Show that if P is an orthogonal projection matrix, then $||Px||\le||x||$ for every x. Use this inequality to prove the Cauchy Schwarz inequality.
I know that if P is an orthogonal projection matrix, then $P^2=P$ and $P^t=P$, but how do I utilize these properties to show $||Px||\le||x||$?
On
I am going to write about the first part of the problem only. It is an easy derivation from the definitions. Recall that if $P$ a is projection, the space, $X$, can be written as a direct sum of $U$ and $V$, where $V$ is the $\mathrm{Ker}\left(P\right)$ and $U$ is the image of $P$. Now $P$ being a perpendicular projection means that you can use the Pythagorean identity for an arbitrary $x$ and write
$$\| x\|^2= \| u+v\|^2=\| u\|^2+\| v\|^2 = \| Px\|^2 + \| v\|^2,$$
where $v \in V$ and $u \in U$. This completes the proof of the first part.
If $P$ is an orthogonal projection then $\Vert x\Vert^2=\Vert Px\Vert^2+\Vert (I-P)x\Vert^2$ for every $x$ by Pythagoras' theorem, so $\Vert x\Vert^2\geq \Vert Px\Vert^2$.
Now, let $u,v$ be two non-zero vectors. Then the orthogonal projection of $u$ into the linear span of $v$ is given by the formula $Pu=\langle u,\frac{v}{\Vert v\Vert}\rangle\, \frac{v}{\Vert v\Vert}\cdot$ Since $\Vert Pu\Vert \leq \Vert u\Vert$ and $\Vert Pu\Vert=\frac{1}{\Vert v\Vert}\, \vert \langle u,v\rangle\vert$ it follows that $\vert \langle u,v\rangle\vert\leq \Vert u\Vert\,\Vert v\Vert$. This inequality is of course also true if either $u$ or $v$ is equal to $0$.