = (x y)TRT z = (x y;RT z) = det(x; y;RT z):

A rotation is isometric, so RT = R1 and detR = 1. Combing the above two equations gives us (RxRy; z) = (R(x y); z). Since z is arbitrary, we must have RxRy = R(x y).

(v)Proof. In the equation det(x; y; z) = (xy; z), we set z = xy. Since the geometrical meaning of det(x; y; z)is the signed volume of a parallelogram determined by x, y, z, and since z = x y is perpendicular to x andy, we have det(x; y; z) = jjxjjjjyjj sin jjzjj, where is the angle between x and y. Then by (x y; z) = jjzjj2,we conclude jjx yjj = jjzjj = jjxjjjjyjj sin .

(vi)Proof.

1 = det241 0 00 1 00 0 1

35 = (24100

3524010

35 ;24001

35):So24100

3524010

35 =24ab1

35. By part (iii), we necessarily have a = b = 0. Therefore, we can conclude24100

3524010

35 =24001

35.(vii)

28

Proof. By Exercise 16 of Chapter 5,

det24a d gb e hc f i

35 = det24a b cd e fg h i

35= aei+ bfg + cdh gec hfa idb= (bf ec)g + (cd fa)h+ (ae db)i

=bf ce cd af ae bd

24ghi

35 :So we have 24ab

c

3524def

35 =24bf cecd afae bd

35 :

I 21. (page 100) Show that in a Euclidean space every pair of vector satises

jju+ vjj2 + jju vjj2 = 2jjujj2 + 2jjvjj2:Proof.

jju+ vjj2 + jju vjj2 = (u+ v; u+ v) + (u v; u v) = (u; u+ v) + (v; u+ v) + (u; u v) (v; u v)= (u; u) + (u; v) + (v; u) + (v; v) + (u; u) (u; v) (v; u) + (v; v) = 2jjujj2 + 2jjvjj2:

8 Spectral Theory of Self-Adjoint Mappings of a Euclidean Spaceinto Itself

The books own solution gives answers to Ex 1, 4, 8, 10, 11, 12, 13.

F Erratum: On page 114, formula (37)0 should be an = maxx6=0 (x;Hx)(x;x) instead of an = minx 6=0(x;Hx)(x;x) .

F Comments:1) In Theorem 4, the eigenvectors of H can be complex (the proof did not show they are real), although

the eigenvalues of H are real.2) The following result will help us understand some details in the proof of Theorem 40 (page 108, It

follows from this easily that we may choose an orthonormal basis consisting of real eigenvectors in eacheigenspace Na.)Proposition 8.1. Let X be a conjugate invariant subspace of Cn (i.e. X is invariant under conjugateoperation). Then we can nd a basis of X consisting of real vectors.Proof. We work by induction. First, assume dimX = 1. 8v 2 X with v 6= 0, we must have Rev 2 X andImv 2 X. At least one of them is non-zero and can be taken as a basis. Suppose for all conjugate invariantsubspaces with dimension no more than k the claim is true. Let dimX = k + 1. 8v 2 X with v 6= 0. If Revand Imv are (complex) linearly dependent, there must exist c 2 C and v0 2 Rn such that v = cv0, and we letY = spanfv0g; if Rev and Imv are (complex) linearly independent, we let Y = spanfv; vg = spanfRev; Imvg.In either case, Y is conjugate invariant. Let Y ? = fx 2 X :Pni=1 xiyi = 0; 8y = (y1; ; yn)0 2 Y g. Thenclearly, X = Y LY ? and Y ? is also conjugate invariant. By assumption, we can choose a basis of Y ?consisting exclusively of real vectors. Combined with the real basis of Y , we get a real basis of X.

29

3) For an elementary proof of Theorem 40 by mathematical induction, see [12, page 297], Theorem5.9.4.

4) Theorem 5 (the spectral resolution representation of self-adjoint operators) can be extended to innitedimensional space and is phrased as any self-adjoint operator can be decomposed as the integral w.r.t.orthogonal projections. See any textbook on functional analysis for details.

5) For the second proof of Theorem 4, compare Spivak[13, page 122], Exercise 5-17 and Keener[5, page15], Theorem 1.6 (the maximum principle).

