Exercise 5.B.11

Proof. Suppose $\alpha$ is an eigenvalue of $p(T)$. Let $c(z-{\lambda }_1)\cdots (z-{\lambda }_n)$ be a factorization of $p(z)-\alpha$. We have

Because $p(T)-\mathit{\alpha I}$ is not injective, it follows that, for some $j$, $T-{\lambda }_{j}I$ is not injective. Therefore ${\lambda }_j$ is an eigenvalue of $T$. Since ${\lambda }_j$ is a root of $p(z)-\alpha$, we have that $p({\lambda }_j)=\alpha$.

The converse is the same as Exercise 10. □