Exercise 7.A.11

Proof. Note that $P^2=P$ implies that $\mathit{Pu}=u$ for $u\in \mathrm{range}<!--\; FUNCTION\; APPLICATION\; -->P$.

Suppose $P=P_U$ for some subspace $U$ of $V$. Let $v_1,v_2\in V$. We have $v_1=u_1+w_1$ and $v_2=u_2+w_2$ for some $u_1,u_2\in U$ and $w_1,w_2\in U^{\text{⊥}}$. Then

Therefore $P=P^{\text{∗}}$.

Conversely, suppose $P$ is self-adjoint. Then, by 7.7 (a), ${(\mathrm{range}<!--\; FUNCTION\; APPLICATION\; -->P)}^{\text{⊥}}=\mathrm{null}<!--\; FUNCTION\; APPLICATION\; -->P^{\text{∗}}=\mathrm{null}<!--\; FUNCTION\; APPLICATION\; -->P$, and so, by 6.47, we have $V=\mathrm{range}<!--\; FUNCTION\; APPLICATION\; -->P\oplus \mathrm{null}<!--\; FUNCTION\; APPLICATION\; -->P$. Let $U=\mathrm{range}<!--\; FUNCTION\; APPLICATION\; -->P$ and $v\in V$. We can write $v=u+w$ for some $u\in \mathrm{range}<!--\; FUNCTION\; APPLICATION\; -->P$ (which equals $U$) and $w\in \mathrm{null}<!--\; FUNCTION\; APPLICATION\; -->P$ (which equals $U^{\text{⊥}}$). Therefore

Hence $P=P_U$. □