How can I show that if $P \in L(V)$ is such that $P^2 =P$ and $\|Pv\| \leq \|v\|$ for every $v\in V$, then $P$ is an orthogonal projection? How can I show that if $P \in L(V)$ is such that $P^2 =P$ and $\|Pv\| \leq \|v\|$ for every $v\in V$, then $P$ is an orthogonal projection? 
The solutions I have references a previous exercise and hence I cannot do this problem in a self contained manner (the solutions works with range and null of $P$). Any tips or ideas would be greatly appreciated! Thanks!
 A: Note that $x=Px+(I-P)x$, and hence we have $x \in \ker P$ iff $ x \in {\cal R} (I-P)$.
In particular, we have $Px - x \in \ker P$ for all $x$.
If $\nu \in \ker P$, then $\|x+\nu \| \ge \|P(x+\nu)\| = \|Px\|$. Since $Px - x \in \ker P$, by letting $\nu = Px - x + \nu'$, we have
$\|Px+\nu' \| \ge \|Px\|$ for all $\nu' \in \ker P$.
Squaring and expanding gives
$\|\nu'\|^2+\|Px\|^2 +2 \operatorname{re} \langle \nu', Px \rangle \ge \|Px\|^2$, or
$\|\nu'\|^2 +2 \operatorname{re} \langle \nu', Px \rangle \ge 0$. Since $\alpha \nu' \in \ker P$ for any $\alpha>0$, we have (after dividing by $\alpha$)
$\alpha \|\nu'\|^2 +2 \operatorname{re} \langle \nu', Px \rangle \ge 0$. Letting $\alpha \downarrow 0$ gives $\operatorname{re} \langle \nu', Px \rangle \ge 0$. Since this inequality holds when $\nu'$ is multiplied by any scalar $\beta$, we conclude that $\langle \nu', Px \rangle = 0$ for all $\nu' \in \ker P$.
Since $\ker P = {\cal R} (I-P)$, we see that $\langle (I-P)y, Px \rangle = 0$ for all $x,y$, hence $(I-P)^* P = 0$, or $P = P^*P$ (equivalently, $\ker P \bot {\cal R} P$). Taking adjoints gives
$P^* = P^* P = P$, hence $P$ is self-adjoint and so is an orthogonal projection.
A: Lemma: Let $V$ be a finite dimensional inner product space and let $P \in \mathcal{L}(V)$ be such that $P^{2}=P,$ then $range(P)\oplus ker(P)=V.$
proof:Suppose $P \in \mathscr{L}(V)$ and $P^2 = P$. Prove that $V = \text{null}P \oplus \text{range} P$
Let U be $range(P).$ First, prove $ker(P)=U^{\perp}.$ Take $v\in ker(P),$ for every $u\in U,$ we have:
$$\|z+v\| \geq \|P(z+v)\| = \|Pz+0\|= \|z\|$$
Square the two sides, $$\|z\|^{2}+\|v\|^{2}+Re\langle z, v\rangle \geq \|z\|^{2}$$
$$\|v\|^{2}+Re\langle z, v\rangle\geq 0$$
Take $v^{\prime}= \alpha v,$ $if \alpha >0,$ $v^{\prime}\in ker(P).$ Thus we have,
$$Re\langle z, v\rangle\geq -\alpha\|v\|^{2},$$ for every $ \alpha > 0.$ Thus, $Re\langle z, v\rangle\geq 0.$ If $\alpha=-1,$ we have $-Re\langle z, v\rangle\geq 0,$ which implies $Re\langle z, v\rangle=0.$ Take $\alpha=i,$ $Im(\langle z,v^{\prime}\rangle)=Im(-i\langle z, v\rangle)=-Re(\langle z, v\rangle)=0.$ Thus, $\langle z, v\rangle=0,$ for every $v\in ker(P), u\in U.$ Thus, $ker(P)=U^{\perp}.$
From the lemma, we have for every $v\in V,$ $v$ can be represented uniquely by $u+w,$ while $u\in U, w\in ker(P)=U^{\perp}.$ $Pv= Pu+Pw,$ while $Pw=\mathbf{0}.$ Thus, we have $Pv=u, Pu=u,$ which is the definition of orthogonal projection onto $U.$
