solutions to axler linear algebra done right pdf

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Solutions to Axler, Linear Algebra Done Right 2nd Ed.Edvard Fagerholmedvard.fagerholm@¦helsinki.[gmail.com¦Beware of errors. I read the book and solved the exercises during spring break (oneweek), so the problems were solved in a hurry. However, if you do nd better orinterestingsolutions to the problems, I'd still like to hear about them. Also please don't put this onthe Internet to encourage copying homework solutions...1 Vector Spaces1. Assuming that C is a eld, write z = a + bi. Then we have 1/z= z/zz = z/[z[2.Plugging in the numbers we get 1/(a + bi) = a/(a2+ b2) - bi/(a2+ b2

) = c + di. Astraightforward calculation of (c+di)(a+bi) = 1 shows that this is indeed an inverse.2. Just calculate ((1 +Ö 3)/2)3.3. We have v + (-v) = 0, so by the uniqueness of the additive inverse (prop. 1.3)-(-v) = v.4. Choose a ,= 0 and v ,= 0. Then assuming av = 0 we get v = a-1

av = a-10 = 0.Contradiction.5. Denote the set in question by A in each part.(a) Let v, w Î A, v = (x1, x2, x3), w = (y1

, y2, y3). Then x1 + 2x2 + 3x3



= 0 andy1 + 2y2 + 3y3 = 0, so that 0 = x1 + 2x2 + 3x3 + y1 + 2y2 + 3y3 = (x1 + y1) +

2(x2 +y2) +3(x3 +y3), so v +w Î A. Similarly 0 = a0 = ax1 +2ax2 +3ay

3, soav Î A. Thus A is a subspace.(b) This is not a subspace as 0 ,Î A.(c) We have that (1, 1, 0) Î A and (0, 0, 1) Î A, but (1, 1, 0)+(0, 0, 1) = (1, 1, 1) ,Î A,so A is not a subspace.(d) Let (x1, x2, x3

), (y1, y2, y3) Î A. If x1 = 5x3



and y1 = 5y3, then ax1 = 5ax3,so a(x1, x2, x3) Î A. Similarly x1 + y1 = 5(x3 + y3), so that (x

1, x2, x3) +(y1, y2, y3) Î A. Thus A is a subspace.

16. Set U = Z2.7. The set ¦(x, x) Î R2[ x Î R¦ È¦(x, -x) Î R2[ x Î R¦ is closed under multiplicationbut is trivially not a subspace ((x, x) + (x, -x) = (2x, 0) doesn't belong to it unlessx = 0).8. Let ¦V

i¦ be a collection of subspaces of V . Set U = ÇiVi. Then if u, v Î U. Wehave that u, v Î Vi for all i because Vi



is a subspace. Thus au Î Vi for all i, so thatav Î U. Similarly u + v Î Vi for all i, so u + v Î U.9. Let U, W ⊂ V be subspaces. Clearly if U ⊂ W or W ⊂ U, then U È W is clarly asubspace. Assume then that U ,⊂ W and W ,⊂ U. Then we can choose u Î U ¸ Wand w Î W ¸ U. Assuming that U È W is a subspace we have u + w Î U È W.Assuming that u + w ⊂ U we get w = u + w -u ⊂ U. Contradiction. Similarly foru + w Î W. Thus U È W is not a subspace.10. Clearly U = U + U as U is closed under addition.11. Yes and yes. Follows directly from commutativity and associativity of vector addition.12. The zero subspace, ¦0¦, is clearly an additive identity. Assumingthat we haveinverses, then the whole space V should have an inverse U such that U + V= ¦0¦.Since V + U = V this is clearly impossible unless V is the trivial vectorspace.13. Let W = R2. Then for any two subspaces U

1, U2 of W we have U1 + W = U2 + W,so the statement is clearly false in general.14. Let W = ¦p Î T(F) [ p =

ni=0

aixi, a2 = a5 = 0¦.15. Let V = R2. Let W = ¦(x, 0) Î R2

[ x Î R¦. Set U1 = ¦(x, x) Î R2[ x Î R¦,U2 = ¦(x, -x) Î R2[ x Î R¦. Then it's easy to see that U



1 + W = U2 + W = R2,but U1 ,= U2, so the statement is false.2 Finite

Dimensional Vector Spaces1. Let un = vn and ui = vi+vi+1, i = 1, . . . , n-1. Now we see that vi

=nj=i ui. Thusvi Î span(u1, . . . , un

), so V = span(v1, . . . , vn) ⊂ span(u1, . . . , un).2. From the previous exercise we know that the span is V . As (v1, . . . , vn

) is a linearlyindependent spanning list of vectors we know that dimV = n. The claim now followsfrom proposition 2.16.23. If (v1 + w, . . . , vn + w) is linearly dependent, then we can write



0 = a1(v1 + w) + . . . + an(vn + w) =n

i=1aivi + wn

i=1ai,where a

i ,= 0 for some i. Now ni=1 ai ,= 0 because otherwise we would get0 =n

i=1a

ivi + wn

i=1ai =n

i=1

aiviand by the linear independence of (vi) we would get ai = 0, ∀i contradicting ourassumption. Thus



w = - _ n

i=1ai

_ -1n

i=1aivi,so w Î span(v1, . . . , vn).4. Yes, multiplying with a nonzero constant doesn't change the degree of a polynomial

and adding two polynomials either keeps the degree constant or makesit the zeropolynomial.5. (1, 0, . . .), (0, 1, 0, . . .), (0, 0, 1, 0, . . .), . . . is trivially linearly independent. Thus F¥isn't nite

dimensional.6. Clearly T(F) is a subspace which is innite dimensional.7. Choose v1 Î V . Assuming that V is innite

dimensional span(v1) ,= V . Thus we can

choose v2 Î V ¸ span(v1). Now continue inductively.8. (3, 1, 0, 0, 0), (0, 0, 7, 1, 0), (0, 0, 0, 0, 1) is clearly linearly independent. Let (x1, x2, x3, x

4, x5) ÎU. Then (x1, x2, x3



, x4, x5) = (3x2, x2, 7x4, x4, x5) = x2(3, 1, 0, 0, 0)+x4(0, 0, 7, 1, 0)+x5(0, 0, 0, 0, 1), so the vectors span U. Hence, they form a basis.9. Choose the polynomials (1, x, x2

-x3, x3). They clearly span T3(F) and form a basisby proposition 2.16.10. Choose a basis (v1, . . . , vn). Let U

i = span(vi). Now clearly V = U1 Å . . . Å Un byproposition 2.19.11. Let dimU = n = dimV . Choose a basis (u1, . . . , un

) for U and extend it to abasis (u1, . . . , un, v1, . . . , vk) for V . By assumption a basis for V has length n, so



(u1, . . . , un) spans V .312. Let U = span(u0, . . . , um). As U ,= Tm(F) we have by the previous exercise thatdimU < dimTm(F) = m + 1. As (p0, . . . , pm) is a spanning list of vectors havinglength m + 1 it is not linearly independent.13. By theorem 2.18 we have 8 = dimR8= dimU + dimW - dim(U Ç W) = 4 + 4 -

dim(U ÇW). Since U +W = R8, we have that dim(U ÇW) = 0 and the claim follows14. Assuming that U Ç W = ¦0¦ we have by theorem 2.18 that9 = dimR9= dimU + dimW -dim(U Ç W) = 5 + 5 -0 = 10.Contradiction.15. Let U1 = ¦(x, 0) Î R2[ x Î R¦, U

2 = ¦(0, x) Î R2[ x Î R¦, U3 = ¦(x, x) Î R2[ x ÎR¦. Then U1Ç U2 = ¦0¦, U

1Ç U3 = ¦0¦ and U2Ç U3 = ¦0¦. Thus the left

hand sideof the equation in the problem is 2 while the right

hand side is 3. Thus we have a



counterexample.16. Choose a basis (ui1, . . . , uini) for each Ui. Then the list of vectors (u11, . . . , umnm) haslength dimU1 + . . . + dimUm and clearly spans U1 + . . . + U

m proving the claim.17. By assumption the list of vectors (u11, . . . , umnm) from the proof of the previous exerciseis linearly independent. Since V = U1+. . .+Un

= span(u11, . . . , umnm) the claim follows.3 Linear Maps1. Any v ,= 0 spans V . Thus, if w Î V , we have w = av for some a Î F. Now T(v) = avfor some a Î F, so for w = bv Î V we have T(w) = T(bv) = bT(v) = bav = aw. ThusT is multiplication by a scalar.2. Dene f by e.g.

f(x, y) = _ x, x = y0, x ,= y ,then clearly f satises the condition, but is non

linear.3. To dene a linear map it's enough to dene the image of the elements of a basis.Choose a basis (u1



, . . . , um) for U and extend it to a basis (u1, . . . , um, v1, . . . , vn) ofV . If S Î L(U, W). Choose some vectors w1, . . . , wn Î W. Now dene T Î L(V, W)by T(ui) = S(ui), i = 1, . . . , m and T(vi) = wi

, i = 1, . . . , n. These relations denethe linear map and clearly T|U = S.4. By theorem 3.4 dimV = dimnull T +dimrange T. If u ,Î null T, then dimrange T >0, but as dimrange T £ dimF = 1 we have dimrange T = 1 and it follows that4dimnull T = dimV -1. Now choose a basis (u1, . . . , un

) for null T. By our assumption(u1, . . . , un, u) is linearly independent and has length dimV . Hence, it's a basis forV . This implies thatV = span(u1, . . . , un, u) = span(u1

, . . . , un) Åspan(u) = null T Å¦au [ a Î F¦.5. By linearity 0 = ni=1 aiT(v



i) = ni=1 T(aivi) = T(

ni=1 aivi). By injectivity wehave ni=1 ai

vi = 0, so that ai = 0 for all i. Hence (T(v1), . . . , T(vn)) is linearlyindependent.6. If n = 1 the claim is trivially true. Assume that the claim is true for n = k.If S

1, . . . , Sk+1 satisfy the assumptions, then S1 Sk is injective by the inductionhypothesis. Let T = S1 Sk. If u ,= 0, then by injectivity of S

k+1 we Sk+1u ,= 0and by injectivity of T we have TSk+1u ,= 0. Hence TSk+1 is injective and the claimfollows.



