Exercise 6.C.10

Proof. Suppose $U$ and $U^{\text{⊥}}$ are both invariant under $T$. Let $v\in V$. Then $v=u+w$ for some $u\in U$ and $w\in U^{\text{⊥}}$. Therefore, by parts (b) and (c) from 6.55, we have

Hence $P_{U}T=T{P}_U$.

Conversely, suppose $P_{U}T=T{P}_U$. Let $u\in U$. Then

Hence $U$ is invariant under $T$. Let $w\in U^{\text{⊥}}$. Then

where the last equality follows from Exercise 5. Therefore $U^{\text{⊥}}$ is invariant under $T$. □