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Hints and Solutions to Selected Exercises

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Hints and Solutions to Selected Exercises Hints and Solutions to Selected Exercises Chapter 1 1.1. Using the formula of Remark 1.4, we obtain II[~ ~]II = J2IUI2+1~J4IUI2+1, II [~ ~] II = max{lul, Ivl}, C?S t - sin t] II = 1 cosh t sinh t] II = {t -t} II [smt cost ' II [sinh t cosh t max e ,e . 1.2. (a) From the definition of the matrix norm, -1 { IBAB-Ixl }{ IAxl } IIBAB II = max Ixl :x t 0 = max IBxl: x t 0 = max {I~~I :x to} = IIAII. (b) The inequality IICAC- I II ~ IICIlIIAIIIIC- I II always holds. 1.3. (a) We have Ixl 2 = x·x by definition, so Since the unit sphere {x E en : Ixl = 1} is closed and bounded, the continuous 2 function x I---> IAxl defined on it attains its supremum, hence 2 (b) The matrix A*A is hermitian and for any x E en, x*A*Ax = IAxI , which is real and non-negative; it also vanishes precisely when Ax = O. By the general theory of hermitian matrices, there is an orthonormal basis {UI, ... ,Un} consisting of eigenvalues AI, ... ,An which are all real. For each j, 303 304 Matrix Groups: An Introduction to Lie Group Theory since ujUj = 1; hence Aj ~ O. If A i= 0 there must be at least one non-zero eigenvalue of A*A since otherwise IAxl2 would always vanish; so in this case we have a largest positive eigenvalue A which we might as well assume is A= AI. Now if Ixl = 1 we can 2 2 write x = tIU1 + ... + tnun, where tj EC and ItII + ... + Itn l = 1. Then 2 x* A*Ax = X*(A1tIU1 + ... + AntnUn) = A11tl12 + + An ltnl ~ A11tII 2 + + A11tnl2 = A1, hence IIAII 2 ~ Al and in fact we have equality since uiA*AU1 = AI. 1.4. (a) Let 1 ( . IIAr+111) £ = 2 1 + r~~ lA::lI < 1. For large enough m we have IIAm + ... + Am+kll ~ IIAmll + ... + II Am+k1l m m m m Am = IIAmll (1 + IIA +l1l + IIA +dIIlA +211 + ... + IIA +lll .. 'II +kll) II Am II IIAmIiIlAm+l1l IIAmll" ·IIAm+k-111 < IIAmll (1 + £ + £2 + ... + £k-l) = II Am II \11-=-£;1-+ I~A_m~1 as k -+ 00. Also, if mo is large enough, then for m > mo, II~:~~ II < £, so IIAmll = ::~:~~il·.I!~~:::IIIAmoll < £m-mOIiAmoll-+ 0 as m -+ 00. Combining these we see that the partial sums of the series form a Cauchy sequence and hence the series converges. (b) Here IIAml1 -+ 00 as m -+ 00 and the series diverges. (c) Any of the usual tests based on absolute convergence works. 1.5. (a) For m, n ~ 1, m m m m n II~ Arll = IIA + Am+! + ... + Am+nll ~ IIAli + IIAll +! + ... + IIAli + A n 1 = IIAli m (1 _II ll + ) -+ 0 as m n -+ 00 (1 -IIAII) ,. So the sequence of partial sums is a Cauchy sequence. This can also be done using the ratio test of Exercise 1.4. (b) For any n ~ 1, (I - A)(I + A + A 2 + ... + An) = 1- An+ 1 = (I + A + A 2 + ... + An)(I - A), so 2 n 1 n II (I - A)(I + A + A + ... + An) - III = IIA + 11 ~ IiAli +l -+ 0 as n -+ 00, and similarly for II(I + A + A 2 + ... + An)(I - A) - III. Hence I - A has inverse 00 r (I - A)-l = LA . r=O Hints and Solutions to Selected Exercises 305 (c) If A k = 0, then k-1 k-1 (I - A)-l = LAr, exp(A) = L ~Ar. T. r=O r=O 1.6. (a) The subset O(n) = {A E Mn(R) : ATA - I = O} ~ Mn(R) is bounded since for A E O(n), IIAII = 1. It is also closed since if a sequence Ar E O(n) converges to A EMn(R) then ATA - I = lim (A;Ar - 1) = lim 0 = O. r-+CX) r-+oo Therefore O(n) is compact. (b) The subset U(n) = {A E Mn(C) : A*A - I = O} ~ Mn(C) is compact by a similar argument to (a). (c) Consider the n x n diagonal matrices A k = diag(k, 11k, 1, ... ,1) (k ~ 1). Then detAk = 1 so Ak E SLn(k) ~ GLn(k). But IIAkll = k -+ 00 as k -+ 00, so this sequence is unbounded. 