What are the eigenvalues of P(A)?

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Homework Statement



αo, α1,..., αd [itex]\inℝ[/itex]. Show that αo + α1λ + α2λ2 + ... + αdλd [itex]\inℝ[/itex] is an eigenvalue of αoI + α1A + α2A2 + ... + αdAd [itex]\inℝ^{nxn}[/itex].


2. The attempt at a solution

If λ is an eigenvalue of A, then |A - Iλ| = 0. Also, λn is an eigenvalue An. So we basically have to somehow prove the following equation (after rearranging):

1(A - Iλ) + α2(A2 - Iλ2) + ... + αd(Ad - Iλd)| = O

3. Relevant equations

I can't seem to get my head around this one. I almost used the triangular inequality to prove it before I realized that these are determinants we are dealing with, not absolute values. :/
 
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So let [itex]P(z)[/itex] be a polynomial. We wish to prove that [itex]P(\lambda)[/itex] is an eigenvalue of [itex]P(A)[/itex].

For each [itex]\mu\in \mathbb{C}[/itex], we can write

[tex]P(z)-\mu=r_0(z-r_1)...(z-r_n)[/tex]

Thus

[tex]P(A)-\mu I = r_0(A-r_1 I)...(z-r_n I)[/tex]

So [itex]P(A)-\mu I[/itex] is invertible iff all [itex](A-r_i I)[/itex] is invertible. What does that imply for the eigenvalues?