Exercise 7.3.13

Let $p(t)$ be the polynomial given in the question. And let $\beta$ be a Jordan basis for $T$. We have ${(T-{\lambda }_i)}^{p_i}(v)=0$ if $v$ is a generalized eigenvector with respect to the eigenvalue ${\lambda }_i$. So $p(T)(\beta )=\left\{0\right\}$. Hence the minimal polynomial $q(t)$ of $T$ must divide $p(t)$ and must be of the form

where $1\le r_i\le p_i$. If $r_i<p_i$ for some $i$, pick the end vector $u$ of the cycle of length $p_i$ in $\beta$ corresponding to the eigenvalue ${\lambda }_i$. This $u$ exist by the definition of $p_i$. Thus ${(T-{\lambda }_i)}^{r_i}(u)=w\ne 0$. Since $K_{{\lambda }_i}$ is $(T-{\lambda }_j)$-invariant and $T-{\lambda }_j$ is injective on $K_{{\lambda }_i}$ for all $j\ne i$ by Theorem 7.1, we know that $q(T)(u)\ne 0$. Hence $r_i$ must be $p_i$. And so $p(t)$ must be the minimal polynomial of $T$.