7. Let w Î W. By surjectivity of T we can nd a vector v Î V suchthat T(v) = w.Writing v = a1v1 + . . . + anvn we get w = T(v) = T(a1v1 + . . . + anvn) = a1T(v1) +. . . + a

nT(vn) proving the claim.8. Let (u1, . . . , un) be a basis for null T and extend it to a basis (u1, . . . , un, v

1, . . . , vk)of V . Let U := span(v1, . . . , vk). Then by construction null T Ç U = ¦0¦ and for anarbitrary v Î V we have that v = a1u1

+ . . . + anun + b1vn + . . . + bk



vk, so thatT(v) = b1T(v1) + . . . + bkT(vk).Hence range T = T(U).9. It's easy to see that (5, 1, 0, 0), (0, 0, 7, 1) is a basis for null T (seeexercise 2.8). Hencedimrange T = dimF4- dimnull T = 4 - 2 = 2. Thus range T = F2, so that T issurjective.10. It's again easy to see that (3, 1, 0, 0, 0), (0, 0, 1, 1, 1) is a basis of null T. Hence, we getdimrange T = dimF

5-dimnull T = 5 -2 = 3 which is impossible.11. This follows trivially from dimV = dimnull T + dimrange T.12. From dimV = dimnull T +dimrange T it follows trivially that if we have a surjectivelinear map T Î L(V, W), then dimV ³ dimW. Assume then that dimV ³ dimW.Choose a basis (v1, . . . , vn) for V and a basis (w1, . . . , w

m) for W. We can denea linear map T Î L(V, W) by letting T(vi) = wi, i = 1, . . . , m, T(vi) = 0, i =m + 1, . . . , n. Clearly T is surjective.513. We have that dimrange T £ dimW. Thus we get dimnull T = dimV -dimrange T ³dimV -dimW. Choose an arbitrary subspace U of V of such that dimU ³ dimV -

dimW. Let (u1, . . . , un) be a basis of U and extend it to a basis (u1, . . . , un, v1



, . . . , vk)of V . Now we know that k £ dimW, so let (w1, . . . , wm) be a basis for W. Denea linear map T Î L(V, W) by T(ui) = 0, i = 1, . . . , n and T(vi) = wi, i = 1, . . . , k.Clearly null T = U.14. Clearly if we can nd such a S, then T is injective. Assume then thatT is injective.Let (v1, . . . , vn) be a basis of V . By exercise 5 (T(v1

), . . . , T(vn)) is linearly inde

pendent so we can extend it to a basis (T(v1), . . . , T(vn), w1, . . . , wm) of W. DeneS Î L(W, V ) by S(T(v

i)) = vi, i = 1, . . . , n and S(wi) = 0, i = 1, . . . , m. ClearlyST = IV .15. Clearly if we can nd such a S, then T is surjective. Otherwise let (v1, . . . , vn

) be abasis of V . By assumption (T(w1), . . . , T(wn)) spans W. By the linear dependencelemma we can make the list of vectors (T(w1), . . . , T(wn



)) a basis by removing somevectors. Without loss of generality we can assume that the rst m vectors formthebasis (just permute the indices). Thus T(w1), . . . , T(wm)) is a basis of W. Denethe map S Î L(W, V ) by S(T(wi)) = wi, i = 1, . . . , m. Now clearly TS = IW.16. We have that dimU = dimnull T + dimrange T £ dimnull T + dimV . SubstitutingdimV = dimnull S + dimrange S and dimU = dimnull ST + dimrange ST we getdimnull ST + dimrange ST £ dimnull T + dimnull S + dimrange S.Clearly dimrange ST £ dimrange S, so our claim follows.17. This is nothing but pencil pushing. Just take arbitrary matrices satisfying the re

quired dimensions and calculate each expression and the equalities easily fall out.

18. Ditto.19. Let V = (x1, . . . ,n ). From proposition 3.14 we have that/(Tv) = /(T)/(v) =

_ ¸

_ a1,1 a

1,n... ... ...am,1 a

m,n _ ¸

_ _ ¸

_ x1.



.

.xn

_ ¸

_ = _ ¸

_ a1,1x1 + . . . + a1,nxn...am,1x

1 + . . . + am,nxn

_ ¸

_ which shows that Tv = (a1,1x1 + . . . + a

1,nxn, . . . , am,1x1 + . . . + am,nxn).20. Clearly dimMat(n, 1, F) = n = dimV . We have that Tv = 0 if

and only if v =0v1 + . . . , 0vn = 0, so null T = ¦0¦. Thus T is injective and hence invertible.621. Let ei denote the n 1 matrix with a 1 in the ith row and 0 everywhere else. Let T



be the linear map. Dene a matrix A := [T(e1), . . . , T(en)]. Now it's trivial to verifythatAei = T(ei).By distributivity of matrix multiplication (exercise 17) we get for an arbitrary v =a1e1 + . . . + anen thatAv = A(a1

e1 + . . . + anen) = a1Ae1 + . . . + anAe

n= a1T(e1) + . . . + anT(en) = T(a1e1

+ . . . + anen),so the claim follows.22. From theorem 3.21 we have ST invertible Û ST bijective Û S and T bijective Û Sand T invertible.23. By symmetry it's sucient to prove this in one direction only. Thus if TS =



I. ThenST(Su) = S(TSu) = Su for all u. As S is bijective Su goes through the whole spaceV as u varies, so ST = I.24. Clearly TS = ST if T is a scalar multiple of the identity.For the other directionassume that TS = ST for every linear map S Î L(V ).25. Let T Î L(F2) be the operator T(a, b) = (a, 0) and S Î L(F2) the operator S(a, b) =(0, b). Then neither one is injective and hence invertible by 3.21. However, T + S isthe identity operator which is trivially invertible. This can be trivially generalizedto spaces arbitrary spaces of dimension ³ 2.26. WriteA :=

_ ¸

_ a1,1

a1,n... ... ...an,1

an,n _ ¸

_, x = _ ¸

_ x1...

xn _ ¸

_.Then the system of equations in a) reduces to Ax = 0. Now A denes a linear mapfrom Mat(n, 1, F) to Mat(n, 1, F). What a) states is now that the map is injectivewhile b) states that it is surjective. By theorem 3.21 these are equivalent.



4 Polynomials1. Let l1, . . . , lm be m distinct numbers and k1, . . . , km > 0 such that their sum is n.Then mi=1(x -li)kiis c

ear

y such a po

ynomia

.72. Let pi(x) =

j=i(x - zj), so that deg pi = m. Now we have that pi(zj) ,= 0 if andon

y if i = j. Let ci

= pi(zi). Denep(x) =m+1

i=1wic-1

i pi(x).Now deg p = m and p c

ear

y p(zi) = wi.3. We on

y need to prove uniqueness as existence is theorem 4.5. Let s



, r

be other suchpo

ynomia

s. Then we get that0 = (s -s

)p + (r -r

) Û r

-r = (s -s

)pWe know that for any po

ynomia

s p ,= 0 ,= q we have deg pq =deg p + deg q.Assuming that s ,= s

we have deg(r

- r) < deg(s - s

)p which is impossib

e. Thuss = s

imp

ying r = r

.4. Let l be a root of p. Thus we can write p(x) = (x -l)q(x). Now we getp

(x) = q(x) + (x -l)q

(x).Thus l is a root of p

if and on

y if l is a root of q i.e. l is a mu

tip

e root. The

statement fo

ows.5. Let p(x) = ni=0 aixi, where ai Î R. By the fundamenta

theorem of ca

cu

us wehave a comp

ex root z. By proposition 4.10 z is a

so a root of p. Then

z and z areroots of equa

mu

tip

icity (divide by (x - z)(x - z) which is rea

). Thus if we haveno rea

roots we have an even number of roots counting mu

tip

icity, but the numberof roots counting mu

tip

icity is deg p hence odd. Thus p has a rea

root.5 Eigenva

ues and Eigenvectors1. An arbitrary e

ement of U1+. . . +U



n is of the form u1+. . . +un where ui Î Ui. Thuswe get T(u1 + . . . + un) = T(u1) + . . . + T(un) Î U1 + . . . + Un by the assumptionthat T(u

i) Î Ui for a

i.2. Let V = ÇiUi where Ui is invariant under T for a

i. Let v Î V , so that v Î Ui for a

i. Now T(v) Î Ui for a

i by assumption, so T(v) Î ÇUi = V . Thus V is invariantunder T.3. The c

ame is c

ear

y true for U = ¦0¦ or U = V . Assume that ¦0¦ ,= U ,=V . Let(u1, . . . , un) be a basis for U and extend it to a basis (u

1, . . . , un, v1, . . . , vm). Byour assumption m ³ 1. Dene a

inear operator by T(ui



) = v1, i = 1, . . . , n andT(vi) = v1, i = 1, . . . , m. Then c

ear

y U is not invariant under T.84. Let u Î nu

(T -lI), so that T(u) -lu = 0. ST = TS gives us0 = S(Tu -lu) = STu -lSu = TSu -lSu = (T -lI)Su,so that Su Î nu

(T -lI).5. C

ear

y T(1, 1) = (1, 1), so that 1 is an eigenva

ue. A

soT(-1, 1) = (1, -1) =-(-1, 1) so -1 is another eigenva

ue. By coro

ary 5.9 these are a

eigenva

ues.6. We easi

y see that T(0,0,1)=(0,0,5), so that 5 is an eigenva

ue. A

so T(1, 0, 0) =(0, 0, 0) so 0 is an eigenva

ue. Assume that l ,= 0 and T(z1, z2, z3) = (2z

2, 0, 5z3) =l(z1, z2, z3). From the assumption l ,= 0 we get z2 = 0, so the equation is of the

form (0, 0, 5z3) = l(z1, 0, z3). Again we see that z1 = 0 so we get the equation(0, 0, 5z3) = l(0, 0, z3

). Thus 5 is the on

y non

zero eigenva

ue.7. Notice that the range of T is the subspace ¦(x, . . . , x) Î Fn[ x Î F¦ and has dimension1. Thus dimrange T = 1, so dimnu

T = n - 1. Assume that T has two distincteigenva

ues l1, l2



and assume that l1 ,= 0 ,= l2. Let v1, v2 be the correspondingeigenvectors, so by theorem 5.6 they are

inear

y independent. Then v1, v2 ,Î nu

T,but dimnu

T = n - 1, so this is impossib

e. Hence T has at most one non

zeroeigenva

ue hence at most two eigenva

ues.Because T is not injective we know that 0 is an eigenva

ue.We a

so see thatT(1, . . . , 1) = n(1, . . . , 1), so n is another eigenva

ue. By the previous paragraph,these are a

eigenva

ues of T.8. Let a Î F. Now we have T(a, a2

, a3, . . .) = (a2, a3, . . .) = a(a, a2, . . .), so every a Î Fis an eigenva

ue.9. Assume that T has k +2 distinct eigenva

ues l1, . . . , l

k+2 with corresponding eiven

vectors v1, . . . , vk+2. By theorem 5.6 these eigenvectors are

inear

y independent. NowTvi = livi

and dimspan(Tv1, . . . , Tvk+2) = dimspan(l1v1, . . . , lk+2



vk+2) ³ k + 1(it's k + 2 if a

li are non

zero, otherwise k + 1). This is a contradiction asdimrange T = k and span(l1v1, . . . , lk+2vk+2) ⊂ range T.10. As T = (T-1)-1we only need to show this in one direction. If T is invertible, then0 is not an eigenvalue. Now let l be an eigenva

ue of T and v the correspondingeigenvector. From Tv = lv we get that T-1

lv = lT-1v = v, so that T-1v = l-1v.11. Let l be an eigenva ue of TS and v the corresponding eigenvector. Then we getSTSv = Slv = lSv, so if Sv ,= 0, then it is is an eigenvector for the eigenva

ue l.IfSv = 0, then TSv = 0, so l = 0. As Sv = 0 we know that S is not injective,so ST

is not injective and it has eigenva

ue 0. Thus if l is an eigenva

ue of TS, then it's aneigenva

ue of ST. The other imp

ication fo

ows by symmetry.912. Let (v1, . . . , vn) be a basis of V . By assumption (Tv1, . . . , Tvn) = (l

