TITLE: Any easier and direct method to prove $\|P\|_2=1$ and $\|Px\|\le\|x\|$ for orthogonal projector $P$
QUESTION [5 upvotes]: I know complicated ways to prove them. But wonder if there is any easier and direct method to prove $\|P\|_2=1$ and $\|Px\|\le\|x\|$ for orthogonal projector $P$ just given the fact $P^T=P$ and $P^2=P$. 
Please help.
Thank you.

REPLY [3 votes]: Here's the upper bound:
$$
\| P x \|^2 
\ = \ x^\mathrm T P^{\, \mathrm T} P x 
\ = \  x^\mathrm T P^2 x
\ = \ \langle x, P x \rangle
\ \leqslant\ \| x \| \cdot \| Px \|, \tag{$\dagger$}
$$
by Cauchy-Schwarz. Hence $\| Px \| \leqslant \|x \|$ for all $x$, or $\| P \| \leqslant 1$.
For a matching lower bound, it suffices to demonstrate an $x \neq 0$ such that equality holds in $(\dagger)$. But equality is possible if and only if $Px = x$. With this in mind, assume that $P$ is nonzero and take $x = Pz$ for some $z$ such that $Pz \neq 0$. Then $Px = P^2 z = Pz = x \neq 0$. Therefore, $\| Px \| \geqslant \| x\|$, and hence $\|P \| \geqslant 1$.