Eigenvalues of a matrix satisfying a polynomial The theorem of Cayley-Hamilton says that a matrix satisfies it's characteristic polynomial. But can we also make a statement about the eigenvalues if a matrix satisfies a monic polynomial in general?
So let's say that we have a matrix $A \in \mathbb{C}^{m,m}$ and a monic polynomial $p$ of degree $n$ (with $n<m$) for which $p(A)=0$. What can be said about the eigenvalues of $A$ with this information?
 A: If $\Bbb  F$ is any field, and $A \in \Bbb F^{m, m}$, and $\lambda \in \Bbb F$ is an eigenvalue of $A$ so that
$Av = \lambda v, \tag{1}$
for $0 \ne v \in \Bbb F^m$, then 
$A^2 v = A(Av) = A(\lambda v) = \lambda Av = \lambda^2 v, \tag{2}$
and if
$A^kv = \lambda^k v, \tag{3}$
then
$A(A^k v) = A(\lambda^k v) = \lambda^k (Av) = \lambda^{k + 1} v, \tag{4}$
which, given (1) and (2) basically completes a very simple inductive proof that 
$A^n v = \lambda^n v \tag{5}$
for all positive $n \in \Bbb Z$.  Taking things one step further, we have
$a_n A^n v = a_n \lambda^n v \tag{6}$
for any $a_n \in \Bbb F$.  From (6) we deduce that if $p(x) \in \Bbb F[x]$ with, say,
$p(x) = \sum_0^N a_i x^i, \tag{7}$
then 
$p(A) v = \sum_0^N a_i A^i v  = \sum_0^N a_i \lambda^i v = p(\lambda) v, \tag{8}$
so that if $p(A)  = 0$, we conclude that $p(\lambda) v = 0$, whence, since $v \ne 0$, we have $p(\lambda)  = 0$.
Hope this helps.  Cheerio,
and as always,
Fiat Lux!!!
A: By applying the polynomial $p[A]$ in$~A$ term by term to an eigenvector of eigenvalue$~\lambda$, one sees that it acts on it as multiplication by $p[\lambda]$. But on the other hand $p[A]=0$, so one must have $p[\lambda]=0$, and $\lambda$ is a root of$~p$. Every eigenvalue of$~A$ must be a root of$~p$ (but not every root of$~p$ needs to be eigenvalue).
This is elementary, and much easier than Cayley-Hamilton.
