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In an euclidean vector space $E$ how can I prove that a projection $p$ is orthogonal if and only if $$||p(x)||\leq ||x||\quad \forall x\in E$$

I can prove that if $p$ is an orthogonal projection then we have the inequality but I'm stuck in the other implication. Any suggestion is welcome.

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    Suppose $p$ isn't an orthogonal projection, i.e. $\mathrm{Im}(p)$ is not orthogonal to $\mathrm{Ker}(p)$. Chose $x$ in the image, and $y$ in the kernel, and draw $p$ in the plane $P=\mathrm{Vect}(x,y)$. It should become clear that for some vectors $v$ in $P$ you will have $||p(v)||>||v||$. Now to prove this, consider the function $t\mapsto ||p(v_t)||^2-||v_t||^2$ where $v_t=tx+y$. What can you say about it? – Olivier Bégassat May 12 '13 at 12:39

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A property of all projections is that $P^2=P$.

Let $v=Pu-u$. Then $$ Pv=P(Pu-u)=P^2u-Pu=0 $$ Next, by assumption $$ \begin{align} \|Pu\|^2 &=\|P(Pu+tv)\|^2\\ &\le\|Pu+tv\|^2\\ &=\|Pu\|^2+2t\langle v,Pu\rangle+t^2\|v\|^2 \end{align} $$ The only way that this can be true for all $t$, is if $\langle v,Pu\rangle=0$. That is, $$ \langle Pu-u,Pu\rangle=0 $$ Therefore, $P$ is orthogonal.

robjohn
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