917 to prove that if g j and f g j each exist then 10

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Unformatted text preview: cj (J − z0 I) j =0 .. j ⎟ ⎟ ⎠ . Buy online from SIAM http://www.ec-securehost.com/SIAM/ot71.html Buy from AMAZON.com 606 Chapter 7 Eigenvalues and Eigenvectors http://www.amazon.com/exec/obidos/ASIN/0898714540 ∞ j =0 cj (J converges. Consequently, it suffices to prove that to f (J ) for a generic k × k Jordan block 1 .. .. .. λ λ J= 0 = λI + N, where N= − z0 I)j converges 1 .. .. .. 0 . k×k It is illegal to print, duplicate, or distribute this material Please report violations to meyer@ncsu.edu ∞ Example 7.9.4 j A standard theorem from analysis states that if j =0 cj (z − z0 ) converges to f (z ) when |z − z0 | < r, then the series may be differentiated term by term to yield series that converge to derivatives of f at points inside the circle of convergence. Consequently, for each i = 0, 1, 2, . . . , f (i) (z ) = i! ∞ cj j =0 j (z − z0 )j −i i j j (λ − z0 )j −1 N + · · · + (λ − z0 )j −(k−1) Nk−1 , 1 k−1 IG R so this together with (7.9.15) produces ∞ ⎛ cj (J − z0 I)j = ⎝ ∞ ⎞ j =0 ⎛ cj (λ − z0 )j ⎠ I + ⎝ Y P j =0 ⎛ + ··· + ⎝ O C (7.9.15) T H We know from (7.9.1) (with f (z ) = z j ) that (J − z0 I)j = (λ − z0 )j I + D E |z − z0 | < r. when ∞ cj j =0 ∞ cj j =0 = f (λ)I + f (λ)N + · · · + ⎞ j (λ − z0 )j −1 ⎠ N 1 ⎞ j (λ − z0 )j −(k−1) ⎠ Nk−1 k−1 f (k−1) (λ)Nk−1 = f (J∗ ). (k − 1)! Note: The result of this example validates the statements made on p. 527. All Matrix Functions Are Polynomials. It was pointed out on p. 528 that if A is diagonalizable, and if f (A) exists, then there is a polynomial p(z ) such that f (A) = p(A), and you were asked in Exercise 7.3.7 (p. 539) to use the Cayley–Hamilton theorem (pp. 509, 532) to extend this property to nondiagonalizable matrices for functions that have an infinite series expansion. We can now see why this is true in general. Problem: For a function f defined at A ∈ C n×n , exhibit a polyno...
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