MATH Homework #2

Transcription

1 MATH Homework #2 September 27, 2012

2 1. Problems Ths contans a set of possble solutons to all problems of HW-2. Be vglant snce typos are possble (and nevtable). (1) Problem 1 (20 pts) For a matrx A R n n we defne a norm by A = / x, where x s a norm n R n. Show that the followng s true: (a) A 0 and f A = 0 then A = 0 (here 0 s the zero matrx n R n n ); (b) αa = α A, α any real number; (c) A + B A + B ; (d) AB A B. (2) Problem 2 (10 pts) Show that a symmetrc n by n matrx wth postve element that s strctly row-wse dagonally domnant s also postve defnte. (3) Problem 3 (10 pts) Let ρ(a) be the spectral radus of the matrx A. Show that for any nteger k > 0, ρ(a k ) = (ρ(a)) k. (4) Problem 4 (10 pts) Prove that A 2 n A for any matrx A R n n. (5) Problem 5 (10 pts) Prove that for any nonsngular matrces A, B R n n, B 1 A 1 B 1 B A cond(a) A where s a matrx norm and cond(a) s the condton number of A w.r.t that norm. (6) Problem 6 (10 pts) Show that any strctly dagonally domnant symmetrc matrx wth postve dagonal elements has only postve egenvalues. (7) Problem 7 (10 pts) Show that f A s symmetrc and postve defnte matrx and B s symmetrc then the egenvalues of AB are real. If n addton B s postve defnte then the egenvalues of AB are postve. (8) Problem 8 (20 pts) Show that (a) A = a, where x = x ; 1n 1n (b) A 1 = 1n a, where x 1 = =1 x. =1 1

3 2 (1) Problem 1: 2. Solutons Proof. (a) If A = 0, then cleary Ax = 0, x R n, so (b) (c) A = = 0. Otherwse a 0, a A. Let e = [0,, 1,, 0] T, then A = Ae e > 0. In summary, A 0, and A = 0 ff A = 0. αa = A + B = αax (A + B)x α = = α + Bx x x + x = α A Ax + Bx = Bx x = A + B (d) Accordng to the defnton of the matrx norm, A = x R n / x we have A x, x R n. So, x R n we have the followng nequaltes: (2) Problem 2: Therefore, AB = ABx A Bx A B x ABx A B x = A B. Proof. A R n n s symmetrc, so a real dagonal matrx Λ and an orthonormal matrx Q such that A = QΛQ T, where the dagonal values of Λ are A s egenvalues λ, = 1,, n and the columns of Q are the correspondng egenvectors. Accordng to Gerschgorn s theorem, A s egenvalues are located n the unon of dsks d = {z C : z a, a, }, = 1,, n.

4 3 Because A s postve and strctly row-wse dagonal domnant, so, d = {z C : 0 < a, a, z a, }, = 1,, n. Therefore all A s egenvalues are postve, then x R n, x 0 we have, x T Ax = x T QΛQ T x = y T Λy > 0, where y = Q T x 0..e. A s postve defnte. (3) Problem 3: Frst, recall that the set of complex numbers σ(a) = {λ : det(a λi) = 0} s called spectrum of A. Proof. Then, λ σ(a), we show that λ k σ(a k ). Indeed, let Au = λu wth u 0. A k u = λa k 1 u = = λ k u. For any µ σ(a k ), we have A k v = µv, v 0. Snce, v belong to the range of A, so at least there exsts one egenvalue λ = µ 1/k s.t. Av = µ 1/k v. Thus, we always have that for any µ where A k u = µu, u 0, there exsts a λ = µ 1/k where Au = λu, u 0; for any λ where Au = λu, u 0, there exsts a µ = λ k = λ k where A k u = µu, u 0. So, ρ(a k ) = (4) Problem 4: µ µ σ(a k ) λ σ(a) λ k = ( λ σ(a) λ )k = (ρ(a)) k. Proof. It follows from the obvous strng of nequaltes A 2 = 2 = ( ( a x ) 2 ) 1/2 x 2 =1 x 2 =1 =1 ( a 2 x 2 ) 1/2 Schwartz nequalty x 2 =1 =1 = ( a 2 ) 1/2 ( ( a ) 2 ) 1/2 (5) Problem 5: =1 (n( =1 a ) 2 ) 1/2 (n A 2 ) 1/2 = n A Proof. Assume the norm satsfes the submultplcatve property,.e. AB A B (subordnate matrx norms have ths property). B 1 A 1 = A 1 B 1 = B 1 (B A)A 1 B 1 B A A 1

