Eigenvalues of a square matrix are the roots of a polynomial.

Question

Let $p(t)$ be a polynomial, $A$ be an $n\times n$ matrix such that $p(A) = 0$. Then the eigenvalues of $A$ are all roots of $p(t)=0$, i.e., $p(\lambda_i) = 0$ for each eigenvalue of $A$.

I know I need to use the Jordan Normal form, but I'm unsure of the rest.

Answer 1

No need for Jordan's form. $Ax=\lambda x$, $x \neq 0$ implies $A^{k} x=\lambda ^{k} x$ for all $k \geq 0$. This gives $p(A) x=p(\lambda) x$. Sinec LHS is $0$ we get $p(\lambda) =0$.