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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
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= 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
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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¦ Ȧ(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
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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
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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
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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
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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
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, 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
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(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
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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
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, . . . , 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 Ŧau [ a Î F¦.5. By linearity 0 = ni=1 aiT(v
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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.
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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
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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
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, . . . , 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
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)) 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.
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.
.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
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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 =
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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.
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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
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, 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
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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
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) = 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
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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
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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
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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
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+ . . . + 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=
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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
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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
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_ ¸
_.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
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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
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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[
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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
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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|
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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
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= (|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|
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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
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¸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
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|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
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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
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) 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
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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
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= 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
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(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)
_
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_ 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,
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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
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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)).
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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
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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
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= 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 =
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(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
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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
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= 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
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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
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this imp
ies that
ni=1[ai[ = 1. Assume that |Tv -lv| < e. If [li -l[ ³ e for a
i, th n|Tv -lv| = |n
i=1(aiTvi -lvi)| = |n
i=1(a
ilivi -lvi)|= |n
i=1a
i(li -l)vi| =n
i=1[ai[[li
-l[|v1|=n
i=1[ai[[l
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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
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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.
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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
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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.
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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
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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
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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
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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
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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
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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
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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
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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
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si¸v,
i) fi| =n
i=1si[ ¸v,
i) [sinc
a
si ar
positiv
. Now c
ar
yà s|v| = à 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
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) 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
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(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
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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
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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
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) 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
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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).
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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
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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
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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
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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
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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
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) 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
_ ¸¸
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¸¸¸
_ .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
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|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 .
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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)
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-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
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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
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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
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) 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
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, 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
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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
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= 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
_
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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
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= ¸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
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. 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
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+ . . . + 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
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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
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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
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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
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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
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pabc,which is the volume of an elli
soid.35