2
$\begingroup$

I had been working on this problem here below, but seem to not know a precise and clean way to show the proof to the question below. I had about a few ways of doing it, but the statements/operations were pretty loosely used. The problem is as follows:

Suppose $\mathbf{A}$ is an $n \times n$ matrix with (not necessarily distinct) eigenvalues $\lambda_{1}, \lambda_{2}, \ldots, \lambda_{n}$. Can it be shown that the polynomial matrix

$p( \mathbf{A} ) = k_{m} \mathbf{A}^{m}+k_{m-1} \mathbf{A}^{m-1}+\ldots+k_{1} \mathbf{A} +k_{0} \mathbf{I} $

has the eigenvalues

$p(\lambda_{j}) = k_{m}{\lambda_{j}}^{m}+k_{m-1}{\lambda_{j}}^{m-1}+\ldots+k_{1}\lambda_{j}+k_{0}$

where $j = 1,2,\ldots,n$ and the same eigenvectors as $\mathbf{A}$.

Thanks.

  • 0
    Looks like homework due to the imperative tone of the question, so here's a hint: Cayley-Hamilton.2011-03-15
  • 2
    @Jerry: 1) I would prefer if you did not give short hints in answers. Either write something substantive or leave a comment. 2) You don't need Cayley-Hamilton for this.2011-03-15

3 Answers 3