# Edu example 736 k n0 s0 3 k 2n0 s0 3 1 3 2 3 s0

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Unformatted text preview: a polynomial in A. In other words, when f (A) exists, there is a polynomial p(z ) such that p(A) = f (A). This is true for all matrices, but the development here is limited to diagonalizable matrices—nondiagonalizable matrices are treated in Exercise 7.3.7. In the diagonalizable case, f (A) exists if and only if f (λi ) exists for each λi ∈ σ (A) = k {λ1 , λ2 , . . . , λk } , and, by (7.3.6), f (A) = i=1 f (λi )Gi , where Gi is the ith spectral projector. Any polynomial p(z ) agreeing with f (z ) on σ (A) does the job because if p(λi ) = f (λi ) for each λi ∈ σ (A) , then k p(A) = i=1 Copyright c 2000 SIAM k p(λi )Gi = f (λi )Gi = f (A). i=1 Buy online from SIAM http://www.ec-securehost.com/SIAM/ot71.html Buy from AMAZON.com 7.3 Functions of Diagonalizable Matrices http://www.amazon.com/exec/obidos/ASIN/0898714540 529 But is there always a polynomial satisfying p(λi ) = f (λi ) for each λi ∈ σ (A)? Sure—that’s what the Lagrange interpolating polynomial from Example 4.3.5 (p. 186) does. It’s given by ⎛ ⎛ ⎞ ⎞ k (z − λj ) ⎟ (A − λj I) ⎟ k⎜ k⎜ j =1 j =1 ⎜ ⎜ ⎟ ⎟ ⎜f (λi ) j=i ⎜f (λi ) j=i ⎟ , so f (A) = p(A) = ⎟. p(z ) = ⎜ ⎜ ⎟ ⎟ k k ⎝ ⎝ ⎠ i=1 i=1 (λ − λ ) (λ − λ ) ⎠ k It is illegal to print, duplicate, or distribute this material Please report violations to [email protected] i Example 7.3.3 j i j =1 j =i j j =1 j =i k that in (7.3.6) yields D E 1 if z = λi , with this representation as well as 0 if z = λi , Using the function gi (z ) = (A − λj I) j =1 j =i k (λi − λj ) = gi (A) = Gi . For example, T H j =1 j =i if σ (An×n ) = {λ1 , λ2 , λ3 }, then f (A) = f (λ1 )G1 + f (λ2 )G2 + f (λ3 )G3 with G1 = (A−λ2 I)(A−λ3 I) , (λ1 −λ2 )(λ1 −λ3 ) G2 = (A−λ1 I)(A−λ3 I) , (λ2 −λ1 )(λ2 −λ3 ) G3 = IG R Below is a summary of these observations. (A−λ1 I)(A−λ2 I) . (λ3 −λ1 )(λ3 −λ2 ) Spectral Projectors Y P If A is diagonalizable with σ (A) = {λ1 , λ2 , . ....
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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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