Let $V$ an inner product space, and $U$ a subspace of $V$. Prove $P^*_U = P_U$ (projection over U).

Question

Edit: I have made progress via the comments and now i wonder if my proof is Ok.

By theoreom we know $V$ = $U$ $\oplus$ $U^\bot$.

Suppose some $v$ $\in$ $V$ such that $v=u+u'$ for some $u \in U$ and some $u' \in U^\bot$.

By inner product we will have:

$<u,u>$ = $<v,P_U^*(u)>$ = $<u+u',P_U^*(u)>$ = $<u,P_U^*(u)>$ + $<u',P_U^*(u)>$

Since $u' \bot P_U^*(u)$ we have $<u,u>$ = $<u,P_U^*(u)>$,

thus resulting in $P_U(u) = u = P_U^*(u)$.

This implements $P_U = P_U^*$.

Does this current proof work?