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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.

Qiang Li
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    Even if Srivatsan's argument is very economic, I think the "right" way to prove this is the Pythagorean theorem: $Px \perp (1-P)x$ (direct computation of the scalar product gives $\langle Px, (1-P)x \rangle = \langle x, P^T(1-P)x \rangle = 0$)and $x = Px + (1-P)x$, so $|x|^2 = |Px|^2 + |(1-P)x|^2$ shows that $|Px| \leq |x|$ and equality holds if and only if $(1-P)x = 0$, that is $x = Px$, so $|P| = 1$ whenever $P \neq 0$. – t.b. Dec 23 '11 at 20:04
  • @t.b.: Yes, I like this approach better. :) – Qiang Li Dec 23 '11 at 20:29

1 Answers1

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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 $\|P \| \geqslant 1$.

Srivatsan
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