Polynomial of Matrix is zero Matrix then Eigenvalues are root

Question

I'm struggling with the following problem:

Let $p$ be a polynomial, then if $p(A)=0$, where $0$ stands for the zero matrix, then the eigenvalues of $A$ are roots of the polynomial.

What I did/tried:

I attempted to calculate the polynomial of $Av$ where v is an eigenvector because then I guessed that the eigenvalues would appear, but I wasn't successful.

Answer 1

If $p(x)=\sum a_kx^k$ and $v$ is eigenvector of$A$ with eigenvalue $\lambda$, then $$p(A)v=(\sum a_kA^k)v=\sum a_kA^kv = \sum a_k\lambda^k v=(\sum a_k\lambda^k) v=p(\lambda)v.$$ As $p(A)=0$ and $v\ne 0$, we conclude $p(\lambda)=0$.

Answer 2

If $P(\lambda)=0$, for an eigenvalue, $\lambda$, then $P(A)=0$.

On the other hand $P(A)=0$ does not necessarily means that every root of $P(\lambda )$ is an eigenvalue of $A.$

You may multiply the characteristic polynomial of A by any polynomial, and get a polynomial $P^*(A)$ of higher degree which still satisfies $P^*(A)=0$ with some roots which are not eigenvalues of $A$