1.7. (a) If A, BE H then there are sequences Ar ,Br in H with Ar -+ A and Br -+ B as T -+ 00. Hence ArBr -+ AB and so AB E H since each ArBr E H. Similarly, A;:-l -+ A-I, showing that A-I E H. (b) Here is an alternative proof in terms of open sets which applies to any topological group. Let u, v E H and consider uv. If uv f/. H, the open set G - H contains uv; notice that G - H is the biggest open subset of G which does not intersect H. Since the product map mult is continuous, there is an open set of the form U x V ~ G x G contained in the open set mult-1(G - H) with u E U and v E V. Notice that UnH = 0 = vnH by the above remark. Now take u' E UnH and v' E VnH and note that u'v' = mult(u', v') E H and also mult(u', v') E G - H, giving a contradiction. So uv E H. Also, if wE H, suppose that w-1 E G -H. Then inv-1(G -H) ~ G is open, where inv is the inverse map. But as w E inv-1 (G - H), there must be an element hE H n inv-1 (G - H) and so h-1 E G - H which is impossible. Hence H ~ G. (c) Consider any sequence of diagonal matrices A k = diag(ak' 1, ... ,1) (k ~ 1) where ak ~ 1 and ak -+ v'2 as k -+ 00 (v'2 could be replaced by any other irrational number greater than 1). Then Ak -+ diag(v'2, 1, ... ,1) f/. r. We must have r = GLn(lR) since for every B = [bij] E GLn(lR) we can find sequences b~7) E Q satisfying b~7) -+ bij as k -+ 00. 1.8. By Proposition 1.26, G ~ Mn(k) is a closed subset of GLn(lR). 1.9. Suppose that G ~ GLn(lR) ~ Mn(lR) for some n. Then H ~ Mn(lR) is closed and bounded since G is compact. For each h E H the set {h} ~ H is open and these form 306 Matrix Groups: An Introduction to Lie Group Theory an open cover of H by disjoint open sets. By the Hein~Borel Theorem 1.20 there must be finitely many of these sets that cover H, so H is itself finite. 1.10. In each case, we have subgroups Gn+l ~ GLn+l(k) (where k = R or C) and Gn ~ GLn(k) where Gn+! is closed in GLn+l(k) and Gn = Gn+l n GLn(k). As GLn(k) is closed in GLn+l(k), Gn is closed in Gn +!. 1.11. (a) If A E SymP2m(R), then ATJ2mA = J2m and as detJ2m = 1, we have detA2 = 1, so detA = ±1. (b) Let A = [~ ~] E SymP2(R). By definition, [~ ~][_~ ~][~ ~]=[_adO+bC adobc]=[_~ ~], so A E SymP2(R) if and only if detA = Ii hence Symp2(R) ~ SL2(R). 1.12. Clearly Pn U(n) ~ pn GLn(C). If Z = [Zr.] E Mn(C) where Zr. = Xr• +Yr.i with Xr.,Yr. E R, then Pn(Z·) = Pn([X.r] - [Y.riJ) = Pn(Z)T. Hence Z E U(n) if and only if Pn(Z) E O(2n). If A E O(2n), then A E SymP2n(R.) if and only if Ahn = hnA and by Proposition 1.44 this is equivalent to A E pn U(n). This shows that pn U(n) = O(2n)nSymP2n(R).1f BE Symp2n(JR) then Bhn = hnB T if and only if B = B. This shows that pn GLn(C)nSymP2n(R) = O(2n)nSymP2n(R). The second collection of equalities is proved in a similar way. 1.13. (a) The triangle inequality for P follows from the following sequence of inequal­ ities valid for any (XI,X2), (YI,Y2), (ZI,Z2) E Xl X X2, P((XI, X2), (Zl, Z2)) =JPI (Xl, ZI)2 + P2(X2, Z2)2 ~J(PI(XI' YI) + PI(YI, zdF + (p2(X2, Y2) + P2(Y2, Z2))2 [by the triangle inequalities for PI and P2] [by the usual triangle inequality for R2] The other properties required for p to be a metric are straightforward to check. If UI x U2 ~ Xl X X2 with UI ~ Xl and U2 ~ X2 open, then for each Xl E UI and X2 E U2 there are open discs NX1>Pl(Xlirl) ~ UI and NX2,P2(X2ir2) ~ U2i for r = min{rl,r2}, we have NX1XX2,P((XI,x2)jr) ~ UI x U2, so UI X U2 is open with respect to p. Conversely, for an open disc NX1XX2,P((XI,x2);r) we have NX1,Pl(xl;r/v'2) x NX2,P2 (X2i r/v'2) ~ NX1XX2'P((XI,x2)jr). Hence every subset of Xl x X2 open with respect to P is a union of products of open discs in Xl and X 2 • So the topology associated with p has the same open sets as the product topology. (b) If (XI,r,X2,r) -. (XI,X2) say, then for i = 1,2, Pi(Xi,r, Xi) ~ P((XI,r,X2,r), (XI,X2)), Hints and Solutions to Selected Exercises 307 SO Xi,r .....
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