F Supplementary notesIn view of the spectral theorem (Theorem 7 of Chapter 6, p.70), the diagonalization of a self-adjoint

matrix A is reduced to showing that in the decomposition

Cn = Nd1(1)M

M

Ndk(k);

we must have di(i) = 1, i = 1; ; k. Indeed, assume for some , d() > 1. Then for any x 2 N2()nN1(),we have

(I A)x 6= 0; (I A)2x = 0:But the second equation implies

((I A)x; (I A)x) = ((I A)2x; x) = 0:

A contradiction. So we must have d() = 1. This is the substance of the proof of Theorem 4, part (b).

I 1. (page 102) Show thatRe(x;Mx) = (x;Msx):

Proof.x;M +M

2x

=

1

2[(x;Mx) + (x;Mx)] =

1

2[(x;Mx) + (Mx; x)] =

1

2[(x;Mx) + (x;Mx)] = Re(x;Mx):

I 2. (page 104) We have described above an algorithm for diagonalizing q; implement it as a computerprogram.Solution. Skipped for this version.I 3. (page 105) Prove that

p+ + p0 = maxdimS; q 0 on Sand

p + p0 = maxdimS; q 0 on S.Proof. We prove p+ + p0 = maxq(S)0 dimS. p + p0 = maxq(S)0 dimS can be proved similarly. We shalluse representation (11) for q in terms of the coordinates z1, , zn; suppose we label them so that d1, ,dp are nonnegative where p = p+ + p0, and the rest are negative. Dene the subspace S1 to consist ofall vectors for which zp+1 = = zn = 0. Clearly, dimS1 = p and q is nonnegative on S1. This showsp+ + p0 = p maxq(S)0 dimS. If < holds, there must exist a subspace S2 such that q(S2) 0 anddimS2 > p = p++p0. Dene P : S2 ! S3 := fz : zp+1 = zp+2 = = zn = 0g by P (z) = (z1; ; zp; 0; ; 0).Since dimS2 > p = dimS3, there exists some z 2 S2 such that z 6= 0 and P (z) = 0. This impliesz1 = = zp = 0. So q(z) =

Ppi=1 diz

2i +

Pni=p+1 diz

2i =

Pni=p+1 diz

2i < 0, contradiction. Therefore, our

assumption is not true and < cannot hold.I 4. (page 109) Show that the columns of M are the eigenvectors of H.

30

Proof. Write M in the column form M = [c1; ; cn] and multiply M to both sides for formula (24), we get

HM = [Hc1; ;Hcn] = MD = [c1; ; cn]

26641 0 00 2 0 0 0 n

3775 = [1c1; ; ncn];where 1, , n are eigenvalues of M , including multiplicity. So we have Hci = ici, i = 1; ; n. Thisshows the columns of M are eigenvectors of H.I 5. (page 118) (a) Show that the minimum problem (47) has a nonzero solution f .

(b) Show that a solution f of the minimum problem (47) satises the equation

Hf = bMf;

where the scalar b is the value of the minimum (47).(c) Show that the constrained minimum problem

min(y;Mf)=0

(y;Hy)

(y;My)

has a nonzero solution g.(d) Show that a solution g of the minimum problem (47)0 satises the equation

Hg = cMg;

where the scalar c is the value of the minimum (47)0.Proof. The essence of the generalization can be summarized as follows: hx; yi = (x;My) is an inner productand M1H is self-adjoint under this new inner product, hence all the previous results apply.

Indeed, hx; yi is a bilinear function of x and y; it is symmetric since M is self-adjoint; and it is positivesince M is positive. Combined, we can conclude hx; yi is an inner product.

Because M is positive, Mx = 0 has a unique solution 0. So M1 exists. Dene U = M1H and wecheck U is self-adjoint under the new inner product h; i. Indeed, 8x; y 2 X,

hUx; yi = (Ux;My) = (M1Hx;My) = (Hx; y) = (x;Hy) = (x;MM1Hy) = (x;MUy) = hx;Uyi:

Applying the second proof of Theorem 4, with (; ) replaced by h; i and H replaced by M1H, we canverify claims (a)-(d) are true.I 6. (page 118) Prove Theorem 11.Proof. This is just Theorem 4 with (; ) replaced by h; i and H replaced by M1H, where h; i is denedin the solution of Exercise 5.I 7. (page 118) Characterize the numbers bi in Theorem 11 by a minimax principle similar to (40).Solution. This is just Theorem 11 with (; ) replaced by h; i and H replaced byM1H, where h; i is denedin the solution of Exercise 5.I 8. (page 119) Prove Theorem 110.Proof. Under the new inner product h; i = (;M ), U = M1H is selfadjoint. By Theorem 4, all theeigenvalues ofM1H are real. If H is positive andM1Hx = x, then hx; xi = hx;M1Hxi = (x;Hx) > 0for x 6= 0, which implies > 0. So under the condition that H is positive, all eigenvalues of M1H arepositive.I 9. (page 119) Give an example to show that Theorem 11 is false if M is not positive.

31

Solution. Skipped for this version.I 10. (page 119) Prove Theorem 12. (Hint: Use Theorem 8.)Proof. By Theorem 8, we can assume N has an orthonormal basis v1; ; vn consisting of genuine eigen-vectors. We assume the eigenvalue corresponding to vj is nj . Then by letting x = vj , j = 1; ; n andby the denition of jjN jj, we can conclude jjN jj max jnj j. Meanwhile, 8x 2 X with jjxjj = 1, there exista1; ; an 2 C, so that

P jaj j2 = 1 and x =P ajvj . SojjNxjjjjxjj = jj

Xajnjvj jj =

qXjajnj j2 max

1jnjnj j

qXjaj j2 = max jnj j:

This implies jjN jj max jnj j. Combined, we can conclude jjN jj = max jnj j.Remark 10. Compare the above result with formula (48) and Theorem 18 of Chapter 7.I 11. (page 119) We dene the cyclic shift mapping S, acting on vectors in Cn, by S(a1; a2; ; an) =(an; a1; ; an1).

(a) Prove that S is an isometry in the Euclidean norm.(b) Determine the eigenvalues and eigenvectors of S.(c) Verify that the eigenvectors are orthogonal.

Proof. jS(a1; ; an)j = j(an; a1; ; an1)j = j(a1; ; an)j. So S is an isometry in the Euclidean norm.To determine the eigenvalues and eigenvectors of S, note under the canonical basis e1; ; en, S correspondsto the matrix

A =

0BBBB@0 0 0 0 11 0 0 0 00 1 0 0 0 0 0 0 1 0

1CCCCA ;whose characteristic polynomial is p(s) = jAsIj = (s)n+(1)n+1. So the eigenvalues of S are the solutionsto the equation sn = 1, i.e. k = e 2kn i, k = 1; ; n. Solve the equation Sxk = kxk, we can obtain the gen-eral solution as xk = (n1k ; n2k ; ; k; 1)0. After normalization, we have xk = 1pn (n1k ; n2k ; ; k; 1)0.Therefore, for i 6= j,

(xi; xj) =1

n

nXk=1

k1i k1j =

1

n

nXk=1

(ij)k1 =

1

n

1 (ij)n1 ij

= 0:

I 12. (page 120) (i) What is the norm of the matrix

A =

1 20 3

in the standard Euclidean structure?

(ii) Compare the value of jjAjj with the upper and lower bounds of jjAjj asked for in Exercise 19 of Chapter7.

(i)

Solution. AA =1 02 3

1 20 3

=

1 22 13

, which has eigenvalues 7 p40. By Theorem 13, jjAjj =p

7 +p40 3:65.

(ii)Solution. This is consistent with the estimate obtained in Exercise 19 of Chapter 7: 3 jjAjj 3:7417.

32

I 13. (page 120) What is the norm of the matrix1 0 12 3 0

in the standard Euclidean structures of R2 and R3.

Solution.24 1 20 31 0

351 0 12 3 0

=

24 5 6 16 9 01 0 1

35, which has eigenvalues 0, 1.6477, and 13.3523 By Theorem13, the norm of the matrix is approximately

p13:3523 3:65.

9 Calculus of Vector- and Matrix- Valued FunctionsThe books own solution gives answers to Ex 2, 3, 6, 7.

F Erratum: In Exercise 6 (p.129), we should have det eA = etrA instead of det eA = eA.

F Comments: In the proof of Theorem 11, to see why sI Ad = Qd10 () holds, see Lemma 3 ofAppendix 6, formula (9) (p.369).