1v, . . . , lnvn).We need to show that li = lj



for a

i, j. Choose i ,= j, then by our assumptionvi + vj is an eigenvector, so T(vi + vj) = livi + ljvj = l(vi + vj) = lvi + lv

j. Thismeans that (li - l)vi + (lj - l)vj = 0. Because (vi, v

j) is

inear

y independent weget that li -l = 0 = lj -l i.e. li = lj.13. Let v1

Î V and extend it to a basis (v1, . . . , vn) of V . Now Tv1 = a1v1



+ . . . + anvn.Let Ui be the subspace generated by the vectors vj, j ,= i. By our assumption eachUi is an invariant subspace. Let Tv1 = a1v1 + . . . + anvn. Now v1

Î Ui for i > 1. So

et j > 1, then Tv Î Uj imp

es aj = 0. Thus Tv1 = a1v1

. We see that v1 was aneigenvector. The resu

t now fo

ows from the previous exercise.14. C

ear

y (STS-1)n= STnS-1, so

n

i=0ai(STS-1)i=



n

i=0aiSTiS-1= S

_ n

i=0aiTi

_ S-1.15. Let l be an eigenva

ue of T and v Î V a corresponding eigenvector. Let p(x) =

ni=0 aixi. Then we havep(T)v =n

i=0a

iTiv =n

i=0ailiv = p(l)v.Thus p(l) is an eigenva

ue of p(T). Then

et a be an eigenva

ue of p(T) and

v Î Vthe corresponding eigenvector. Let q(x) = p(x) -a. By the fundamenta

theorem ofa

gebra we can write q(x) = c

ni=1(x -li). Now q(T)v = 0 and q(T) = c



ni=1(T -liI). As q(T) is non

injective we have that T -liI is non

injective for some i. Henceli is an eigenva

ue of T. Thus we get 0 = q(li) = p(li) -a, so that a = p(li).16. Let T Î L(R2) be the map T(x, y) = (-y, x). On page 78 it was shown that it hasno eigenva

ue. However, T2(x, y) = T(-y, x) = (-x, -y), so -1 it has an eigenva

ue.

17. By theorem 5.13 T has an upper

triangu

ar matrix with respect to some basis (v1, . . . , vn).The c

aim now fo

ows from proposition 5.12.18. Let T Î L(F2) be the operator T(a, b) = (b, a) with respect to the standard basis.Now T2

= I, so T is c

ear

y invertib

e. However, T has the matrix _ 0 11 0

_ .1019. Take the operator T in exercise 7. It is c

ear

y not invertib

e, but has the matrix

_ ¸

_ 1 1

..

. ... ...1 1



_ ¸

_.20. By theorem 5.6 we can choose basis (v1, . . . , vn) for V where vi is an eigenvector ofT corresponding to the eigenva

ue li. Let l be the eigenva

ue of S corresponding tovi. Then we getSTvi = Slivi = li

Svi = lilvi = llivi = lTvi = Tlv

i = TSvi,so ST and TS agree on a basis of V . Hence they are equa

.21. C

ear

y 0 is an eigenva

ue and the corresponding eigenvectors are nu

T= W ¸ ¦0¦.Now assume l ,= 0 is an eigenva

ue and v = u + w is a corresponding eigenvector.Then PU,W(u + w) = u = lu + lw Û (1 -l)u = lw. Thus lw Î U, so lw = 0, butl ,= 0, so we have w = 0 which imp

ies (1 -l)u = 0. Because v is an eigenvector we

have v = u + w = u ,= 0, so that 1 - l = 0. Hence l = 1. As we can choose free

your u Î U, so that it's non

zero we see that the eigenvectors corresponding to 1 areu = U ¸ ¦0¦.22. As dimV = dimnu

P + dimrange P, it's c

ear

y sucient to prove that nu

P Çrange P = ¦0¦. Let v Î nu

P Ç range P. As v Î range P, we can nd a u Î V suthat Pu = v. Thus we have that v = Pu = P2



u = Pv, but v Î nu

P, so Pv = 0.Thus v = 0.23. Let T(a, b, c, d) = (-b, a, -d, c), then T is c

ear

y injective, so 0 is not an eigenva

ue.If l ,= 0 is an eigenva

ue and l(a, b, c, d) = (-b, a, -d, c). We c

ear

ysee that a ,=0 Û b ,= 0 and simi

ar

y with c, d. By symmetry we can assume thata ,= 0 (aneigenvector is non

zero). Then we have la = -b and lb = a. Substituting we getl2b = -b Û (l2+ 1)b = 0. As b ,= 0 we have l2+ 1 = 0, but this equation has noso

utions in R. Hence T has no eigenva

ue.24. If U is an invariant subspace of odd degree, then by theorem 5.26 T|U has an eigenva

uel with eigenvector v. Then l is an eigenva

ue of T with eigenvector v against ourassumption. Thus T has no subspace of odd degree.

6 Inner

Product Spaces1. From the

aw of cosines we get |x - y|2= |x|2+ |y|2- 2|x||y| cos q. Solving weget|x||y| cos q = |x|2+|y|

2-|x -y|22 .11Now let x = (x1, x2), y = (y1, y2). A straight calculation shows that

|x|2+|y|2-|x-y|2= x21+x



22+y21 +y22-(x1-y1)2-(x2-y2)2= 2(x1y1

+x2y2),so that|x||y| cos q = |x|2+|y|2-|x -y|2

2 = 2(x1y1 + x2y2)2 = ¸x, y) .

2. If ¸u, v) = 0, then ¸u, av) = 0, so by the Pythagorean theorem |u| £ |u| + |av| =|u+av|. Then assume that |u| £ |u+av| for all a Î F. We have |u|2£ |u+av|2Û¸u, u) £ ¸u + av, u + av). Thus we get¸u, u) £ ¸u, u) +¸u, av) +¸av, u) +¸av, av) ,so that-2Re a ¸u, v) £ [a[



2|v|2.Choose a = -t ¸u, v) with t > 0, so that2t ¸u, v)2£ t2¸u, v)2|v|2Û 2 ¸u, v)2£ t ¸u, v)2|v|2.If v = 0, then clearly ¸u, v) = 0. If not choose t = 1/|v|2, so that we get2 ¸u, v)

2£ ¸u, v)2.Thus ¸u, v) = 0.3. Let a = (a1,Ö 2a2, . . . ,Ö

nan) Î Rnand b = (b1, b2/Ö 2, . . . , bn/

Ö n) Î Rn. Thise uality is then simply ¸a, b)2£ |a|2|b|2



which follows directly from the Cauchy

Schwarz ine

uality.4. From the parallelogram e

uality we have |u + v|2+ |u - v|2= 2(|u|2+ |v|2).Solving for |v| we get |v| = Ö 17.5. Set e.g. u = (1, 0), v = (0, 1). Then |u| = 1, |v| = 1, |u + v| = 2, |u - v| = 2.Assuming that the norm is induced by an inner

product, we would have by theparallelogram ine

uality8 = 22+ 22= 2(1

2+ 12) = 4,which is clearly false.6. Just use |u| = ¸u, u) and simplify.7. See previous exercise.128. This exercise is a lot trickier than it might seem. I'll prove it for R, the proof for Cis almost identical except for the calculations that are longer and more tedious. Allnorms on a nite

dimensional real vector space are e

uivalent. This gives us

limn®¥|rnx + y| = |lx + y|when rn ® l. This probab

y doesn't make any sense, so check a book on topo

ogyor functiona

ana

ysis or just assume the resu

t.Dene ¸u, v) by¸u, v) = |u + v|2

-|u -v|24 .Trivia

y we have |u|2= ¸u, u), so positivity deniteness fo

ows. Now4(¸u + v, w) -¸u, w) -¸v, w)) = |u + v + w|2-|u + v -w|



2-(|u + w|2-|u -w|2)-(|v + w|2-|v -w|2)= |u + v + w|2-|u + v -w|2+(|u -w|2+|v -w|2)-(|u + w|2+|v + w|2

)We can app

y the para

e

ogram equa

ity to the two parenthesis getting4(¸u + v, w) -¸u, w) -¸v, w)) = |u + v + w|2-|u + v -w|2+ 12(|u + w -2w|2+[u -v[2

)- 12(|u + v + 2w|2+|u -v|2)= |u + v + w|2-|u + v -w|2

+ 12|u + w -2w|2- 12|u + v + 2w|2



= (|u + v + w|2+|w|2) + 12|u + v -2w|2- (|u + v -w|2+|w|2) - 12|u + v + 2w|2App

ying the para

e

ogram equa

ity to the two parenthesis we and simp

ifying weare

eft with 0. Hence we get additivity in the rst s

ot. It's easy to see that¸-u, v) = -¸u, v), so we get ¸nu, v) = n¸u, v) for a

n Î Z. Now we get¸u, v) = ¸nu/n, v) = n¸u/n, v) ,

13so that ¸u/n, v) = 1/n¸u, v). Thus we have ¸nu/m, v) = n/m¸u, v). It fo

ows thatwe have homogeneity in the rst s

ot when the sca

ar is rationa

. Now

et l Î R andchoose a sequence (rn) of rationa

numbers such that rn ® l. This gives usl¸u, v) =

imn®¥rn¸u, v) =

im

n®¥¸rnu, v)=

imn®¥14(|rnu + v|2-|r

nu -v|2)= 14(|lu + v|2-|lu -v|



2)= ¸lu, v)Thus we have homogeneity in the rst s

ot. We trivia

y a

so have symmetry, so wehave an inner

product. The proof for F = C can be done by dening the inner

productfrom the comp

ex po

arizing identity. And by using the identities¸u, v)0 = 14(|u + v|2-|u -v|2) ⇒ ¸u, v) = ¸u, v)0 +¸u, iv)0 i.and using the properties just proved for ¸, )0

