**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
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7.11 Minimum Polynomials and Krylov Methods
http://www.amazon.com/exec/obidos/ASIN/0898714540 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 [email protected] (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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