I 1. (page 122) Prove the fundamental lemma for vector valued functions. (Hint: Show that for every vectory, (x(t); y) is a constant.)Proof. Following the hint, note ddt (x(t); y) = ( _x(t); y)+(x(t); _y) = 0. So (x(t); y) is a constant by fundamentallemma for scalar valued functions. Therefore (x(t) x(0); y) = 0, 8y 2 Kn. This implies x(t) x(0).I 2. (page 124) Derive formula (3) using product rule (iii).Proof. A1(t)A(t) = I. So 0 = ddt

A1(t)A(t)

= ddtA

1(t) A(t) + A1(t) _A(t) and ddtA1(t) = ddtA1(t) A(t)A1(t) = A1(t) _A(t)A1(t).I 3. (page 128) Calculate

eA+B = exp

0 11 0

:

Solution.0 11 0

2=

1 00 1

. So

0 11 0

n=

8>:I22 if n is even 0 1

1 0

!if n is odd:

Therefore, we have

expfA+Bg =1Xn=0

1

n!

0 11 0

n=

1Xk=0

I22(2k)!

+1Xk=0

1

(2k + 1)!

0 11 0

=e+ e1

2I22 +

e e12

0 11 0

:

I 4. (page 129) Prove the proposition stated in the Conclusion.

33

Proof. For any " > 0, there exists M > 0, so that 8m M , supt jj _Em(t) F (t)jj < ". So 8m M , 8t; h, 1h [Em(t+ h) Em(t)] F (t)

=

1hZ t+ht

[ _Em(s) F (s)]ds+ 1h

Z t+ht

F (s)ds F (t)

R t+ht

jj _Em(s) F (s)jjdsh

+

1hZ t+ht

F (s)ds F (t)

< "+

1hZ t+ht

F (s)ds F (t) :

Under the assumption that F is continuous, we have

limh!0

1h [E(t+ h) E(t)] F (t) = limh!0 limm!1

1h [Em(t+ h) Em(t)] F (t)

"+ limh!0

1hZ t+ht

F (s)ds F (t) = ":

Since " is arbitrary, we must have limh!0 1h [E(t+ h) E(t)] = F (t).I 5. (page 129) Carry out the details of the argument that _Em(t) converges.

Proof. By formula (12), _Em(t) =Pm

k=1

Pk1i=0

1k!A

i(t) _A(t)Aki1(t). So for m and n with m < n,

jj _Em(t) _En(t)jj nX

k=m+1

k1Xi=0

jjAi(t) _A(t)Aki1(t)jjk!

=

nXk=m+1

k1Xi=0

jjA(t)jjk1jj _A(t)jjk!

=nX

k=m+1

jjA(t)jjk1(k 1)! jj

_A(t)jj = jj _A(t)jj[en(jjA(t)jj) em(jjA(t)jj)]! 0

as m;n!1. This shows ( _Em(t))1m=1 is a Cauchy sequence, hence convergent.I 6. (page 129) Apply formula (10) to Y (t) = eAt and show that

det eA = etrA:

Proof. Apply formula (10) to Y (t) = eAt, we have ddt log detY (t) = tr(eAteAtA) = trA. Integrating from 0to t, we get log detY (t) log detY (0) = ttrA. So detY (t) = ettrA. In particular, det eA = etrA.I 7. (page 129) Prove that all eigenvalues of eA are of the form ea, a an eigenvalue of A. Hint: Use Theorem4 of Chapter 6, along with Theorem 6 below.Proof. Without loss of generality, we can assume A is a Jordan matrix. Then eA is an upper triangularmatrix and its entries on the diagonal line have the form ea, where a is an eigenvalue of A. So all eigenvaluesof eA are the exponentials of eigenvalues of A.I 8. (page 142) (a) Show that the set of all complex, self-adjoint n n matrices forms N = n2-dimensionallinear space over the reals.

(b) Show that the set of complex, self-adjoint n n matrices that have one double and n 2 simpleeigenvalues can be described in terms of N 3 real parameters.

(a)

34

Proof. The total number of free entries is n(n+1)2 . The entries on the diagonal line must be real. So thedimension is n(n+1)2 2 n = n2.

