In other words let t h ig r y xj 1 xj j rj and

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Unformatted text preview: ion of preceding Krylov vectors (say k−1 k−1 Ak b = j =0 αj Aj b ), then v (x) = xk − j =0 αj xj (or v (x) = 1 when b = 0 ) is the minimum polynomial for b relative to A. sot labor,” and he is one of a few mathematicians to have a lunar feature named in his honor—on the moon there is the “Crater Krylov.” Copyright c 2000 SIAM Buy online from SIAM Buy from 7.11 Minimum Polynomials and Krylov Methods 647 So is the minimum polynomial for a matrix related to minimum polynomials for vectors? It seems intuitive that knowing the minimum polynomial of b relative to A for enough different vectors b should somehow lead to the minimum polynomial for A. This is indeed the case, and here is how it’s done. Recall that the least common multiple (LCM) of polynomials v1 (x), . . . , vn (x) is the unique monic polynomial l(x) such that It is illegal to print, duplicate, or distribute this material Please report violations to (i) each vi (x) divides l(x); (ii) if each vi (x) also divides q (x), then l(x) divides q (x). Minimum Polynomial as LCM D E T H Let A ∈ C n×n , and let B = {b1 , b2 , . . . , bn } be any basis for C n×1 . If vi (x) is the minimum polynomial for bi relative to A, then the minimum polynomial m(x) for A is the least common multiple of v1 (x), v2 (x), . . . , vn (x). (7.11.5) IG R Proof. The strategy first is to prove that if l(x) is the LCM of the vi (x) ’s, then m(x) divides l(x). Then prove the reverse by showing that l(x) also divides m(x). Since each vi (x) divides l(x), it follows that l(A)bi = 0 for each i. In other words, B ⊂ N (l(A)), so dim N (l(A)) = n or, equivalently, l(A) = 0. Therefore, by property (7.11.4) on p. 645, m(x) divides l(x). Now show that l(x) divides m(x) . Since m(A)bi = 0 for every bi , it follows that deg[vi (x)] < deg[m(x)] for each i, and hence there exist polynomials qi (x) and ri (x) such that m(x) = qi (x)vi (x) + ri (x), w...
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