.9. This is rea

y an exercise in ca

cu

us. We have integra

s of the types¸sinnx, sinmx) =

_ p-psinnxsinmxdx¸cos nx, cos mx) =

_ p-pcos nxcos mxdx¸sinnx, cos mx) =

_ p-psinnxcos mxdxand they can be evaluated using the trigonometric identitiessinnxsinmx = cos((n -m)x) -cos((n + m)x)2cos nxcos mx = cos((n -m)x) + cos((n + m)x)2cos nxsinmx = sin((n -m)x) -sin((n + m)x)

2 .10. e1 = 1, then e2 = x-x,11x-x,11. Where



¸x, 1) = _ 10x = 1/2,14so that|x -1/2| = ¸x -1/2, x -1/2) =¸

_ 10(x -1/2)2dx =

_ 1/12,which gives e2 = Ö 12(x - 1/2) = Ö 3(2x - 1). Then to continue

encil

ushing we

have¸x2, 1

_ = 1/3,

_ x2,Ö 3(2x -1)

_ = Ö 3/6,so that x2-¸x2, 1

_ 1 -

x̧2,Ö 3(2x -1)

_ Ö 3(2x -1) = x2-1/3 -1/2(2x -1). Now



|x2-1/3 -1/2(2x -1)| = 1/(6Ö 5)giving e3 = Ö 5(6x2-6x+1). The orthonormal basis is thus (1,Ö 3(2x-1),Ö 5(6x2-6x + 1).11. Let (v1, . . . , vn) be a linearly de

endent list. We can assume that (v

1, . . . , vn-1) islinearly inde

endent and vn Î s

an(v1, . . . , vn). Let P denote the

rojection on thesubs

ace s

anned by (v1

, . . . , vn-1). Then calculating en with the Gram

Schmidtalgorithm gives usen = vn -P(vn

)|vn -Pvn|,but vn = Pv



n by our assum

tion, so en = 0. Thus the resulting list will sim

lycontain zero vectors. It's easy to see that we can extend the algorithm to workforlinearly de

endent lists by tossing away resulting zero vectors.12. Let (e1, . . . , en-1) be an orthogonal list of vectors. Assume that (e1, . . . , en) and(e1, . . . , en-1, e

n) are orthogonal and s

an the same subs

ace. Then we can write

en = a1e1 + . . . + anen. Now we have ¸e

n, ei) = 0 for all i < n, so that ai = 0 fori < n. Thus we have e

n = anen

and from |e

n| = 1 we have an = ±1.Let (e1, . . . , en



) be the orthonormal base

roduced from (v1, . . . , vn) by Gram

Schmidt.Then if (e

1, . . . , e

n) satises the hy

othesis from the

roblem we have by the

revious

aragra

h that e

i = ±ei. Thus we have 2n

ossible such orthonormal lists.13. Extend (e1, . . . , em

) to a basis (e1, . . . , en) of V . By theorem 6.17v = ¸v, e1) e1 + . . . +¸v, en) en

,so that|v| = | ¸v, e1) e1 + . . . +¸v, em) en|= | ¸v, e1

) e1| + . . . +| ¸v, em) en|= [ ¸v, e1) [ + . . . +[ ¸v, e



n) [.Thus we have |v| = [ ¸v, e1) [ + . . . +[ ¸v, em) [ if and only if ¸v, ei) = 0 for i > m i.e.v Î s

an(e1, . . . , em).1514. It's easy to see that the dierentiation o

erator has a u

er

triangular matrix in theorthonormal basis calculated in exercise 10.15. This follows directly from V = U ÅU^.16. This follows directly from the

revious exercise.17. From exercise 5.21 we have that V = range P Å null P. Let U := range P, then by

assum

tion null P ⊂ U^. From exercise 15, we have dimU = dimV - dimU =dimnull P, so that null P = U^. An arbitrary v Î V can be written as v = Pv +(v -Pv). From P2= P we have that P(v - Pv) = 0, so v - Pv Î null P = U^. Hencethe decomposition v = Pv + (v -Pv) is the unique decomposition in U ÅU^

. NowPv = P(Pv +(v -Pv)) = Pv, so that P is the identity on U. By denition P = PU.18. Let u Î range P, then u = Pv for some v Î V hence Pu = P2v = Pv = u, so P isthe identity on range P. Let w Î null P. Then for a Î F|u|2= |P(u + aw)|2£ |u + aw|

2.By exercise 2 we have ¸u, w) = 0. Thus null P ⊂ (range P)^ and from the dimensionequality null P = (range P)^. Hence the claim follows.19. If TPU



= PUTPU clearly U is invariant. Now assume that U is invariant. Then foru Î U we have PUTPUu = PUTu = Tu = TPUu.20. Let u Î U, then we have Tu = TPUu = PUTu Î U. Thus U is invariant. Then letw Î U^. Now we can write Tw = u+u

where u Î U and u

Î U^. Now PUTw = u,but u = PUTw = TPUw = T0 = 0, so Tw Î U^. Thus U

̂is also invariant.21. First we need to nd an orthonormal basis for U. With Gram

Schmidt we get e1 =(1/Ö 2, 1/Ö 2, 0, 0), e2 = (0, 0, 1/

Ö 5, 2/Ö 5). Let U = span(e1, e2). Then we haveu = PU



(1, 2, 3, 4) = ¸(1, 2, 3, 4), e1) e1 +¸(1, 2, 3, 4), e2) e2 = (3/2, 3/2, 11/5, 22/5).22. If p(0) = 0 and p

(0) = 0, then p(x) = ax2+ bx3. Thus we want to nd theprojection of 2 + 3x to the subspace U := span(x2, x3). With Gram

Schmidt we getthe orthonormal basis (Ö 3x

2, _ 420/11(x3- 12x2)). We getPU

(2+3x) = _ 2 + 3x,Ö 3x2

_ Ö 3x2+

_ 2 + 3x,

_ 420/11(x3- 12x2)

_



_ 420/11(x3-12x2).Here

_ 2 + 3x,Ö 3x2

_ = 1712Ö 3,

_ 2 + 3x,

_

420/11(x3- 12x2)

_ = 47660Ö

1155,so thatPU(2 + 3x) = -7122x2+ 32922 x

3.1623. There's is nothing special with this exercise, compared to the previous two, exceptthat it takes ages to calculate.24. We see that the map T : T2(R) ® R dened by p ® p(1/2) is linear. From exercise10 we get that (1,



Ö 3(2x - 1),Ö 5(6x2- 6x + 1) is an orthonormal basis for T2(R).From theorem 6.45 we see thatq(x) = 1 +Ö 3(2 1/2 -1)Ö 3(2x -1) +Ö 5(6 (1/2)2-6 1/2 + 1)Ö 5(6x2-6x + 1)= -3/2 + 15x -15x2

.25. The map T : T2(R) ® R dened by p ®

_ 10 p(x) cos(px)dx is clearly linear. Againwe have the orthonormal basis (1,Ö 3(2x -1),Ö 5(6x

2-6x + 1), so thatq(x) =

_ 10cos(px)dx +

_ 10Ö 3(2x -1) cos(px)dxÖ

3(2x -1)+ _ 10Ö 5(6x2-6x + 1) cos(px)dxÖ



5(6x2-6x + 1)= 0 - 4Ö 3p2Ö 3(2x -1) + 0 = -12p2(2x -1).26. Choose an orthonormal basis (e1, . . . , en) for V and let F have the usual basis. Thenby

ro

osition 6.47 we have/(T, (e1, . . . , e

n)) = [¸e1, v) ¸en, v)] ,so that/(T, (e1, . . . , en

)) = _ ¸

_ ¸e1, v)...¸en, v)

_ ¸ _ and nally Ta = (a¸e1, v), . . . , a¸en, v)).



27. with the usual basis for Fnwe get/(T) =

_ ¸¸¸¸

_ 0 1 0... ...... 10 0

_ ¸¸

¸̧ _ so by

ro

osition 6.47/(T) =

_ ¸¸¸¸

_

0 01 ...... ...0 1 0

_

¸̧

¸¸

_ .17Clearly T(z



1, . . . , zn) = (z2, . . . , zn, 0).28. By additivity and conjugate homogeneity we have (T -lI) = T-lI. From exercise31 we see that T-lI is non

injective if and on

y if T -lI is non

injective. That isl is an eigenva

ue of T if and on

y if l is an eigenva

ue of T.29. As (U^)^ = U and (T

) = T it's enough to show on

y one of the imp

ications. LetU be invariant under T and choose w Î U^. Now we can write Tw = u + v, whereu Î U and v Î U^, so that0 = ¸Tu, w) = ¸u, T

w) = ¸u, u + v) = ¸u, u) = |u|2.Thus u = 0 which comp

etes the proof.30. As (T) = T it's sucient to prove the rst part. Assume that T is injective, thenby exercise 31 we havedimV = dimrange T = dimrange T

,but range T is a subspace of V , so that range T = V .31. From proposition 6.46 and exercise 15 we getdimnu

T = dim(range T)^ = dimW -dimrange T= dimnu

T + dimW -dimV.For the second part we havedimrangeT



= dimW -dimnu

T= dimV -dimnu

T= dimrange T,where the second equa

ity fo

ows from the rst part.32. Let T be the operator induced by A. The co

umns are the images of the basis vectorsunder T, so the generate range T. Hence the dimension of the spanof the co

umnvectors equa

dimrange T. By proposition 6.47 the span of the row vectors equa

srange T. By the previous exercise dimrange T = dimrange T, so the c

aim fo

ows.7 Operators on Inner

Product Spaces1. For the rst part,

et p(x) = x and q(x) = 1. Then c

ear

y 1/2 = ¸Tp, q) ,= ¸p, Tq) =¸p, 0) = 0. For the second part it's not a contradiction since the basis is not orthog

ona

.182. Choose the standard basis for R

2. Let T and S the the operators dened by thematrices _ 0 11 0

_ ,

_ 0 00 1

_ .

Then T and S as c

ear

y se

f

adjoint, but TS has the matrix _ 0 10 0

_ .so TS is not se

f

adjoint.3. If T and S are se

f

adjoint, then (aT +bS) = (aT)+(bS)

= aT +bS for any a, b Î R,so se

f

adjoint operators form a subspace. The identity operator I is se

f

adjoint, butc

ear

y (iI) = -iI, so iI is not se

f

adjoint.4. Assume that P is se

f

adjoint. Then from proposition 6.46 we have nu

P = nu

P =



(range P)^. Now P is c

ear

y a projection to range P. Then assume that P is aprojection. Let (u1, . . . , un) an orthogona

basis for range P and extend it to a basisof V . Then c

ear

y P has a diagona

matrix with respect to this basis, so P isse

f

adjoint.5. Choose the operators corresponding to the matricesA =

_ 0 -11 0

_ , B =

_ 1 11 0

_ .

