O c in particular if re i 0 for every i a then

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Unformatted text preview: ize of the largest Jordan block associated with an eigenvalue λ is k (i.e., if index (λ) = k ), then f (λ), f (λ), . . . , f (k−1) (λ) must exist in order for f (J) to make sense. D E Matrix Functions For A ∈ C n×n with σ (A) = {λ1 , λ2 , . . . , λs } , let ki = index (λi ). • • T H A function f : C → C is said to be defined (or to exist) at A when f (λi ), f (λi ), . . . , f (ki −1) (λi ) exist for each λi ∈ σ (A) . .. Suppose that A = PJP−1 , where J = .J IG R .. is in Jordan form . with the J ’s representing the various Jordan blocks described on p. 590. If f exists at A, then the value of f at A is defined to be ⎛ f (A) = Pf (J)P−1 = P ⎝ Y P .. . ⎞ f (J ) ⎠ P−1 , .. . (7.9.3) where the f (J ) ’s are as defined in (7.9.2). O C We still need to explain why (7.9.3) produces a uniquely defined matrix. The following argument will not only accomplish this purpose, but it will also establish an alternate expression for f (A) that involves neither the Jordan form J nor the transforming matrix P. Begin by partitioning J into its s Jordan segments as described on p. 590, and partition P and P−1 conformably as P = P1 | · · · | Ps , ⎛ J=⎝ J(λ1 ) .. ⎠, . J(λs ) ⎞ Q1 ⎜.⎟ = ⎝ . ⎠. . ⎛ ⎞ and P−1 Qs Define Gi = Pi Qi , and observe that if ki = index (λi ), then Gi is the projector onto N (A − λi I)ki along R (A − λi I)ki . To see this, notice that Li = J(λi ) − λi I is nilpotent of index ki , but J(λj ) − λi I is nonsingular when Copyright c 2000 SIAM Buy online from SIAM http://www.ec-securehost.com/SIAM/ot71.html Buy from AMAZON.com 602 Chapter 7 Eigenvalues and Eigenvectors http://www.amazon.com/exec/obidos/ASIN/0898714540 i = j, so (A − λi I) = P(J − λi I)P−1 ⎞ ⎛ J (λ ) − λ I 1 i .. ⎜ . ⎜ Li = P⎜ ⎝ .. It is illegal to print, duplicate, or distribute this material Please report violations to meyer@ncsu.edu . J(λs ) − λi I ⎟ ⎟ −1 ⎟P ⎠ (7.9.4) is a core-nilpo...
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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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