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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.

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    @t.b.: Yes, I like this approach better. :)2011-12-23

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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 $\|P \| \geqslant 1$.

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    Thanks Jonas and tb for your comments. For some reason, I had missed the other direction at first. :-)2011-12-23