Could it be shown that the polynomial matrix $p(\mathbf{A})$ has eigenvalues and same eigenvectors as $\mathbf{A}$? 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.
 A: Here's another hint: If $X$ is an eigenvector of $A$, say $AX=\lambda X$, then you can use that to simplify $p(A)X$ into $(\text{some scalar value})X$, and that thing in front of $X$ is then an eigenvalue of $p(A)$, corresponding to the eigenvector $X$.
A: Hint: write $A$ in upper triangular form. Locate the eigenvalues of $A$. Show that $p(A)$ is upper triangular as well and compute its diagonal. Conclude.
A: You can start by writing $\mathbf{A} = P \mathbf{\Lambda} P^{-1}$, where $P$ is the matrix of eigenvectors and $\mathbf{\Lambda}$ is the diagonal matrix of eigenvalues. Therefore, you can write $\mathbf{A}^k$ as $P\mathbf{\Lambda}^k P^{-1}$, where $\mathbf{\Lambda}^k$ is simply the diagonal matrix made up of the $n^{th}$ powers of the eigenvalues of $\mathbf{A}$.
Using this fact, you can write
\begin{equation}
p( \mathbf{A} ) =  P(k_{m}\mathbf{\Lambda}^{m}+k_{m-1} \mathbf{\Lambda}^{m-1}+\ldots+k_{1} \mathbf{\Lambda} +k_{0} \mathbf{I})P^{-1} \
\end{equation}
This shows that $P(\mathbf{A})$ has eigenvalues given by $p(\lambda_i)$ for $i = 1,2,\ldots,n$ and the same eigenvectors as $\mathbf{A}$
EDIT: My proof works only for diagonalizable matrices as Theo Buehler has commented below. 