(b)Proof. Similar to the argument in the text, the total number of complex parameters that determine theeigenvectors is (n1)+ +2 = n(n1)2 1. This is equivalent to n(n1)2 real parameters. The numberof distinct (real) eigenvalues is n 1. So the dimension = n2 n 2 + n 1 = n2 3.I 9. (page 142) Choose in (41) at random two self-adjoint 10 10 matrices M and B. Using availablesoftware (MATLAB, MAPLE, etc.) calculate and graph at suitable intervals the 10 eigenvalues of B + tMas functions of t over some t-segment.Solution. See the Matlab/Octave program aoc.m below.function aoc

%AOC illustrates the avoidance-of-crossing phenomenon% of the neighboring eigenvalues of a continuous% symmetric matrix. This is Exercise 9, Chapter 9% of the textbook, Linear Algebra and Its Applications,% 2nd Edition, by Peter Lax.

% Initialize global variablesmatrixSize = 10;lowerBound = 0.01; %lower bound of t's rangeupperBound = 3; %upper bound of t's rangestepSize = 0.1;t = lowerBound:stepSize:upperBound;

% Generate random symmetric matrixtemp1 = rand(matrixSize);temp2 = rand(matrixSize);M = temp1+temp1';B = temp2+temp2';

% Initialize eigenvalue matrix to zeros;% use each column to store eigenvalues for% a given parametereigenval = zeros(matrixSize,numel(t));for i = 1:numel(t)

eigenval(:,i) = eig(B+t(i)*M);end

% Plot eigenvalues according to values of parameterhold off;disp(['There are ', num2str(matrixSize), ' eigenvalue curves.']);disp(' ');for j = 1:matrixSize

disp(['Eigenvalue curve No. ', num2str(j),'. Press ENTER to continue...']);plot(t, eigenval(j,:));xlabel('t');ylabel('eigenvalues');title('Matlab illustration of Avoidance of Crossing');hold on;

35

pause;endhold off;

10 Matrix InequalitiesThe books own solution gives answers to Ex 1, 3, 4, 5, 6, 7, 10, 11, 12, 13, 14, 15.

F Erratum: In Exercise 6 (p.152), Imx > 0 should be Imz > 0.

I 1. (page 146) How many square roots are there of a positive mapping?Solution. Suppose H has k distinct eigenvalues 1, , k. Denote by Xj the subspace consisting ofeigenvectors of H pertaining to the eigenvalue j . Then H can be represented as H =

Pkj=1 jPj were Pj

is the projection to Xj . Let A be a positive square root of H, we claim A has to bePk

j=1

pjPj . Indeed, if

is an eigenvalue of A and x is an eigenvector of A pertaining to , then 2 is an eigenvalue of H and x isan eigenvector of H pertaining to 2. So we can assume A has m distinct eigenvalues 1, , m (m k)with i =

pi (1 i m). Denote by Yi the subspace consisting of eigenvectors of A pertaining to i.

Then Yi Xi. Since H =Lm

i=1 Yi =Lk

j=1Xj , we must have m = k and Yi = Xi, otherwise at lease oneof the in the sequence of inequalities dimH =Pmi=1 dimYi Pmi=1 dimXi Pki=1 dimXi = dimH is 0 suciently small and(1; 0)0 to (1; 0)0 with 1 > 0 suciently large. Let B be a mapping that maps the vector (1; 1)0 to (1; 1)0with 1 > 0 suciently small and (1; 1)0 to (2; 2)0 with 2 > 0 suciently large. Then both A and Bare positive mappings, and we can nd x between (1; 1)0 and (0; 1)0 so that (Ax;Bx) < 0. By the analysis inthe paragraph below formula (14)0, AB + BA is not positive. More precisely, we have A =

1 00 2

and

B = 12

1 + 2 1 21 2 1 + 2

.

36

I 4. (page 151) Show that if 0 < M < N , then (a) M1/4 < N1/4. (b) M1/m < N1/m, m a power of 2. (c)logM logN .Proof. By Theorem 5 and induction, it is easy to prove (a) and (b). For (c), we follow the hint. If M hasthe spectral resolution M =Pki=1 iPi, logM is dened as

logM =kXi=1

logiPi =kXi=1

limm!1m(

1mi 1)Pi = limm!1m

kXi=1

1mi Pi

kXi=1

Pi

!= lim

m!1m(M1m I):

So logM = limm!1m(M 1m I) limm!1m(N 1m I) = logN .I 5. (page 151) Construct a pair of mappings 0 < M < N such that M2 is not less than N2. (Hint: UseExercise 3).Solution. (from the textbooks solution, pp. 291) Choose A and B as in Exercise 3, that is positive matriceswhose symmetrized product is not positive. Set

M = A;N = A+ tB;

t suciently small positive number. Clearly, M < N .

