m along with initial conditions y 0 y 1

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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 ⎟⎜ ⎟ ⎜ ⎟ ⎟⎜ ⎟ ⎜ ⎟ . ⎟⎜ . ⎟ ⎜ ⎟ . . . ⎟⎜ . ⎟ = ⎜ ⎟. . . ⎟⎜ . ⎟ ⎜ ⎟ ⎟⎜ ⎟ ⎜ ⎟ ⎟⎜ ⎟ ⎜ ⎟ ⎟⎜...
View Full Document

Ask a homework question - tutors are online