Then we easi

y see that AA = AA and BB = BB. However, A + B doesn'tdene a norma

operator which is easy to check.6. From proposition 7.6 we get nu

T = nu

T. It fo

ows from proposition 6.46 that

range T = (nu

T)^ = (nu

T)^ = range T.7. C

ear

y nu

T ⊂ null Tk. Let v Î null Tk

. Then we have _ TTk-1v, TTk-1



v _ =

_ TTkv, Tk-1v

_ = 0,so that TTk-1v = 0. Now

_ Tk-1v, Tk-1v

_ = _ TTk-1v, Tk-2v

_ = 0,so that v Î null T

k-1. Thus null Tk⊂ null Tk-1and continuing we get null Tk⊂null T.Now let u Î range Tk, then we can nd a v Î V such that u = Tk

v = T(Tk-1v), sothat range Tk⊂ range T. From the rst part we get dimrange Tk= dimrange T, sorange Tk



= range T.198. The vectors u = (1, 2, 3) and v = (2, 5, 7) are both eigenvectors corresponding todierent eigenvalues. A self

adjoint operator is normal, so by corollary 7.8 if T isnormal that would imply orthogonality of u and v. Clearly ¸u, v) ,= 0, so Tcan't beeven normal much less self

adjoint.9. If T is normal, we can choose a basis of V consisting of eigenvectors of T. Let A bethe matrix of T corresponding to the basis. Now T is self

adjoint if and only if theconjugate transpose of A equals A that is the eigenvalues of T are real.10. For any v Î V we get 0 = T9v -T8v = T8(Tv -v). Thus Tv -v Î null T8= null Tby the normality of T. Hence T(Tv -v) = T

2v -Tv = 0, so T2= T. By the spectraltheorem we can choose a basis of eigenvectors for T such that T has a diagonal matrixwith l1, . . . , ln on the diagona

. Now T2has the diagona

matrix with l

21, . . . , l2n onthe diagona

and from T2= T we must have l2i = li

for a

i. Hence li = 0 or li = 1,so the matrix of T equa

s its conjugate transpose. Hence T is se

f

adjoint.11. By the spectra

theorem we can choose a basis of T such that the matrix of Tcorresponding to the basis is a diagona

matrix. Let l



1, . . . , ln be the diagona

e

ements. Let S be the operator corresponding to the diagona

matrix having thee

ements Ö l1, . . . ,Ö ln on the diagona

. C

ear

y S2= T.12. Let T the operator corresponding to the matrix

_ 0 -11 0

_ . ThenT

2+ I = 0.13. By the spectra

theorem we can choose a basis consisting of eigenvectors of T. ThenT has a diagona

matrix with respect to the basis. Let l1, . . . , ln be the diago

na

e

ements. Let S be the operator corresponding to the diagona

matrix having3Ö

l1, . . . , 3Ö ln on the diagona

. Then c

ear

y S3= T.14. By the spectra

theorem we can choose a basis (v1, . . . , v

n) consisting of eigenvectorsof T. Let v Î V be such that |v| = 1. If v = a1v1 + . . . + anvn



i -l[ ³ en

i=1[ai[ = e,which is a contradiction. Thus w can nd an ig nva

u li such that [l -li[ < e.2015. If such an inn r

product xists w hav a basis consisting of

ig

nv

ctors by th

sp ctra

th or m. Assum th n that (v1, . . . , vn) is basis such that th matrix of Tis a matrix consisting of

ig

nv

ctors of T. D n

an inn

r

product by¸vi

, vj) =

_ 0, i ,= j1, i = j .Ext

nd this to an inn

r

product by bi

in

arity and homog

n

ity. Th

nw hav aninn

r

product and c

ar

y T is s

f

adjoint as (v1, . . . , vn

) is an orthonorma

basis ofU.16. L

t T Î L(R2) b th op rator T(a, b) = (a, a + b). Th n span((1, 0)) is invariantund r T, but it's orthogona

comp

m nt span((0, 1)) is c

ar

y not.17. L

t T, S b

two positiv

op

rators. Th

n th

y ar

s

f

adjointand (S + T)

=S

+T = S +T, so S +T is s

f

adjoint. A

so ¸(S + T)v, v) = ¸Sv, v) +¸Tv, v) ³ 0.Thus, S + T is positiv .18. C

ar

y Tkis s

f

adjoint for

v

ry positiv

int

g

r. For k = 2 w

hav

¸T2



v, v _ =¸Tv, Tv) ³ 0. Assum that th r su

t is tru for a

positiv k < n and n ³ 2. Th n¸Tnv, v) =¸Tn-2Tv, Tv

_ ³ 0,by hypoth sis and s

f

adjointn ss. H nc th r su

t fo

ows by induction.19. C

ar

y ¸Tv, v) > 0 for a

v Î V ¸ ¦0¦ imp

i

s that T is inj

ctiv

h inv rtib

.So assum

that T is inv

rtib

. Sinc

T is s

f

adjoint w

can choos

an orthonorma

basis (v1, . . . , vn) of

ig

nv

ctors such that T has a diagona

matrix with r

sp

ct to

th

basis. L

t th

m

nts on th

diagona

b

l1, . . . , ln and by assumption li > 0for a

i. L

t v = a1v1 + . . . + an

vn Î V ¸ ¦0¦, so that¸Tv, v) = [a1[2l1 + . . . +[an[2

ln > 0.20.

_ sinq cos qcos q -sinq

_ is a s

uar root for v ry q Î R, which shows that it has innit

ymany s

uar roots.



21. L t S Î L(R2) b d n d by S(a, b) = (a + b, 0). Th n |S(1, 0)| = |(1, 0)| = 1 and|S(0, 1)| = 1, but S is c

ar

y not an isom try.22. R3is an odd

dim nsiona

r a

v ctor spac . H nc S has an ig nva

u

l and acorr sponding ig nv ctor v. Sinc |Sv| = [l[|v| = |v| w hav l2= 1. H nc

S2v = l2v = v.2123. Th

matric

s corr

sponding to T and T ar

/(T) = _ _ 0 0 1

2 0 00 3 0 _ _ , /(T) =

_ _ 0 2 00 0 31 0 0

_

_ .Thus w

g

t/(TT) =

_ _ 4 0 00 9 00 0 1

_ _

,so that TT is th op rator (z1, z2, z3) ® (4z



1, 9z2, z3). H

nc

Ö TT(z1, z2, z3) =(2z1, 3z2, z3). Now w just n d to p rmut th indic s which corr sponds to th

isom

try S(z

1, z2, z3) = (z3, z1, z2).24. C

ar

y T

T is positiv , so by proposition 7.26 it has a uni

u positiv s

uar root.Thus it's suci

nt to show that R2= TT. Now w hav T = (SR) = RS

= RSby s

f

adjointn ss of R. Thus R2= RIR = RSSR = TT.



25. Assum that T is inv rtib

. Th n Ö TT must b inv rtib

(po

ar d composition).H

nc

S = TÖ TT-1, so S is uni

u

y d

t

rmin

d. Now assum

that T is notinv rtib

. Th n rang

Ö TT is not inv rtib

(h nc not surj ctiv ).Now assum

that T is not inv

rtib

, so that Ö TT is not inv

rtib

. By th

sp

ctra

th or m w can choos a basis (v1

, . . . , vn) of ig nv ctors of Ö TT and w canassum

that v1 corr

sponds to th

ig

nva

u

0. Now

t T = SÖ T

T. D

n

Uby Uv1 = -Sv1, Uvi = Svi, i > 1. C

ar

y U ,= S and T = UÖ T

T. Now choos

u, v Î V . Th n it's asy to v rify that ¸Uu, Uv) = ¸Su, Sv) = ¸u, v), so is anisom try.26. Choos a basis of ig nv ctors for T such that T has a diagona

matrix with ig nva

u

s l1, . . . , ln



on th diagona

. Now TT = T2corr sponds to th diagona

matrixwith l21, . . . , l2n on th

diagona

and th

s

uar

root Ö TT corr

sponds to th

diag

ona

matrix having [l1[, . . . , [ln[ on th diagona

. Th s ar th singu

ar va

u s, soth

c

aim fo

ows.27. L t T Î L(R2

) b

th

map T(a, b) = (0, a). Th

n from th

matrix r

pr

s

ntationw

s

that TT(a, 0) = (a, 0), h

nc

Ö TT = TT and 1 is c

ar

y a singu

ar va

u

.How v r, T2= 0, so

_ (T2)T2= 0, so 12= 1 is not a singu

ar va

u of T2.28. Th composition of two bij ctions is a bij ction. If T=S

Ö TT, th n sinc S is anisom try w hav that T is bij ctiv h nc inv rtib

if and on

y if Ö TT is bij ctiv .But



Ö TT is inj ctiv h nc bij ctiv if and on

y if 0 is not an ig nva

u . Thus Tisbij

ctiv

if and on

y if 0 is not a singu

ar va

u

.2229. From th

po

ar d

composition th

or

m w

s

that dimrang

T = dimrang

Ö TT.Sinc

Ö TT is s

f

adjoint w can choos a basis of ig nv ctors of Ö TT such thatth

matrix ofÖ T

T is a diagona

matrix. C

ar

y dimrang

Ö TT

ua

s th

numb

rof non

z ro

m nts on th diagona

i. . th numb r of non

z ro singu

ar va

u

s.30. If S is an isom try, th n c

ar

y Ö SS = I, so a

singu

ar va

u s ar 1. Now assum

that a

singu

ar va

u

s ar

1. Th

n Ö SS is a s

f

adjoint (positiv ) op rator witha

ig nva

u s

ua

to 1. By th sp ctra

th or m w can choos a basis ofV suchthat Ö SS has a diagona

matrix. As a

ig

nva

u

s ar

on

, this m

ans that th

matrix of Ö SS is th id ntity matrix. Thus Ö SS = I, so S is an isom try.31. L t T



1 = S1

_ T1T1 and T2 = S2

_ T2T2. Assum

that T1 and T2 hav

th

sam

singu

ar va

u

s s1, . . . , sn. Th n w can choos bas s of ig nv ctors of _ T1T1 and

_

T2T2 such that th y hav th sam matrix. L t th s bas s b (v1, . . . , vn) and(w1, . . . , w

n). L t S th op rator d n d by S(wi) = vi, th n c

ar

y S is an isom tryand

_ T



2T2 = S-1

_ T1T1S. Thus w

g

t T2 = S2

_ T2T2 = S2S

-1 _ T1T1S. WritingS3 = S2S

-1S-11 , which is c

ar

y an isom try, w g tT2 = S3S1

_ T

1T1S = S3T1S.Now

t T



2 = S1T1S2. Th

n w

hav

T2T2 = (S1T1S2)(S1T1

S2) = S1T1T1S1 =

S-11 T1T1S1. L t l b an ig nva

u of T1

T1 and v a corr

sponding

ig

nv

ctor.B caus S1 is bij ctiv w hav a u Î V such that v = S1u. This giv s usT



2T2u = S-11 T1T1S1u = S-11 T1T1v = S-11

lv = lu,so that l is an

ig

nva

u

of T2T2. L t (v1, . . . , vn) b an orthonorma

basis of ig n

v

ctors of T

1T1, th

n (S-11 v1, . . . , S-1vn) is an orthonorma

basis for V of ig nv ctors

of T2T2. H nc T2T2



and T1T1 hav

th

sam

ig

nva

u

s. Th

singu

ar va

u

s ar

simp

y th positiv s

uar roots of th s ig nva

u s, so th c

aim fo

ows.32. For th

rst part, d

not

by S th

map d n

d by th

formu

a. S

tA := /(S, (f1, . . . , fn), (

1, . . . ,

n)), B := /(T, (

1, . . . ,

n), (f1, . . . , fn

)).C

ar

y A = B and is a diagona

matrix with s1, . . . , sn on th

diagona

. A

th

siar

positiv

r

a

numb

rs

, so that B = B = A. It fo

ows from proposition 6.47that S = T.