N2 = A2 + t(AB +BA) + t2B2;

for t small the term t2B is negligible compared with the linear term. Therefore for t small N2 is not greaterthan M2.I 6. (page 151) Verify that (19) denes f(z) for a complex argument z as an analytic function, as well asthat Imf(z) > 0 for Imz > 0.

Proof. For f(z) = az + b R10

dm(t)z+t , we have

f(z +z) f(z) = az +z Z 10

dm(t)

(z +z + t)(z + t):

So if we can show limz!0R10

dm(t)(z+z+t)(z+t) exists and is nite, f(z) is analytic by denition. Indeed, if

Imz > 0, for z suciently small, we have 1z +z + t 1jz + tj jzj 1Imz jzj 2Imz :

So by Dominated Convergence Theorem, limz!0R10

dm(t)(z+z+t)(z+t) exists and is equal to

R10

dm(t)(z+t)2 , which

is nite. To see Imf(z) > 0 for Imz > 0, we note

Imf(z) = aImz ImZ 10

dm(t)

Rez + t+ iImz = Imza+

Z 10

dm(t)

(Rez + t)2 + (Imz)2:

Remark 11. This exercise can be used to verify formula (19) on p.151.I 7. (page 153) Given m positive numbers r1, , rm, show that the matrix

Gij =1

ri + rj + 1

is positive.

37

Proof. Consider the Euclidean space L2(1; 1], with the inner product (f; g) := R 11 f(t)g(t)dt. Choosefj = e

rj(t1), j = 1; ;m, then the associated Gram matrix is

Gij = (fi; fj) =

Z 11

e(ri+rj)t

eri+rjdt =

1

ri + rj:

Clearly, (fj)mj=1 are linearly independent. So G is positive.I 8. (page 158) Look up a proof of the calculus result (35).Proof. We apply the change of variable formula as followsZ 1

1ez

2

dz =

sZR2ex2y2dxdy =

sZ 20

d

Z 10

er2rdr =

r2 1

2=p:

I 9. (page 162) Extend Theorem 14 to the case when dimV = dimU m, where m is greater than 1.Proof. The extension is straightforward, just replace the paragraph (on page 161) If S is a subspace of V ,then T = S and dimT = dimS. ... It follows that

dimS 1 dimTas asserted. with the following one: Let T = S \ V and T1 = S \ V ?, where V ? stands for the complimentof V in U . Then dimT + dimT1 = dimS. Since dimT1 dimV ? = n (nm) = m, dimT dimS m.

The rest of the proof is the same as the proof of Theorem 14 and we can conclude that

p(A)m p(B) p(A):

I 10. (page 164) Prove inequality (44)0.Proof. For any x, (x; (N M dI)x) = (x; (N M)x) djjxjj2 jjN M jjjjxjj2 djjxjj2 = 0. Similarly,(x; (M N dI)x) = (x; (M N)x) djjxjj2 jjM N jjjjxjj2 djjxjj2 0.I 11. (page 166) Show that (51) is largest when ni and mj are arranged in the same order.Proof. Its easy to see the problem can be reduced to the case k = 2. To prove this case, we note if m1 m2and np1 np2 , we have

m2np1 +m1np2 m2np2 m1np1 = (m2 m1)(np1 np2) 0:

I 12. (page 168) Prove that if the self-adjoint part of Z is positive, then Z is invertible, and the self-adjointpart of Z1 is positive.Proof. Assume Z is not invertible. Then there exists x 6= 0 such that Zx = 0. In particular, this implies(x; Zx) = (x;Zx) = 0. Sum up these two, we get (x; (Z + Z)x) = 0. Contradictory to the assumptionthat the selfadjoint part of Z is positive. For any x 6= 0, there exists y 6= 0 so that x = Zy. So

(x; (Z1 + (Z1))x) = (x; Z1x) + (x; (Z1)x)= (Zy; y) + (Z1x; x)= (y; Zy) + (y; Zy)= (y; (Z + Z)y) > 0:

This shows the selfadjoint part of Z1 is positive.

