xj span q1 qj span r0 r1

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Unformatted text preview: here deg[ri (x)] < deg[vi (x)]. But Y P O C 0 = m(A)bi = qi (A)vi (A)bi + ri (A)bi = ri (A)bi insures ri (x) = 0, for otherwise ri (x) (when normalized to be monic) would be an annihilating polynomial for bi of degree smaller than the minimum polynomial for bi , which is impossible. In other words, each vi (x) divides m(x), and this implies l(x) must also divide m(x). Therefore, since m(x) and l(x) are divisors of each other, it must be the case that m(x) = l(x). The utility of this result is illustrated in the following development. We already know that associated with n × n matrix A is an nth -degree monic polynomial—namely, the characteristic polynomial c(x) = det (xI − A). But the reverse is also true. That is, every nth -degree monic polynomial is the characteristic polynomial of some n × n matrix. Copyright c 2000 SIAM Buy online from SIAM http://www.ec-securehost.com/SIAM/ot71.html Buy from AMAZON.com 648 Chapter 7 Eigenvalues and Eigenvectors http://www.amazon.com/exec/obidos/ASIN/0898714540 Companion Matrix of a Polynomial For each monic polynomial p(x) = xn + αn−1 xn−1 + · · · + α1 x + α0 , the companion matrix of p(x) is defined (by G. Frobenius) to be It is illegal to print, duplicate, or distribute this material Please report violations to meyer@ncsu.edu ⎛0 C= • Proof. ⎜1 ⎜. . ⎝. 0 0 0 0 .. . ··· 0 ··· ··· .. . 1 ··· −α0 ⎞ −α1 ⎟ . ⎟. . . ⎠ −αn−2 −αn−1 n×n 0 0 0 1 (7.11.6) D E The polynomial p(x) is both the characteristic and minimum polynomial for C (i.e., C is nonderogatory). T H To prove that det (xI − C) = p(x), write C = N − ceT , where n ⎛ ⎞ ⎛ ⎞ 0 α0 ⎜ 1 ... ⎟ ⎜ α1 ⎟ ⎟ and c = ⎝ . ⎠ , N=⎜ . .. ⎝ ⎠ . . . . IG . 1 0 R Y and use (6.2.3) on p. 475 to conclude that αn−1 −1 det (xI − C) = det (xI − N)(1 + eT det (xI − N) n = xn 1 + eT n P c) N I N Nn−1 + 2 + 3 + ··· + xx x xn 2 c = xn + αn−1 xn−1 + αn−2 xn−2 + · · · + α0 O C = p(x). The fact that p(x) is also the minimum polynomial for C is a consequence of (7.11.5). Set B = {e1 , e2 , ....
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