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Let $X$ be a Hilbert space and $P \in B(X)$ a projector. Then for any $x\in X$:

$$\langle Px,x\rangle=\|Px\|^2.$$

My proof:

$$\|Px\|^{2}=\langle Px,Px\rangle=\langle P^{*}Px,x\rangle=\langle P^2x,x\rangle=\langle Px,x\rangle.$$

Is ok ? Thanks :)

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Yes, seems ok. Is self-adjointness included in the definition of projector? – Berci Feb 7 '13 at 11:39
yes :) $P$ is a projector if $P^2=P$ and $P^{*}=P.$ – Iuli Feb 7 '13 at 11:41
up vote 2 down vote accepted

Yes, that is all. $$ \quad \quad $$

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