38

I 13. (page 170) Let A be any mapping of a Euclidean space into itself. Show that AA and AA have thesame eigenvalues with the same multiplicity.Proof. Exercise Problem 14 has proved the claim for non-zero eigenvalues. Since the dimensions of the spacesof generalized eigenvectors of AA and AA are both equal to the dimension of the underlying Euclideanspace, we conclude by Spectral Theorem that their zero eigenvalues must have the same multiplicity.I 14. (page 171) Let A be a mapping of a Euclidean space into another Euclidean space. Show that AAand AA have the same nonzero eigenvalues with the same multiplicity.Proof. Suppose a is a non-zero eigenvalue of AA and x is an eigenvector of AA pertaining to a: AAx = ax.Applying A to both sides, we get AA(Ax) = aAx. Since a 6= 0 and x 6= 0, Ax 6= 0 by AAx = ax.Therefore, a is an eigenvalue of AA with Ax an eigenvector of AA pertaining to a. By symmetry, weconclude AA and AA have the same set of non-zero eigenvalues.

Fix a non-zero eigenvalue a, and suppose x1, , xm is a basis for the space of generalized eigenvectorsof AA pertaining to a. Since a 6= 0, we can claim Ax1, , Axm are linearly independent. Indeed,assume not, then there must exist 1, , m not all equal to 0, such that

Pmi=1 iA

xi = 0. This impliesa(Pm

i=1 ixi) =Pm

i=1 iAAxi = A(

Pmi=1 iA

xi) = 0, which further implies x1, , xm are linearlydependent since a 6= 0. Contradiction.

This shows the dimension of the space of generalized eigenvectors of AA pertaining to a is no greaterthan that of the space of generalized eigenvectors of AA pertaining to a. By symmetry, we conclude thespaces of generalized eigenvectors of AA and AA pertaining to the same nonzero eigenvalue have the samedimension. Combined, we can conclude AA and AA have the same non-zero eigenvalues with the same(algebraic) multiplicity.Remark 12. The multiplicity referred to in this problem is understood as algebraic multiplicity, which isequal to the dimension of the space of generalized eigenvectors.I 15. (page 171) Give an example of a 2 2 matrix Z whose eigenvalues have positive real part but Z +Zis not positive.

Solution. Let Z =1 + bi 3

0 1 + bi

where b could be any real number. Then the eigenvalue of Z, 1 + bi,

has positive real part. Meanwhile, Z + Z =2 33 2

has characteristic polynomial p(s) = (2 s)2 9 =

(s 5)(s+ 1). So Z + Z has eigenvalue 5 and 1, and therefore cannot be positive.I 16. (page 171) Verify that the commutator (50) of two self-adjoint matrices is anti-self-adjoint.Proof. Suppose A and B are selfadjoint. Then for any x and y,

(x; (AB BA)y) = ((AB BA)x; y) = (ABx; y) (BAx; y) = (x;BAy) (x;ABy) = (x;(AB BA)y):

So (AB BA) = (AB BA).

11 Kinematics and DynamicsThe books own solution gives answers to Ex 1, 5, 6, 8, 9.

I 1. (page 174) Show that if M(t) satises a dierential equation of form (11), where A(t) is antisymmetricfor each t and the initial condition (5), then M(t) is a rotation for every t.Proof. We note ddt (M(t)M(t)) = _M(t)M(t) + M(t) _M(t) = A(t) + A(t) = 0. So M(t)M(t) M(0)M(0) = I. Also, f(t) = detM(t) is continuous function of t and takes values either 1 or -1 bythe isometry property of M(t). Since f(0) = 1, we have f(t) 1. By Theorem 1, M(t) is a rotation forevery t.

39

I 2. (page 174) Suppose that A is independent of t; show that the solution of equation (11) satisfying theinitial condition (5) is

M(t) = etA:

Proof. limh!0 M(t+h)M(t)h = limh!0etA(ehAI)

h = AetA, i.e. _M(t) = AM(t). Clearly M(0) = I.

