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# m along with initial conditions y 0 y 1

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Unformatted text preview: mial p(z ) such that f (A) = p(A). Copyright c 2000 SIAM Buy online from SIAM http://www.ec-securehost.com/SIAM/ot71.html Buy from AMAZON.com 7.9 Functions of Nondiagonalizable Matrices http://www.amazon.com/exec/obidos/ASIN/0898714540 607 Solution: Suppose that σ (A) = {λ1 , λ2 , . . . , λs } with index (λi ) = ki . The trick is to ﬁnd a polynomial p(z ) such that for each i = 1, 2, . . . , s, p(λi ) = f (λi ), p (λi ) = f (λi ), p(ki −1) (λi ) = f (ki −1) (λi ) ..., (7.9.16) because if such a polynomial exists, then (7.9.9) guarantees that It is illegal to print, duplicate, or distribute this material Please report violations to [email protected] s ki −1 p(A) = i=1 j =0 p(j ) (λi ) (A − λi I)j Gi = j! s ki −1 i=1 j =0 f (j ) (λi ) (A − λi I)j Gi = f (A). j! D E s Since there are k = i=1 ki equations in (7.9.16) to be satisﬁed, let’s look for a polynomial of the form T H p(z ) = α0 + α1 z + α2 z 2 + · · · + αk−1 z k−1 by writing the equations in (7.9.16) as the following k × k linear system Hx = f : ⇒ ⎛ 1 ⎜. ⎜. ⎜. ⎜ ⇒ ⎜1 ⎜ ⎜ ⎜ ⎜. . ⎜. . ⎜. . ⎜ p (λi ) = f (λi ) ⇒ ⎜ 0 ⎜ ⎜ . ⎜. . ⎜. . ⎜. ⎜ ⎜ ⎜ . ⎜. . ⎜. . ⎜. p (λi ) = f (λi ) ⇒ ⎜ 0 ⎜ ⎜ . ⎜. . ⎜. . ⎜. ⎜ ⎜ ⎝ . . . . . . p(λ1 ) = f (λ1 ) . . . p(λs ) = f (λs ) λ1 . . . λs λ2 1 . . . λ2 s . . . 1 . . . . . . 2λi . . . . . . 3λ2 i . . . . . . 0 . . . . . . 2 . . . . . . 6λi . . . . . . . . . . . . IG R λ3 1 . . . λ3 s Y P O C ··· ··· ··· ··· ⎞⎛ α ⎞ ⎛ ⎞ 0 f (λ1 ) ⎟⎜ ⎟ ⎜ ⎟ . ⎟⎜ ⎟ ⎜ ⎟ . ⎟ ⎜ α1 ⎟ ⎜ ⎟ . ⎟⎜ ⎟ ⎜ ⎟ ⎟⎜ ⎟ ⎜ f (λs ) ⎟ ⎟⎜ ⎟ ⎜ ⎟ ⎟ ⎜ α2 ⎟ ⎜ ⎟ ⎟⎜ ⎟ ⎜ ⎟ ⎟⎜ ⎟ ⎜ ⎟ . . ⎟⎜ ⎟ ⎜ ⎟ . . ⎟⎜ α ⎟ ⎜ ⎟ . . 3⎟ ⎟⎜ ⎜ ⎟ k −2 ⎟⎜ ⎟ ⎜ f (λi ) ⎟ (k − 1)λi ⎟⎜ ⎟ ⎜ ⎟ ⎟⎜ ⎟ ⎜ ⎟ . ⎟⎜ . ⎟ ⎜ ⎟ . . . ⎟⎜ . ⎟ = ⎜ ⎟. . . ⎟⎜ . ⎟ ⎜ ⎟ ⎟⎜ ⎟ ⎜ ⎟ ⎟⎜ ⎟ ⎜ ⎟ ⎟⎜...
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