can be expressed as a polynomial in a of at most

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Unformatted text preview: that the spectral projectors are known, any function defined at A can be evaluated. For example, if f (z ) = z 1/2 , then √⎞ ⎛ 5 1 7 − 2√ 2 √ √ √ √ 1 f (A) = A = 2G1 + 4G2 + (1/2 4)(A − 4I)G2 = ⎝ −1 3 5 −√ 2 ⎠ . 4 2 00 22 This technique illustrated above is rather ad hoc, but it always works if a sufficient number of appropriate functions are used. For example, using f (z ) = z p for p = 0, 1, 2, . . . will always produce a system of equations that will yield the component matrices Zij given in (7.9.14) because for f (z ) = 1 : I= for f (z ) = z : A= λi Zi0 + 2 λ2 Zi0 i 2 for f (z ) = z : . . . D E Zi0 , A= Zi1 , + T H 2λi Zi1 + 2Zi2 , and this can be considered as a generalized Vandermonde linear system (p. 185) ⎛ 10 ⎞ ⎛ 1 ··· 1 ⎞ Z. ⎞ ⎛ . I ⎜.⎟ 1 ··· 1 ⎜ λ1 · · · λ s ⎟ ⎜ Zs0 ⎟ ⎜ A ⎟ ⎜2 ⎟ ⎜ Z11 ⎟ ⎜ 2 ⎟ λ1 · · · λ2 2λ1 · · · 2λs 2 · · · 2 ⎜ ⎟⎜ . ⎟ = ⎜A ⎟ s ⎜. ⎟⎜ . ⎟ ⎜ 3 ⎟ . . . . . ⎝. ⎠ ⎜ Z.s1 ⎟ ⎝ A ⎠ . . . . . . . . . . . ⎝Z ⎠ . 21 . . . ··· ··· ··· ··· . IG R Y P . that can be solved for the Zij ’s. Other sets of polynomials such as {1, (z − λ1 )k1 , (z − λ1 )k2 (z − λ2 )k2 , . . . (z − λ1 )k1 · · · (z − λs )ks } O C will generate other linear systems that yield solutions containing the Zij ’s. Example 7.9.3 ∞ j Series Representations. Suppose that j =0 cj (z − z0 ) converges to f (z ) at each point inside a circle |z − z0 | = r, and suppose that A is a matrix such that |λi − z0 | < r for each eigenvalue λi ∈ σ (A) . ∞ j =0 cj (A Problem: Explain why − z0 I)j converges to f (A). Solution: If P−1 AP = J is in Jordan form as described on p. 601, then it’s ∞ j not difficult to argue that j =0 cj (A − z0 I) converges if and only if ⎛ ∞ P−1 ∞ cj P−1 (A−z0 I)j P = cj (A−z0 I)j P = j =0 Copyright c 2000 SIAM ∞ j =0 j =0 ⎜ ⎝ cj (J − z 0 I) j = ⎜ .. ⎞ ....
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