I 3. (page 174) Show that when A depends on t, equation (11) is not solved by

M(t) = eR t0A(s)ds;

unless A(t) and A(s) commute for all s and t.Proof. The reason we need commutativity is that the following equation is required in the calculation ofderivative:

1

h(M(t+ h)M(t)) = 1

h

eR t+h0

A(s)ds eR t0A(s)ds

=

1

h

eR t0A(s)ds+

R t+ht

A(s)ds AR t0A(s)ds

=

1

heR t0A(s)ds

eR t+ht

A(s)ds I;

i.e. eR t0A(s)ds+

R t+ht

A(s)ds = eR t0A(s)dse

R t+ht

A(s)ds. So when this commutativity holds,

_M(t) = limh!0

1

h(M(t+ h)M(t)) = M(t)A(t):

I 4. (page 175) Show that if A in (15) is not equal to 0, then all vectors annihilated by A are multiples of(16).Proof. If f = (x; y; z)T satises Af = 0, we must have8>:

ay + bz = 0

ax+ cz = 0bx cy = 0:

By discussing various possibilities (a; b; c = 0 or not), we can check f is a multiple of (c; b;a)T .I 5. (page 175) Show that the two other eigenvalues of A are ipa2 + b2 + c2.Proof.

det(sI A) = det0@ a ba cb c

1A = (2 + c2) a(a+ bc) + b(ac+ b) = 3 + (c2 + b2 + a2):Solving it gives us the other two eigenvalues.I 6. (page 176) Show that the motion M(t) described by (12) rotation around the axis through the vectorf given by formula (16). Show that the angle of rotation is t

pa2 + b2 + c2. (Hint: Use formula (4)0.)

40

Proof. Since A =0@ 0 a ba 0 cb c 0

1A is anti-symmetric, M(t)M(t) = etAetA = etAetA = I. By Exercise 7 ofChapter 9, all eigenvalues of eAt has the form of eat, where a is an eigenvalue of A. Since the eigenvaluesof A are 0 and ik with k = pa2 + b2 + c2 (Exercise 5), the eigenvalues of eAt are 1 and eikt. Thisimplies det eAt = 1 eikt eikt = 1. By Theorem 1, M = eAt is a rotation. Let f be given by formula(16). From Af = 0 we deduce that eAtf = f ; thus f is the axis of the rotation eAt. The trace ofeAt is 1 + eikt + eikt = 2 cos kt + 1. According to formula (4)0, the angle of rotation of eAt satises2 cos + 1 = treAt. This shows that = kt =

pa2 + b2 + c2.

I 7. (page 177) Show that the commutator

[A;B] = AB BA

of two antisymmetric matrices is antisymmetric.Proof. (AB BA) = (AB) (BA) = BA AB = (B)(A) (A)(B) = BAAB = (AB BA).I 8. (page 177) Let A denote the 3 3 matrix (15); we denote the associated null vector (16) by fA.Obviously, f depends linearly on A.

(a) Let A and B denote two 3 3 antisymmetric matrices. Show that

trAB = 2(fA; fB);

where (; ) denotes the standard scalar product for vectors in R3.Proof. See the solution in the textbook, on page 294.I 9. (page 177) Show that the cross product can be expressed as

fjA;Bj = fA fB :

Proof. See the solution in the textbook, page 294.I 10. (page 184) Verify that solutions of the form (36) form a 2n-dimensional linear space.Proof. It suces to note that the set of 2n functions, f(cos cjt)hj ; (sin cjt)hjgnj=1, are linearly independent,since any two of them are orthogonal when their subscripts are distinct.

12 ConvexityThe books own solution gives answers to Ex 2, 6, 7, 8, 10, 16, 19, 20.

F Comments:1) The following results will help us understand some details in the proofs of Theorem 6 and Theorem

10.Proposition 12.1. Let S be an arbitrary subset of X and x an interior point of S. For any real linearfunction l dened on X, if l 6 0, then l(x) is an interior point of = l(S) in the topological sense.Proof. We can nd y 2 X so that l(y) 6= 0. Then for t suciently small, l(x) + tl(y) = l(x+ ty) 2 . So contains an interval which contains l(x), i.e. l(x) is an interior point of under the topology of R1.Corollary 1. If K is an open convex set and l is a linear function with l 6 0, = l(K) is an open interval.