Proof that $|| P_v(x) ||^2 = \left< x,P_v(x) \right>$ $(X,\left< \cdot,\cdot \right>)$ is euclides space, $V \subset X$ is linear subspace and $P_v(x)$ is orthogonal projection $x \in X$ to subspace $V$. 
Proof that
$$|| P_v(x) ||^2 = \left< x,P_v(x) \right>$$ 
My try
Thesis is equal to
$$\left< P_v(x),P_v(x) \right> = \left< x,P_v(x) \right>$$ 
But in my first opinion it doesn't seem to be true.

Let X - space of all lines in $ \mathbb R^2$. Let $$V = \left\{[c,0] : c\in \mathbb R \right\}$$
let $\vec{x} = [\sqrt{2},\sqrt{2}]$. Its projection to V will be $[\sqrt{2},0]$
 Let $\left< \cdot,\cdot \right>$ as standard scalar product. Then
$$4=\left< P_v(x),P_v(x) \right> = \left< x,P_v(x) \right>=2$$ 
so that is false. But probably there is mistake because thesis should be true...
 A: Since $P_v$ is projection, therefore $P_v^T=P_v$ and $P_v^2=P_v$. Thus,
\begin{align*}
\langle P_v(x),P_v(x) \rangle &= \langle x,P_v^TP_v(x) \rangle\\
&= \langle x,P^2_v(x) \rangle\\
&= \langle x,P_v(x) \rangle\\
\end{align*}
A: You are making a mistake in calculation. $\|P_vx\|^{2}=2$ in your
 example. 
To prove the result note that $P_v$ is self adjoint and idempotent . Hence $\langle P_v x, P_v x \rangle =\langle P_v^{2} x, x \rangle =\langle P_v x, x \rangle =\langle x, P_v x \rangle$
A: In your example we have $\left< P_v(x),P_v(x) \right>=2$ and not $=4.$
For the proof of the thesis I write $P$ instead of $P_v.$ Observe that $P=P^2=P^T$.
Then:
$$\left< P(x),P(x) \right>=\left< x,P^TP(x) \right>=\left< x,P^2(x) \right>=\left< x,P(x) \right>.$$
A: Just write


*

*$x = P_V(x) + (x- P_V(x))$ and note that $x- P_V(x) \perp P_V(x)$ since $P_V$ is an orthogonal projector.


It follows:
$$<x,P_V(x)> = <P_V(x) + (x- P_V(x)),P_V(x)>$$ $$ = <P_V(x) ,P_V(x)> + \underbrace{<x- P_V(x),P_V(x)>}_{= 0} = ||P_V(x)||^2 $$
