Solution
= Solution
If $Av=\lambda v$, then $p(A)v=p(\lambda)v$, proving one inclusion. Conversely, if $\mu$ is an <eigenvalue> of $p(A)$, factor $p(z)-\mu=c\prod_j(z-\lambda_j)$. If none of the $\lambda_j$ were <eigenvalues> of $A$, every $A-\lambda_jI$ would be invertible, making $p(A)-\mu I$ invertible, a contradiction. Thus $\mu=p(\lambda)$ for some <eigenvalue> $\lambda$ of $A$.