Math 510 Assignment 6 Due date: Nov. 26, 2012
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1 1 If A M is Hermitia, show that Math 510 Assigmet 6 Due date: Nov 6, 01 tr A rak (Atr (A with equality if ad oly if A = ap for some a R ad orthogoal projectio P M (ie P = P = P Solutio: Let A = A If A = O, the tr A = 0 = rak (Atr (A ad A = 0P for every orthogoal projectio P M Suppose A 0, the tr A = tr A A > 0 Let r = rak A ad λ 1,, λ r the o-zero eigevalues of A The there exists a orthoormal set {u 1,, u } of eigevectors of A such that A λ λr 0 0 = u 1 u 0 0 u 1 u = r λ iu i u i λ 1 0 = u 1 u r λr By the Cauchy-Schwartz iequality, we have u 1 u r (tr A = ( r λ i = ( r 1 λ i ( r 1 ( r λ i = r ( r λ i = rak (Atr A rak (A (tr A tr A with equality holds if ad oly if (λ 1,, λ r = a(1,, 1 for some a R A = au 1 u r I r u 1 u r = au 1 u r u 1 u r Note that for U = u 1 u r M, r with orthoormal colums, P = UU is Hermitia ad satisfies P = UU UU = UI r U = UU = P So P is a orthogoal projectio Suppose A, B H have eigevalues a 1 a ad b 1 b (a Give uitary matrices U, V, let C = U AU + V BV have eigevalues c 1 c Prove that a 1 + b 1 c 1 a 1 + b, a + b 1 a 1 + b, a + b 1 c a + b (1 a 1 + a + b 1 + b = c 1 + c 1
2 Solutio: Note that λ i (A = λ i (U AU ad λ i (B = λ i (V BV for 1 j Applyig Theorem 437 (i our order with (j, k = (1, 1, (1,, (, 1, (,, we have the iequalities i (1 The equality follows from c 1 + c = tr (C = tr (U AU + V BV = tr (A + tr (B = a 1 + a + b 1 + b (b Suppose C H has eigevalues c 1 c satisfyig (1 Show that there exist uitary matrices U, V such that C = U AU + V BV a1 0 (Hit: First cosider the case whe A = ad B = Show that 0 a 0 b there exists t such that cos t si t cos t si t A + B si t cos t si t cos t has eigevalues c 1, c Deduce the geeral case from this a1 0 Solutio: First cosider the case whe A = ad B = Let ( 0 a 0 b cos t si t U(t = ad C(t = U(t si t cos t AU(t + B The U(t is uitary for all t ad we have ( a1 + b C(0 = 1 0 ad 0 a + b Sice ( π C = ( a + b a 1 + b ( π λ 1 (C(0 = a 1 + b 1 c 1 max{a 1 + b, a + b 1 } = λ 1 (C, there exists t 0, π such that λ1 (C(t = c 1 The λ (C(t = tr (C(t λ 1 (C(t = tr A + tr B c 1 = a 1 + a + b 1 + b c 1 = c Hece, there exists a uitary U 1 such that U1 c1 0 C(tU 1 = 0 c C = U c1 0 U 0 c for some uitary U The Suppose C = U U 1 C(tU 1 U = U U 1 (U(t AU(t + BU 1 U = U AU + V BV, where U = U(tU 1 U ad V = U 1 U For the geeral case, we ca get uitaries V 1, V such that V1 a1 0 AV 1 = 0 a The we have 0 b, V BV = C = U U 1 C(tU 1 U = U U 1 (U(t (V 1 AV 1 U(t + V BV U 1 U = U AU + V BV, where U = V 1 U(tU 1 U ad V = V U 1 U
3 3 A matrix S = (s i j is said to be doubly stochastic (ds if s i j 0, s i j = s i j = 1 for all 1 i, j Prove that if S M is ds, the for Sx x for all x R (You ca use the hit i Problem 9 o page 199, Hor ad Johso but be careful with the differece i orderig You eed to write dow more details tha those give i the hit Solutio: We prove a slightly more geeral result Suppose s i j 0, s i j 1 for all 1 i, j The Sx w x For 1 k, let w (k j Let Sx = y 1 y s ij The 0 w (k j 1 ad w(k j s i j k Sice y i = s ijx j, we have k x i k y i x i k s ijx j k x i k s ijx j x k (k w(k j x j w(k j (1 w(k j x j x j x k ( k j=k+1 w(k j (1 w(k j j=k+1 w(k j x j x k ( k (1 w(k j (x j x k + j=k+1 w(k j (x k x j 0 because x j x k for j k ad x k x j for j k + 1 Hece, Sx w x If, i additio, k=1 s k j = 1 for all 1 j The Hece, Sx x y i = s ij x j = s ij x j = (1 w(k j j=k+1 w(k j x j 4 Suppose A = a ij is a ormal matrix with eigevalues λ i, i = 1,,, such that λ 1 λ λ Let (d 1, d,, d = diag (A Prove that ( d 1, d,, d w ( λ 1, λ,, λ Solutio: We may assume that d 1 d d ad A = U d 1 d = u ij λ 1 λ Let u ij λ 1 λ = (b 1, b,, b ( λ 1, λ,, λ 3 b 1 b λ 1 O O λ The we have U The
4 Therefore, for 1 k, d i = u ij λ j u ij λ j = b i λ i 5 Suppose A = a ij H has eigevalues λ 1 (A λ (A λ (A Let 1 r 1 Prove that (a if r a ii = r λ A1 0 i(a the A = where A 0 A 1 H r has eigevalues λ 1 (A λ (A λ r (A Solutio: Firstly, we follow the hit ad prove the followig: a11 a Claim: Suppose A = 1 a 1 a such that (U AU 11 > a 11 H ad a 1 0 The there exists a uitary U M If a 1 0, let a 1 = a 1 e it for some t R The a 1 = a 1 = a 1 e it Choose 0 < θ < π such that cot θ > (a 11 a a 1 si θ > (a 11 a si θ Let U = e it si θ e it si θ a 1 The (U AU 11 = e it a11 a si θ 1 a 1 a e it si θ = a 11 cos θ + a si θ + a 1 si θ > a 11 cos θ + a si θ + (a 11 a si θ = a 11 A11 A Suppose 1 r < ad A = 1 with A A 1 A 11 H r such that r a ii = r λ i(a To show that A 1 = 0 Suppose the cotrary that there exist 1 i r ad r + 1 j such ( that a ij 0 By the above claim, there exists a uitary matrix U 1 such that U1 aii a ij U 1 > a ii Let U = (u st be give by a ji a jj 11 u st = (U 1 11 if (s, t = (i, i (U 1 1 if (s, t = (i, j (U 1 1 if (s, t = (j, i (U 1 if (s, t = (j, j δ st otherwise The U is uitary ad (U AU ii > a ii ad (U AU kk = a kk for all 1 k r ad k i So we have r k=1 (U AU kk > r k=1 a kk = r k=1 λ i(a, a cotradictio to Theorem 43 4
5 (b if r a ii = r λ +1 i(a, the A = λ +1 r (A λ + r (A λ (A A1 0 0 A where A 1 H r has eigevalues Solutio: r a ii = r λ +1 i(a r ( a ii = r λ +1 i(a = r λ i( A The apply the result i part (a to A 5
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