**Unformatted text preview: **istic equation, and this guarantees that An+j
(j = 0, 1, 2, . . .) can be expressed as a polynomial in A of at most
degree n − 1. Since f (A) is always a polynomial in A, the Cayley–
Hamilton theorem insures that f (A) can be expressed as a polynomial
in A of at most degree n − 1. Such a polynomial can be determined
whenever f (j ) (λi ), j = 0, 1, . . . , ai − 1 exists for each λi ∈ σ (A) ,
where ai = alg mult (λi ) . The strategy is the same as that in Example
7.9.4 except that ai is used in place of ki . If we can ﬁnd a polynomial
p(z ) = α0 + α1 z + · · · + αn−1 z n−1 such that for each λi ∈ σ (A) , IG
R Y
P p(λi ) = f (λi ), p (λi ) = f (λi ), ..., p(ai −1) (λi ) = f (ai −1) (λi ), then p(A) = f (A). Why? These equations are an n × n linear system
with the αi ’s as the unknowns, and, for the same reason outlined in
Example 7.9.4, a solution is always possible.
(a) What advantages and disadvantages does this approach have
with respect to the approach in Example 7.9.4?
(b) Use this method to ﬁnd a polynomial p(z ) such that p(A) = eA O
C for A = 7.9.17. Show that if f
⎛
αβ
A=⎝0 α
00 3
−3
−3 2
−2
−2 1
−1
−1 . Compare with Exercise 7.9.15. is a function deﬁned at
⎞
γ
β ⎠ = α I + β N + γ N2 ,
α ⎛ where then f (A) = f (α)I + βf (α)N + γ f (α) + Copyright c 2000 SIAM 0
N = ⎝0
0 1
0
0 ⎞
0
1⎠,
0 β 2 f (α) 2
N.
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7.9 Functions of Nondiagonalizable Matrices
http://www.amazon.com/exec/obidos/ASIN/0898714540 615 7.9.18. Composition of Matrix Functions. If h(z ) = f (g (z )), where f
and g are functions such that g (A) and f g (A) each exist, then
h(A) = f g (A) . However, it’s not legal to prove this simply by saying
“replace z by A. ” One way to prove that h(A) = f g (A) is to
demonstrate that h(J ) = f g (J ) for a generic Jordan block and then
invoke (7.9.3). Do this for a 3 × 3 Jordan block—the generalization to
k × k blocks is similar...

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