where n 1 2 n n prove that if n converges

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Unformatted text preview: and ρ(K) < 1. (7.10.31) Now that we know when limk→∞ Ak exists, let’s describe what limk→∞ Ak looks like. We already know the answer when p = 0—it’s 0 (because ρ (A) < 1). But when p is nonzero, limk→∞ Ak = 0, and it can be evaluated in a couple of different ways. One way is to partition P = P1 | P2 and P−1 = Q1 , and Q2 use (7.10.5) and (7.10.31) to write Y P lim Ak ×n = lim P n O C k→∞ k→∞ = P1 | P2 Ip×p 0 Ip×p 0 0 Kk P−1 = P Ip×p 0 0 0 P−1 (7.10.32) 0 0 Q1 Q2 = P1 Q1 = G. Another way is to use f (z ) = z k in the spectral resolution theorem on p. 603. If σ (A) = {λ1 , λ2 , . . . , λs } with 1 = λ1 > |λ2 | ≥ · · · ≥ |λs |, and if index (λi ) = ki , where k1 = 1, then limk→∞ k λk−j = 0 for i ≥ 2 (see p. 618), and i j s ki −1 Ak = i=1 j =0 s k k−j λ (A − λi I)j Gi ji ki −1 = G1 + i=2 j =0 Copyright c 2000 SIAM k k−j λ (A − λi I)j Gi → G1 ji as k → ∞. Buy online from SIAM http://www.ec-securehost.com/SIAM/ot71.html Buy from AMAZON.com 630 Chapter 7 Eigenvalues and Eigenvectors http://www.amazon.com/exec/obidos/ASIN/0898714540 In other words, limk→∞ Ak = G1 = G is the spectral projector associated with λ1 = 1. Since index (λ1 ) = 1, we know from the discussion on p. 603 that R (G) = N (I − A) and N (G) = R (I − A). Notice that if ρ(A) < 1, then I − A is nonsingular, and N (I − A) = {0}. So regardless of whether the limit is zero or nonzero, limk→∞ Ak is always the projector onto N (I − A) along R (I − A). Below is a summary of the above observations. It is illegal to print, duplicate, or distribute this material Please report violations to [email protected] Limits of Powers ρ(A) = 1, where λ = 1 is the only eigenvalue on the unit circle, and λ = 1 is semisimple. When it exists, D E lim Ak = the projector onto N (I − A) along R (I − A). (7.10.34) For A ∈ C n×n , limk→∞ Ak exists if and only if ρ(A) < 1 or else (7.10.33) T H k→∞ IG R With each scal...
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This document was uploaded on 03/06/2014 for the course MA 5623 at City University of Hong Kong.

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