For th

s

cond part obs

rv

that a

in

ar map S d

n

d by th

formu

a maps fito s-1i

i which is mapp d by T to fi. Thus TS = I. Th c

aim now fo

ows from

x

rcis

3.23.23

33. By d

nition Ö TT is positiv . Thus a

singu

ar va

u s of T ar positiv . From th

singu

ar va

u d composition th or m w can choos bas s of V such that|Tv| = |n

i=1



si¸v,

i) fi| =n

i=1si[ ¸v,

i) [sinc

a

si ar

positiv

. Now c

ar

yÃ s|v| = Ã sn

i=1[ ¸v,

i

) [ £n

i=1si[ ¸v,

i) [ = |Tv|and simi

ar

y for s|v|.34. From th triang

ua

ity and th pr vious x rcis w g t|(T

+ T

)v| £ |T

v| +|T

v| £ (s

+ s

)|v|.Now

t v b

th

ig

nv

ctor of _

(T

+ T

)(T

+ T



) corr sponding to th singu

arva

u

s. L

t T + T

= S _ (T

+ T

)(T

+ T

), so that|(T

+ T

)v| = |S _ (T

+ T

)(T

+ T

)v| = |Ssv| = s|v| £ (s

+ s

)|v|,

so that s £ s

+ s

.8 Op rators on Comp

x V ctor Spac s1. T is not inj ctiv , so 0 is an ig nva

u of T. W s that T2= 0, so that any v Î Vis a g n ra

iz d ig nv ctor corr sponding to th ig nva

u 0.2. On pag 78 it's shown that ±i ar th ig nva

u s of T. W hav dimnu

(T -iI)2+

dimnu

(T + iI)2= dimC2= 2, so that dimnu

(T - iI)2= dimnu

(T + iI)2= 1.B caus (1, -i) is an ig nv ctor corr sponding to th ig nva

u i and



(1, i) is an

ig

nv

ctor corr

sponding to th

ig

nva

u

-i. Th

s

t of a

g

n

ra

iz

d

ig

nv ctorsar simp

y th spans of th s corr sponding ig nv ctors.3. L t m-1i=0 aiTiv = 0, th n 0 = Tm-1(

m-1i=0 aiTiv = a0

Tm-1v, so that a0 = 0. Ap

p

ying r

p

at

d

y Tm-ifor i = 2, . . . w

s

that ai = 0 for a

i. H

nc

(v, . . . , Tm-1v)is

in ar

y ind p nd nt.

4. W

s

that T3= 0, but T2,= 0. Assum that S is a s

uar root of T, th n Sis ni

pot nt, so by coro

ary 8.8 w hav SdimV= S3= 0, so that 0 = S4= T2

.Contradiction.5. If ST is ni

pot

nt, th

n ∃n Î N such that (ST)n= 0, so (TS)n+1= T(ST)nS = 0.24



6. Assum that l ,= 0 is an ig nva

u of N with corr sponding ig nv

ctor v. Th

nNdimVv = ldimVv ,= 0. This contradicts coro

ary 8.8.7. By

mma 8.26 w

can nd a basis of V such that th

matrix of N is upp

r

triangu

arwith z

ro

s on th

diagona

. Now th

matrix of N is th

conjugat

transpos

. Sinc

N = N, th matrix of N must b z ro, so th c

aim fo

ows.8. If NdimV -1,= NdimV, th

n dimnu

Ni-1< dimnu

Nifor a

i £ dimV by propo

sition 8.5. By coro

ary dimnu

NdimV= dimV , so th c

aim c

ar

y fo

ows.9. rang

Tm= rang

Tm+1imp

i

s dimnu

Tm= dimnu

Tm+1i.

. nu

Tm

= nu

Tm+1.From proposition 8.5 w

g

t dimnu

Tm= dimnu

Tm+kfor a

k ³ 1. Again thisimp

i s that dimrang Tm= dimrang Tm+kfor a

k ³ 1, so th c

aim fo

ows.

10. L

t T b

th

op

rator d

n

d in

x

rcis

1. C

ar

y nu

T Ç rang

T ,= ¦0¦,so th

c

aim is fa

s

.11. W hav dimV = dimnu

Tn+dimrang Tn, so it's suci nt to prov that nu

TnÇ



rang Tn= ¦0¦. L t v Î nu

TnÇ rang Tn. Th n w can nd a u Î V such thatTnu = v. From 0 = Tv = Tn+1u w

s

that u Î nu

Tn+1= nu

Tnwhich imp

i

sthat v = Tnu = 0.12. From th

or

m 8.23 w

hav

V = nu

TdimV, so that TdimV= 0. Th

n

t T Î L(R

3)b th op rator T(a, b, c) = (-b, a, 0). Th n c

ar

y 0 is th on

y ig nva

u ,but T isnot ni

pot nt.13. From nu

Tn-2,= nu

Tn-1and from proposition 8.5 w

s

that dimnu

Tn³ n-1.Assum that T has thr di r nt ig nva

u s 0, l

1, l2. Th

n dimnu

(T -liI)n³ 1,so from th or m 8.23n ³ dimnu

Tn+dimnu

(T -l1

I)n+dimnu

(T -l2I)n³ n -1 +1 +1 = n +1,which is impossib

, so T has at most two di r nt ig nva

u s.14. L t T Î L(C4



) b d n d by T(a, b, c, d) = (7a, 7b, 8c, 8d) from th matrix of T it's

asy to s

that th

charact

ristic po

ynomia

is (z -7)2(z -8)2.15. L t d1 b th mu

tip

icity of th ig nva

u 5 and d2 th mu

tip

icity of th ig nva

u

6. Th

n d1, d2 ³ 1 and d1 + d2 = n. It fo

ows that d1, d2 £ n - 1, so that

(z - 5)d1(z - 6)d2divid s (z - 5)n-1(z - 6)n-1. By th Cay

y

Hami

ton th or m(T -5I)d

1(T -6I)d2= 0, so that (T -5I)n-1(T -6I)n-1= 0.16. If v ry g n ra

iz d ig nv ctor is an ig nv ctor, th n by th or m 8.23 T has a basisof

ig

nv

ctors. If th

r

's a g

n

ra

iz

d

ig

nv

ctor that is not an

ig

nv

ctor, th n

w

hav

an

ig

nva

u

l such that dimnu

(T -lI)dimV,= dimnu

(T -lI). Thus if25l1, . . . , lm ar th ig nva

u s of T, th n



mi=1 dimnu

(T - liI) < dimV , so th r

do

sn't

xists a basis of

ig

nv

ctors.17. By

mma 8.26, choos a basis (v1, . . . , vn) such that N has an upp r

triangu

armatrix. App

y Gram

Schmidt orthogona

ization to th

basis. Th

n N

1 = 0 andassum that N

i Î span(

1, . . . ,

i), so thatN

i+1 = N

_ vi+1 -¸vi+1,

1)

1 -. . . -¸vi+1,

i

)

i|vi+1 -¸vi+1,

1)

1 -. . . -¸vi+1,

i)

i|

_ Î span(

1, . . . ,

i+1).



Thus N has an upp r

triangu

ar matrix in th basis (

1, . . . ,

n).18. Continuing in th

proof of

mma 8.30 up to j = 4 w

s

that a4 = -5/128. So thatÖ I + N = I + 12N - 18N2+ 116N3- 5

128N419. Just r

p

ac

th

Tay

or

po

ynomia

in

mma 8.30 with th

Tay

or po

ynomia

of3Ö 1 + x and copy th

proof of th

mma and th

or

m 8.32.20. L t p(x) = mi=0 a

ixib

th

minima

po

ynomia

of T. If a0 ,= 0, th n p(x) =x

mi=1 aix

i-1. Sinc T is inv rtib

w must hav

mi=1 aiTi-1= 0 which contradicts



th minima

ity of p. H nc a0 ,= 0, so so

ving for I in p(T) w g t-a-10 (a1 + . . . + amTm-1)T = I.H

nc

s

tting

(x) = -a-10 (a1 + . . . + amxm-1) w hav

(T) = T-1

.21. Th

op

rator d n

d by th

matrix _ _ 0 0 10 0 00 0 0

_ _ is c

ar

y an

xamp

.22. Choos th matrixA =

_

¸̧

_ 1 0 0 00 1 1 00 0 1 00 0 0 0

_ ¸¸

_ .It's asy to s that A(A-I)

2= 0. Thus, th minima

po

ynomia

divid s z(z -1)2.How v r, th who

po

ynomia

is th on

y factor that annihi

at s A, so z(z -1)2isth minima

po

ynomia

.2623. L t l



1, . . . , lm b th ig nva

u s of T with mu

tip

icity d1, . . . , dm. Assum

that Vhas a basis of ig nv ctors (v1, . . . , vn). L t p(z) =

mi=1(z -li). Th n w g tp(T)vi =

_ _

j=i(T -ljI)

_ _ (T -liI)vi = 0,so that th minima

po

ynomia

divid s p and h nc has no doub

root.

