Exercise 5.B.3

Proof. Let $v\in V$. Then

which implies $(T-I)v\in \mathrm{null}<!--\; FUNCTION\; APPLICATION\; -->(T+I)$. We but need to prove that $T+I$ is injective and we will have $\mathit{Tv}=v$. Let $u,w\in V$ such that $(T+I)u=(T+I)w$. Then $\mathit{Tu}+u=\mathit{Tw}+w$ which implies $T(u-w)=w-u=-1(u-w)$. But $-1$ is not an eigenvalue of $T$, thus $u-w=0$, implying $u=w$, as desired. □