Let $(X, \langle \cdot, \cdot\rangle )$ be an inner-product space. Consider $V\subset X$ a subspace with $P_V(x)$ theh orthogonal projection of $x$ on $V$.
Then:
a. $x-P_V(x)$ is orthogonal projection of $x$ on subspace $V^{\perp}$.
b. if $P_V(P_V(x))=x$ then $x\in V$
c. if $v\in V$ and $||v-x|| \le ||P_v(x)-x||$ then $v=P_V(x)$.
My approach
a. We know that $x=v+w$, where $w\in V^{\perp}$ and $P_V(x)=v$.
So $x-P_v(x) = w$. We can see that $v$ is orthogonal projection of $x$ on $V$, and analogously $w$ is orthogonal projection of $x$ on $V^{\perp}$. Hence, a. is true.
b. I know that if $x\in V$ then $P_V(x)=x$, but I am not sure this one.
c. I know theorem:
$||x-P_Z(x)|| \le ||x-z||$ - moreover we have equality if only and only $z=P_Z(x)$.
Because of $||a-b||=||b-a||$ then using assumption in c. we have that $||v-x||=||P_V(x)-x||$. Therefore c. is true.
Can you check my reasoning a.,c. and help with solving b. please ?
For b, if $P_Vx=x$, then $$ x-P_V(x)=P_V(x)-P_V(x)=0. $$ So b follows from a.
Note that $P_VP_Vx=P_Vx$, because $P_V$ is a projection (it is the identity on its image).