Now assum

that th

minima

po

ynomia

has no doub

roots. L

t th minima

po

ynomia

b

p and

t (v1, . . . , vn) b a Jordan

basis of T. L t A b th

arg stJordan

b

ock of T. Now c

ar

y p(A) = 0. How

v

r, by

x

rcis

29, th minima

po

ynomia

of (A - lI) is zm+1wh

r

m is th

ngth of th

ong

st cons

cutiv

string of 1's that app ar just abov th diagona

, so th minima

po

ynomia

of Ais (z - l)m+1. H nc m = 0, so that A is a 1 1 matrix and th basis v ctorcorr sponding to A is an ig nva

u . Sinc A was th

arg st Jordan

b

ock it fo

owsthat a

basis v ctors ar ig nva

u s (corr sponding to a 1 1 matrix).24. If V is a comp

x v ctor spac , th n V has a basis of ig nv ctors. Now

t l



1, . . . , lmb th distinct ig nva

u s. Now c

ar

y p =

mi=1(z -li) annihi

at

s T, so that th

minima

po

ynomia

b ing a factor of p do sn't hav a doub

root.Now assum

that V is a r

a

v

ctor spac

. By th

or

m 7.25 w

can nd a basis of Tsuch that T has a b

ock diagona

matrix wh

r

ach b

ock is a 1 1 or 2 2 matrix.L

t l1, . . . , lm b

th

distinct

ig

nva

u

s corr

sponding to th

1 1 b

ocks. Th

np(z) = mi=1

(z - li) annihi

at s a

but th 2 2 b

ocks of th matrix. Now it'ssuci

nt to show that

ach 2 2 b

ock is annihi

at

d by a po

ynomia

which do

sn'thav r a

roots.By th

or

m 7.25 w

can choos

th

2 2 b

ocks to b

of th

form _ a -bb a

_ wh r b > 0 and it's asy to s that

_ a -b

b a _ -2a

_ a -bb a

_ + (a2+ b2)

_

1 00 1

_ = 0.C

ar

y this po

ynomia

has a n gativ discriminant, so th c

aim fo

ows.25. L t

b th minima

po

ynomia

. Writ

= sp + r, wh r d gr < d

g p, so that0 =

(T)v = s(T)p(T)v +r(T)v = r(T)v. Assuming r ,= 0, w can mu

tip

y th bothsid s with th inv rs of th high st co ci nt of r yi

ding a monic po

ynomia



r2of d gr

ss than p such that r2(T)v = 0 contradicting th minima

ity of p. H nc

r = 0 and p divid

s

.2726. It's

asy to s

that no prop

r factor of z(z -1)2(z -3) annihi

at

s th

matrixA =

_ ¸¸

_ 3 1 1 10 1 1 00 0 1 00 0 0 0

_ ¸¸

_ ,

so th

it's minima

po

ynomia

is z(z - 1)2(z - 3) which by d nition is a

so th

charact

ristic po

ynomia

.27. It's asy to s that no prop r factor of z(z -1)(z -3) annihi

at s th matrixA =

_ ¸¸

_ 3 1 1 10 1 1 00 0 1 0

0 0 0 0 _ ¸¸

_ ,but A(A-I)(A-3I) = 0, so th c

aim fo

ows.28. W

s

that T(

i) =

i+1 for i < n. H

nc

(

1

, T

1, . . . , Tn-1

1) = (

1, . . . ,

n



) is

in

ar

y ind

p

nd

nt. Thus for any non

z

ro po

ynomia

p of d

gr

ss than n w

hav p(

i) ,= 0. H

nc

th

minima

po

ynomia

has d

gr

n, so that th

minima

po

ynomia

ua

s th

charact

ristic po

ynomia

.Now from th matrix of T w s that Tn(

i) = -a0-a1T(

1) -. . . -an-1Tn-1. S tp(z) = a

0 +a1z +. . . +an-1zn-1+zn. Now p(T)(

1) = 0, so by

x

rcis

25 p divid

sth minima

po

ynomia

. How v r, th minima

po

ynomia

is monic and has d gr

n, so p is th minima

po

ynomia

.29. Th

bigg

st Jordan

b

ock of N is of th

form _ ¸¸¸¸¸

_ 0 10 1.

.

. ...0 10

_ ¸¸



¸¸¸

_ .Now c

ar

y Nm+1= 0, so th

minima

po

ynomia

divid

s zm+1. L

t vi, . . . , vi+m b

th basis

m nts corr sponding to th bigg st Jordan

b

ock. Th n Nm

i+m =

i ,=0, so th

minima

po

ynomia

is zm+1.

30. Assum

that V can't b

d

compos

d into two prop

r subspac

s. Th

n T has on

yon ig nva

u . If th r is mor than on Jordan

b

ock, th n L t (v1, . . . , vm) b

th v ctors corr sponding to a

but th

ast Jordan

b

ock and (vm+1, . . . , vn) th

28v ctors corr sponding to th

ast Jordan

b

ock. Th n c

ar

y V = span(v1, . . . , vm) Åspan(vm+1, . . . , vn). Thus, w hav on

y on Jordan

b

ock and T - lI is ni

pot ntand has minima

po

ynomia

zdimV

by th

pr

vious

x

rcis

. H

nc

T has minima

po

ynomia

(z -l)dimV.Now assum that T has minima

po

ynomia

(z -l)dimV. If V = U ÅW,

t p1 b th

minima

po

ynomia

of T



|U and p2 th minima

po

ynomia

of T|W. Now (p1p2)(T) =0, so that (z - l)dimVdivid s p1p2. H nc d g p1p2 = d g p1 + d g p2

³ dimV , butd

g p1+d

g p2 £ dimU +dimW = dimV . Thus d

g p1p2 = dimV . This m

ans thatp1(z) = (z - l)

dimUand p2(z) = (z - l)dimW. Now if n = max¦dimU, dimW¦ w

hav (T|U - lI)n= (T|W - lI)

n= 0. This m ans that (T - lI)n= 0 contradictingth fact that (z -l)dimVis th

minima

po

ynomia

.31. R v rsing th Jordan

basis simp

y r v rs s th ord r of th Jordan

b

ocksand achb

ock n ds to b r p

ac d by its transpos .



9 Op rators on R a

V ctor Spac s1. L

t a b

th

va

u

in th

upp

r

ft corn

r. Th

n th

matrix must hav

th form

_ a 1 -a1 -a a

_ and it's

asy to s

that _ 11

_ is an ig nv ctor corr sponding to th ig nva

u 1.2. Th

charact

ristic po

ynomia

of th

matrix is p(x) = (x - a)(x - d) - bc. Now th

discriminant

ua

s (a + d)2-4(ad -bc) = (a -d)2+ 4bc. H

nc

th

charact

risticpo

ynomia

has a root if and on

y if (a - d)2+ 4bc ³ 0. If p(x) = (x - l1

)(x - l2),th

n p(T) = 0 imp

i

s that

ith

r A - l1I or A - l2I is not inv

rtib

h

nc

A hasan ig nva

u . If p do sn't hav a root th n assuming that Av = lv w g t0 = p(A)v = p(l)v,so that v = 0. H nc A has no ig nva

u s. Th c

aim fo

ows.3. S

th

proof of th

n

xt prob

m, though in this sp

cia

cas

th

proof is trivia

.

4. First

t l b

an

ig

nva

u

of A. Assum

that l is not an

ig

nva

u

ofA1, . . . , Am,so that Ai -lI is inv rtib

for ach i. Now

t B b th matrix _ ¸

_ (A

1 -lI)-1...0 (Am -lI)



-1 _ ¸

_.29It's now

asy to v

rify that B(A-lI) is a diagona

matrix with a

1s on th

diagona

,h nc it's inv rtib

. How v r, this is impossib

, sinc A -lI is not. Thus, l s an

ig nva

u of som Ai.Now

t l b

an

ig

nva

u

of Ai. L

t T b

th

op

rator corr

sponding to A - lIin L(Fn). L t Ai corr spond to th co

umns j, . . . , j + k and

t (

1, . . . ,

n) b th

standard basis. L

t (a

1, . . . , ak) b

th

ig

nv

ctor of Ai corr

sponding to l. S

t v =a1

j +. . . +ak

j+k. Th n it's asy to s that T maps th spac span(

1, . . . ,

j-1, v)to th spac span(

1, . . . ,

j-1). H nc T is not inj ctiv , so that l is an ig nva

u

of A.5. Th proof is id ntica

to th argum nt us d in th so

ution of x rcis 2.

6. L

t (v1, . . . , vn) b th basis with th r sp ct to which T has th giv n matrix.App

ying Gram

Schmidt to th

basis th

matrix trivia

y has th

sam

form.7. Choos a basis (v1, . . . , v



n) such that T has a matrix of th

form in th

or

m 9.10.Now if vj corr sponds to th s cond v ctor in a pair corr sponding to a 2 2 matrixor corr

sponds to a 1 1 matrix, th

n span(v1, . . . , vj) is an invariant subscap

. Th

s cond posibi

ity m ans that vj corr sponds to th rst v ctor in a pair corr spondingto a 2 2 matrix, so that span(v1, . . . , vj+1) is an invariant subspac

.8. Assuming that such an op rator xist d w wou

d hav basis (v1, . . . , v7) such thatth

matrix of T wou

d hav

a matrix with

ig

npair (1, 1)

dimnu

(T2+ T + I)dimV2 = 7/2tim s on th diagona

. This wou

d contradict th or m 9.9 as 7/2 isnot a who

numb r. It fo

ows that such an op rator do sn't xist.9. Th

uation x2+ x + 1 = 0 has a so

ution in C. L

t l b

a so

ution. Th

n th

op rator corr sponding to th matrix lI c

ar

y is an xamp

.

10. L

t T b

th

op

rator in L(R2k) corr

sponding to th

b

ock diagona

matrix wh

r

ach b

ock has charact

ristic po

ynomia

x2+ax +b. Then

y theorem 9.9 we h

vek = dimnull(T2+ aT + bI)2k2 ,

so th

t dimnull(T2+ aT + bI)2kis even. However, the minim

l polynomi

l of Tdivides (T2+aT +bI)2k

nd h

s degree less th

n 2k, so th

t (T



2+aT +bI)k= 0.It follows th

t dimnull(T2+ aT + bI)2k= dimnull(T2+ aT + bI)k

nd the cl

imfollows.3011. We see th

t (a, b) is

n eigenp

ir

nd from the nilpotency of T2+ aT + bI

ndtheorem 9.9 we h

vedimnull(T2+ aT + bI)dimV2

= (dimV )/2it follows th

t dimV is even. For the second p

rt we know from the C

yley-H

miltontheorem th t the minim l polynomi l p(x) of T h s degree less th n dimV . Thus weh

ve p(x) = (x2+ ax + b)k

nd from deg p £ dimV we get k £ (dimV )/2. Hence(T + aT + bI)(dimV )/2= 0.

