Hey I'm stuck on this question, I'll be glad to get some help.
$A$ is a matrix, $f(x)$ is a polynomial such that $f(A)=0$. Show that every eigenvalue of $A$ is a root of $f$.
Well, I thought of something but I got stuck: we know that if $t$ is an eigenvalue of $A$, then $f(t)$ is an eigenvalue of $f(A)$, so letting $v$ be an eigenvector for $t$: $$f(A)=0\implies f(t)v=f(A)v=0\implies (v\ne 0)\implies f(t)=0$$ although I think that the last step is not true. Any help?
Thanks
Assuming $A$ is diagonalizable, you can write it as $A=PDP^{-1}$ and transcribe your equation into
$f(A)=P( \rm{diag\,}{f(\lambda)}) P^{-1}$
If $f(A)$ is to be zero, and $P$ is nonsingular, then the diagonal matrix on the right must be zero, which explicitly states that all the eigenvalues are roots of $f$.
However, this may assume too much. Depending on how rigorous a proof you need.