Exercise 8.B.4

Proof. By the same reasoning used in the proof of 8.4, it follows that $\dim <!--\; FUNCTION\; APPLICATION\; -->\mathrm{null}<!--\; FUNCTION\; APPLICATION\; -->T^{n-1}\ge n-1$. But $\mathrm{null}<!--\; FUNCTION\; APPLICATION\; -->T^{n-1}\subset \mathrm{null}<!--\; FUNCTION\; APPLICATION\; -->T^n=G(0,T)$ (see 8.2 and 8.11). Thus $\dim <!--\; FUNCTION\; APPLICATION\; -->G(0,T)\ge n-1$ and $0$ is an eigenvalue of $T$. If $\dim <!--\; FUNCTION\; APPLICATION\; -->G(0,T)=n$, 8.26 shows that $0$ is the only eigenvalue of $T$. If $\dim <!--\; FUNCTION\; APPLICATION\; -->G(0,T)=n-1$, there is only space for one more eigenvalue with multiplicty $1$. □