Exercise 8.C.17

Proof. The characteristic polynomial of $T$, call it $q$, has degree $\dim <!--\; FUNCTION\; APPLICATION\; -->V$ (see 8.36) and is a polynomial multiple of the minimal polynomial of $T$ (see 8.48), call it $p$. Because $q$ is a multiple of $q$, it follows that $q=\mathit{ps}$ for some polynomial $s$ and, because $\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->q=\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->p$, it follows that $\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->s=0$. Hence $s$ is a constant. Since they’re both monic, $s=1$ and they should be equal. □