12. By theorem 9.9 we c

n choose

sis such th

t the m

trix of T h

s the form of 9.10.Now we h

ve the m

trices [5]

nd [7]

t le

st once on the di

gon

l. Assuming th

tT h

s

n eigenp

ir we would h

ve

t le

st one 2 2 m

trix on the di

gon

l. This isimpossi

le

s the di

gon

l h

s only length 3.13. By proposition 8.5 dimnull Tn-1³ n-1. Lik in th pr vious x rcis w know thatth matrix [0] is at

ast n-1 tim s on th diagona

and it's impossib

to t a 2 2matrix in th

on

y p

ac

ft.14. W s that A satis s th po

ynomia

p(z) = (z - a)(z - d) - bc no matt r if

isR or C. If F = C, th n b caus p is monic, has d gr 2 and annihi

at s A it fo

owsthat p is th charact ristic po

ynomia

.Assum th n that F = R. Now if A has no ig nva

u s, th n p is th charact risticpo

ynomia

by d nition. Assum

th

n that p(z) = (z - l1)(z - l2



) which imp

i sthat p(T) = (T -l1I)(T -l2I) = 0. If n

ith

r T -l1I or T -l2I is inj

ctiv

, th

nby d nition p is th minima

po

ynomia

. Assum th n that T - l2I is inv rtib

.Th

n dimnu

(T - l1I) = 2, so that T - l1I = 0. H

nc

w

g

t c = b = 0 anda = d = l1. H nc p(z) = (z - l1)2

, so that p is th

charact

ristic po

ynomia

byd nition.15. S is norma

, so th r 's an orthonorma

basis of V such that S has a b

ock diagona

matrix with r sp ct to th basis and ach b

ock is a 1 1 or 2 2 matrix and th

22 b

ocks hav

no

ig

nva

u

(th

or

m 7.25). From SS = I w

s

that

ach b

ockAi satisfy Ai

Ai = I, so that th

y ar

isom

tri

s. H

nc

a 2 2 b

ock is of th

form _ cos q -sinqsinq cos q

_ .W s that T2+aT +bI is not injective if

nd only if A2i

+aAi+bI is not injectivefor some 22 m

trix Ai. Hence x2+ax+b must equ

l the ch

r

cteristic polynomi

lof some Ai



, so

y the previous exercise it's of the form(x -cos q)(z -cos q) + sin2q = x2-2xcos q + cos2q + sin2qso that b = cos2q + sin2q = 1.3110 Trac and D t rminant1. Th

map T ® /(T, (v1, . . . , vn)) is bij

ctiv

and satis

s ST ® /(ST, (v1, . . . , v

n)) =/(S, (v1, . . . , vn))/(T, (v1, . . . , vn)). H nc

ST = I Û /(ST, (v1

, . . . , vn)) = /(S, (v1, . . . , vn))/(T, (v1, . . . , vn)) = I.Th

c

aim now fo

ows trivia

y.2. Both matric s r pr s nt an op rator in L(F

n). Th c

aim now fo

ows from x rcis

3.23.3. Choos a basis (v1, . . . , vn) and

t A = (aij) b th matrix corr sponding to th



basis. Th n Tv1 = a11v1 + . . . + an1vn. Now (v1, 2v2, . . . , 2vn) is a

so a basis and w

hav

by our assumption Tv = a11v1 + 2(a21v

2 + . . . + an1vn). W

thus g

ta21v2 + . . . + an1v

n = 2(a21v2 + . . . + an1vn),which imp

i s (a21v

2 + . . . + an1vn = 0. By

in ar ind p nd nc w g t a21 = . . . =an1



= 0, so that Tv1 = a11v. Th c

aim c

ar

y fo

ows.4. Fo

ows dir

ct

y from th

d

nition of /(T, (u1, . . . , un), (v1, . . . , vn)).5. L t (

1, . . . ,

n) b th standard basis for Cn. Th n w can nd an op rator T ÎL(Cn) such that /(T, (

1, . . . ,

n)) = B. Now T has an upp

r

triangu

ar matrixcorr sponding to som basis (v1, . . . , vn) of V . L t A = /((v1, . . . , vn), (

1, . . . ,

n)).Th nA-1BA = /(T, (v1, . . . , vn))which is upp r

triangu

ar. C

ar

y A is an inv rtib

s

uar matrix.

6. L

t T b

th

op

rator corr

sponding to th

matrix _ 0 -11 0

_ ,

_ 0 -11 0

_



2=

_ -1 00 -1

_ .By th

or

m 10.11 w

hav

trac

(T2) = -2 < 0.7. L t (v1, . . . , vn) b a basis of ig nv ctors of T and l1, . . . , ln b th corr sponding

ig

nva

u

s. Th

n T has th

matrix _ ¸

_ l

1 0...0 ln

_ ¸

_.C

ar

y trac (T2) = l

21 + . . . + l2n ³ 0 by th or m 10.11.328. Ext

nd v/|v| to an orthonorma

basis (v/|v|,

1, . . . ,

n) of V and

t A = /(T, (v/|v|,

1

, . . . ,

n)).L t a, a1, . . . , an d not th diagona

m nts of A. Th n w hav

ai



= ¸T

i,

i) = ¸¸

i, v) w,

i) = ¸0,

i) = 0,so thattrac (T) = a = ¸Tv/|v|, v/|v|) = ¸¸v/|v|, v) w, v/|v|) = ¸w, v) .9. From

x

rcis

5.21 w

hav

V = nu

P Årang

P. Now

t v Î rang

P. Th

nv = Pufor som

u Î V . H

nc

Pv = P2w = Pw = v, so that P is th

id

ntity on rang

P.Now choos a basis (v1, . . . , vm) for rang P and xt nd it with a basis (u1

, . . . , un)of nu

P to g

t a basis for V . Th

n c

ar

y th

matrix for P in this basis consistsof a diagona

matrix with 1s on part of th

diagona

corr

spondingto th v ctors(v1, . . . , vm) and 0s on th

r

st of th

diagona

. H

nc

trac

P = dimrang

P ³ 0.10. L t (v

1, . . . , vn) b

som

basis of V and

t A = /(T, (v1, . . . , vn)). L

t a1, . . . , an b

th diagona

m nts of A. By proposition 6.47 T

has th matrix A, so that th

diagona

m nts of A ar

a1, . . . , an



. By th or m 10.11 w hav

trac

(T) = a1 + . . . + an = a1 + . . . + an = trac

(T).11. A positiv op rator is s

f

adjoint. By th sp ctra

th or mw

can nd a basis(v1, . . . , vn) of ig nv ctors of T. Th n A = /(T, (v1, . . . , vn)) is a diagona

matrixand by positivity of T a

th

diagona

m

nts a

1, . . . , an ar

positiv

. H

nc

a1 +. . . + an = 0 imp

i s a1 = . . . = an = 0, so that A = 0 and h nc T = 0.

12. Th

trac

of T is th

sum of th

ig

nva

u

s. H

nc

l - 48 + 24 = 51 - 40 + 1, sothat l = 36.13. Choos

a basis of (v1, . . . , vn) of V and

t A = /(T, (v1, . . . , vn)). Th

n th

matrixof cT is cA. L t a

1, . . . , an b th diagona

m nts of A. Th n w hav

trac (cT) = ca1 + . . . + can = c(a1



+ . . . + an) = ctrac (T).14. Th xamp

in x rcis 6 shows that this is fa

s . For anoth

r xamp

tak S =T = I.15. Choos a basis (v1, . . . , vn) for V and

t A = /(T, (v1, . . . , vn)). It's suci nt toprov

that A = 0. Now

t ai,j b

th

m

nt in row i, co

umn j of A. L

t B b

th matrix with 1 in row j co

umn i and 0

s wh r . Th n it's asy to s that inBA th on

y non

z ro diagona

m nt is ai,j. Thus trac (BA) = ai,j

. L

t S b

th

op

rator corr

sponding to B. It fo

ows that ai,j = trac

(ST) = 0, so that A = 0.3316. L

t TT

i = a1

1

+ . . . + an

n. Th n w hav that ai = ¸TT

i,

i) = |T

i|2byth ortogona

ity of (

1, . . . ,

n). C

ar

y ai



is th ith diagona

m nt in th matrix/(TT, (

1, . . . ,

n)), so thattrac (TT) = |T

1|2+ . . . +|T

n|2.Th

s

cond ass

rtion fo

ows imm

diat

y.17. Choos a basis (v1, . . . , vn

) such that T has an upp

r triangu

ar matrix B = (bij)with

ig

nva

u

s l1, . . . , ln on th

diagona

. Now th

ith

m

nt on th

diagona

ofBB is n

j=1 bjibji =

nj=1[bji[2

, wh

r

[bii[2= [li[2. H nc

n



i=1[li[2£n

i=1n

j=1[bji[2= trac

(BB) = trac

(AA) =n

k=1n

j=1[ajk[2.18. Positivity and d

nit

n

ss fo

ows from

x

rcis

16 and additivityin th rst s

otfrom coro

ary 10.12. Homog

n

ity in th

rst s

ot is

x

rcis

13 and by

x

rcis 10

¸S, T) = trac

(ST) = trac

((ST)) = trac (TS) = ¸T, S).It fo

ows that th formu

a d n s an inn r

product.19. W hav 0 £ ¸Tv, Tv) - ¸Tv, T

v) = ¸(TT -TT)v, v). H nc TT - TT is a



positiv op rator (it's c

ar

y s

f

adjoint), but trac (TT - TT) = 0, so a

th

ig

nva

u

s ar

0. H

nc

TT -TT = 0 by th

sp

ctra

th

or

m, so T is norma

.20. L t dimV = n and writ A = /(T). Th n w hav

d

t cT = d

t cA =

scas(1),1 cas(n),n = cndet A = cndet T21. Let S = I and T = -I. Then we have 0 = det(S + T) ,= det S + det T =

1 + 1 = 2.22. We can u

e the re

ult of the next exerci

e which mean

that we only need to provethi

for the complex ca

e. Let T Î L(Cn) be the operator corre

ponding to thematrix A. Now each block Aj on the diagonal corre pond to ome ba i vector

(ei, . . . , ei+k

). The

e can be replaced with another

et of vector

, (vi, . . . , vi+k), pan

ning the ame ub pace uch that Aj in thi

new ba

i

i

upper

triangular. We haveT( pan(vi, . . . , vi+k)) ⊂ span(v

i, . . . , vi+k), so after doing this for all blocks we canassume that A is upper

triangular. The claim follows immediately.23. This follows trivially from the formula of trace and determinant for a matrix, becausethe formulas only depends on the elements of the matrix.3424. Let dimV = n and let A = /(T), so that B = /(T) equals the conjugate tran



sposeof A i.e. bij = aji. Then we havedet T = det A =

sas(1),1 as(n),n=

sa1,s(1) an,s(n) =

sa

1,s(1) an,s(n)=

sbs(1),1 bs(n),n = det B = det T.

The r

t equality on the

econd line follow

ea

ily that every term in the upper

umi

repre

ented by a term in the lower

um and vice ver

a. The

econd claim follow

immediately from theorem 10.31.25. Let W = ¦(x, y, z) Î R3[ x2+ y2+ z2

< 1¦ and let T be the operator T(x, y, z) =(ax, by, cz). It's easy to see that T(W) is the ellipsoid. Now W is a circle with radius1. Hence[ det T[volume(W) = abc43p = 43



pabc,which is the volume of an elli

soid.35

solutions to axler linear algebra done right pdf

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