Exercise 8.B.8

Proof. If $0$ is not an eigenvalue of $T$, then $T^j$ is injective and surjective for all integers $j$, which gives the desired result (take $j=n-2$).

Suppose $0$ is an eigenvalue of $T$. Since $3$ and $8$ are also eigenvalues of $T$, by 8.26 the multiplicity of $0$, namely $\dim <!--\; FUNCTION\; APPLICATION\; -->G(0,T)$ which equals $\dim <!--\; FUNCTION\; APPLICATION\; -->\mathrm{null}<!--\; FUNCTION\; APPLICATION\; -->T^n$, is at most $n-2$. By the same reasoning used in the proof of 8.4, we have $\mathrm{null}<!--\; FUNCTION\; APPLICATION\; -->T^{n-2}=\mathrm{null}<!--\; FUNCTION\; APPLICATION\; -->T^n$ (because the $\mathrm{null}<!--\; FUNCTION\; APPLICATION\; -->T^{n-2}\subset \mathrm{null}<!--\; FUNCTION\; APPLICATION\; -->T^n$). Exercise 19 of section 8A implies that $\mathrm{range}<!--\; FUNCTION\; APPLICATION\; -->T^{n-2}=\mathrm{range}<!--\; FUNCTION\; APPLICATION\; -->T$. Now 8.5 completes the proof. □