5 4 A, B are nonsngular matrces, dvde B 1 on both sdes, (6) Problem 6: B 1 A 1 B 1 B A A 1 = B A A 1 A A B A = cond(a) A Proof. For any symmetrx matrx A, all ts egenvalues are real. Accordng to Gerschgorn s theorem, A s egenvalues are located n the unon of dsks.e., d = {z C : z a, d = {z C : a, a, z a, + a, }, = 1,, n. a, }, = 1,, n. Because A s dagonal elements are postve and A s strctly dagonal domnant, so, a, a, = a, a, > 0 a, + a, = a, + a, > 0, = 1,, n. So, all ths dsks are located at the rght sde of the y-axs n the complex plane, so all egenvales of A are postve. (7) Problem 7: Proof. Frst possble soluton: (1) Consder an egenvalue λ and ts egenvector ψ of the matrx AB, that s ABψ = λψ. We do not know whether λ and ψ are real, so we assume that they are complex. Recal that the nner product of two complex vectors ψ and φ s defned as (φ, ψ) = φ ψ, where ψ s the complex conugate to ψ. Recall, that (φ, ψ) = (ψ, φ). For the nner product of the complex vectors ABψ and ψ we get ABψ = λψ BABψ = λbψ, (BABψ, ψ) = λ(bψ, ψ). Now usng the symmetry of A and B we get (BABψ, ψ) = (ψ, BABψ) = (BABψ, ψ), whch means that the complex number (BABψ, ψ) s equal to ts complex conugate,.e. the number s real. In the same way we prove that (Bψ, ψ) s reas as well, from where we conclude that λ s real. Then we conclude that ψ s real as well. (2) If A and B are postve defnte, then (Bψ, ψ) > 0 and (BABψ, ψ) > 0, therefore from (BABψ, ψ) = λ(bψ, ψ) t follows that λ > 0.

6 Another possble soluton (for those wth more advanced knowlegde n lnear algebra): (1) A s SPD, ts square root exsts A 1/2 whch s SPD, then AB A 1/2 ABA 1/2 = A 1/2 BA 1/2 Because both A 1/2 and B are symmetrc, then (A 1/2 BA 1/2 ) T = (A 1/2 ) T B T (A 1/2 ) T = A 1/2 BA 1/2, so A 1/2 BA 1/2 s symmetrc, and the egenvalues of A 1/2 BA 1/2 are real. AB A 1/2 BA 1/2, they have the same spectrum, so all egenvalues of AB are real. (2) If B s also SPD, we can show that A 1/2 BA 1/2 s SPD. The symmetry s shown n (1). Now show postve defnte. x R n, x 0 wth y = A 1/2 x 0 we have (A 1/2 BA 1/2 x, x) = (BA 1/2 x, A 1/2 x) = (By, y) > 0. So A 1/2 BA 1/2 s SPD, and all ts egenvalues are postve. AB A 1/2 BA 1/2, they have the same spectrum, so all egenvalues of AB are postve. 5 we show (b) as well (8) Problem 8: Proof. We prove (a). Frst we show that A a,. Indeed, A = x =1 = x =1 x =1 x =1 a = a x a x x a Next we show aslo that a, A. Assume that for some nteger k [1, n] we have a k = a. By choosng a vector y s.t., { 1, for ak, 0; y = 1, for a k, < 0;

7 6 Then, y = 1 and A = sup Ay = x R n, x =1 Thus, from the nequaltes that A = The nequalty (b) s shown n a smlar way. a k, = a,. a, A a,, t follows a,.