Exercise 5.B.9

Proof. Let $c(z-{\lambda }_n)\cdots (z-{\lambda }_1)$ be a factorization of $p$. Then $c(T-{\lambda }_{n}I)\cdots (T-{\lambda }_{1}I)$ is a factorization of $p(T)$. Since $p$ is the polynomial of smallest degree such that $p(T)v=0$, it follows that $(T-{\lambda }_{j}I)\cdots (T-{\lambda }_{1}I)v\ne 0$, for $j<n$. Therefore, we have that $(T-{\lambda }_{n-1}I)\cdots (T-{\lambda }_{1}I)v\ne 0$ is an eigenvector of $T$ and ${\lambda }_n$ the corresponding eigenvalue. Note that, by 5.20, the order of factorization can be changed, placing any other factor $(T-{\lambda }_j)$ in the beginning. This implies that all $\lambda$’s are indeed eigenvalues of $T$. □