Eigenvalue of a matrix.

1. Jan 14, 2012

chocolatefrog

1. The problem statement, all variables and given/known data

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

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 realised that these are determinants we are dealing with, not absolute values. :/

2. Jan 14, 2012

micromass

Staff Emeritus
So let $P(z)$ be a polynomial. We wish to prove that $P(\lambda)$ is an eigenvalue of $P(A)$.

For each $\mu\in \mathbb{C}$, we can write

$$P(z)-\mu=r_0(z-r_1)...(z-r_n)$$

Thus

$$P(A)-\mu I = r_0(A-r_1 I)...(z-r_n I)$$

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