TITLE: How to prove "eigenvalues of polynomial of matrix $A$ = polynomial of eigenvalues of matrix $A$ "
QUESTION [16 upvotes]: Title looks a little bit twisted. What I want to say is the following:
$A\in\mathbb{R}^{n\times n}$, polynomial of matrix $A$: $P(A)=\displaystyle \sum_{k=0}^{n} c_k A^k$. $\lambda(A)$ is the set of eigenvalues of $A$. 
So, we want to prove $\lambda(P(A))=P(\lambda(A))$
So far, I can prove it when $A$ is diagonalisable, but when $A$ is not diagonalisable, it seems we can't just simply use Jordan normal form to prove it. (Or this statement won't hold when it's not diagonalisable?)
Can anyone suggest me something and help me out? Thanks!

REPLY [4 votes]: As mentioned by Matemáticos Chibchas, the statement is not true if the underlying field is not algebraically closed.
As mentioned by loup blanc, when the underlying field is algebraically closed, the statement can be easily proved by triangularising $A$.
Without using triangularisation, the statement can be proved in two steps as follows (see Graham (2018), Matrix Theory and Applications for Scientists and Engineers, theorems 7.6 and 7.7):
Theorem 1. Let $A\in M_n(F)$ where $F$ is algebraically closed. If the spectrum of $A$ (as a multiset) is $\{\lambda_1,\lambda_2,\ldots,\lambda_n\}$ and $p\in F[x]$ is a polynomial with coefficients in $F$, then
$$
\det\left(p(A)\right)=\prod_{i=1}^np(\lambda_i).
$$
Proof. By assumption, we have $\det(A-xI)=\prod_{i=1}^n(\lambda_i-x)$.
Let $p(x)=c\prod_{k=1}^r(x-c_k)$. Then
\begin{aligned}
\det(p(A))
&=\det\left(c\prod_{k=1}^r(A-c_kI)\right)\\
&=c^n\prod_{k=1}^r\det(A-c_kI)\\
&=c^n\prod_{k=1}^r\prod_{i=1}^n(c_k-\lambda_i)\\
&=\prod_{i=1}^n\left[c\prod_{k=1}^r(c_k-\lambda_i)\right]\\
&=\prod_{i=1}^np(\lambda_i).
\end{aligned}
Theorem 2. Let $A\in M_n(F)$ where $F$ is algebraically closed. If the spectrum of $A$ (as a multiset) is $\{\lambda_1,\lambda_2,\ldots,\lambda_n\}$ and $p\in F[x]$ is a polynomial with coefficients in $F$, then the spectrum of $p(A)$ is $\{p(\lambda_1),p(\lambda_2),\ldots,p(\lambda_n)\}$.
Proof. Let $y$ be an indeterminate independent of $x$ and $K$ be the algebraic closure of $F[y]$. Then $f(x)=y-p(x)$ is a polynomial with coefficients in $K$. By theorem 1,
$$
\det(yI-p(A))
=\det(f(A))
=\prod_{i=1}^nf(\lambda_i)
=\prod_{i=1}^n\left(y-p(\lambda_i)\right)
$$
over $K$. Therefore, as polynomials in $y$ over $F$, we have $\det(yI-p(A))=\prod_{i=1}^n\left(y-p(\lambda_i)\right)$ too. Hence $\{p(\lambda_1),p(\lambda_2),\ldots,p(\lambda_n)\}$ is the spectrum of $p(A)$.