Let $W \subset V$ and $V = W \oplus W^{\perp}$. Denote by $w \in W$ and $w^{\perp} \in W^{\perp}$. Let $P: V \to V$ be a (linear) operator and satisfies $$P^{2} = P,$$ $$P|_{w} = id|_{w}, (i.e. P[w] = w),$$ $$\langle P[u],v\rangle = \langle u,P[v]\rangle.$$ Prove $P[w + w^{\perp}] = w$, i.e. $P$ is an orthogonal projection operator.
Proof. I have $$P[w + w^{\perp}]= P[w ] + P[w^{\perp}].$$ For any $w' \in W$, $$\langle P[w^{\perp}],w'\rangle = \langle w^{\perp},P[w']\rangle = \langle w^{\perp},w'\rangle = 0.$$ ...
I know I actually need to prove $P[w^{\perp}] = 0$, but I can only do this far. This is an exercise that I wrote down from my class (may not 100% accurate), but I really cannot solve it. Hope to get some help here. Thanks a lot!
Using matrix notation since we must have $Pw = w$ this implies that $Pv = P(w + w^\perp)= Pw + Pw^\perp = w + Pw^\perp$ together with $Pv = w$ gives us $w= w + Pw^\perp$ and therefore $Pw^\perp = 0$ as required.