Prove that if $p(A)=0$ where A is a matrix of a linear operator ($A \in L(V)$), $p(\lambda)=0$ if $\lambda \in \sigma(A)$ I think it's all in the title. $p$ is some random polynomial.
I don't know how to approach this one. I've tried taking the roots of $p$, placing them on the diagonal of a new matrix and reasoning that the linear operator of the new matrix has a common specific value (if it's called that in English) with A, but I have no idea how to prove that that specific value is also a root of $p$.
 A: Considering $\sigma(A)$ denotes the set of characteristic(eigen) values of A
As $\lambda$ is an eigen value, $\exists 0\ne v\in V$ such that $Av=\lambda v$ 
Note that $A^kv=\lambda^k v,\forall k\in N$(easy to prove using Induction)
Let $p(x)=\sum_{i=0}^{n}a_ix^i,a_i\in R$
Then we have ,
$p(A)=\sum_{i=0}^{n}a_iA^i$
So we have ,
$p(A)v=\sum_{i=0}^{n}a_iA^iv=\sum_{i=0}^{n}a_i\lambda^iv=p(\lambda)v$
Given $p(A)=0$
$\Rightarrow 0=p(A)v=p(\lambda)v$
As $v\ne 0\Rightarrow p(\lambda)=0$
A: Hint: if $v$ is an eigenvector for eigenvalue $\lambda$, what is $p(A) v$ for polynomials $p$?  
A: Another way to look at this problem is noticing that $p$ has to be a (polynomial) multiple of the minimal polynomial of $A$. Now what happens to the minimal polynomial if you apply it on an Eigenvalue of $A$? What happens consequently to $p$ ?
A: I'll give you an example ($p(x)=x^2+x+1$) and leave the general case to you.
Suppose that $\lambda\in\sigma(A)$ with corresponding eigenvector $u\neq0$. Then
$$
0=p(A)u=A^2u+Au+u=A(\lambda u)+\lambda u+u=
\\
\lambda Au+\lambda u+u=\lambda^2u+\lambda u+u=(\lambda^2+\lambda+1)u=p(\lambda)u.
$$
Since $u\neq0\Rightarrow p(\lambda)=0$.
