Orthogonal projection using Cauchy Schwarz Inequality

Question

For this example, let v, x, y and L be as above, and let z be any vector running parallel to L, i.e. z = av, for some number a. Show that x−y and y − z are orthogonal, regardless of the value of a. Show that:

$||x-z||^2 = ||x-y||^2+||y-z||^2$

and thus, $||x-y||\leqslant||x-z||$

How would I start this? I did the following, but I believe it's incorrect:

$||x−z||^2=(x−z)∗(x−z)$

$x∗(x−z)+z(x−z)$

$x∗x+x∗z+z∗x+x∗z$

$||x||^2+2(x−z)+||z||^2$

$||x||^2+||z||^2$

$||x||\leqslant\sqrt{||x||^2+||z||^2}=||x−z||$

Please let me know where I went wrong here.

Answer 1

We don't need Cauchy-Schwarz inequality to prove Pythagorean's theorem. In general, for any two mutually orthogonal vectors $u$ and $w$, we have \begin{aligned} \|u+w\|^2 &=\langle u+w,u+w\rangle\\ &=\langle u,u\rangle + \langle u,w\rangle + \langle w,u\rangle + \langle w,w\rangle\\ &=\langle u,u\rangle + \langle w,w\rangle\\ &=\|u\|^2+\|w\|^2. \end{aligned} In your case, the result follows by putting $u=x-y$ and $w=y-z$.

Answer 2

A more general prove using the orthogonal projection matrix $P$ and the Cauchy-Schwarz inequality.
We have $x$ is the original vector, $P$ is the orthogonal projection matrix and $I$ is the identity matrix. Then, $y = Px$ is the projected vector. Now, with $|<x,y>|^{2} \; \leqslant \;<x,x>\cdot<y,y>$, the Cauchy-Schwarz inequality to prove: $$<y,y> \; \leqslant \; <x,x>$$ By definition: $$<Px,(I-P)x> \; = 0$$ Then, $<y, (x-y)> \;= 0$, which is $<x, y> \;= \;<y, y> \;\geqslant 0$.
Now by the Cauchy-Schwarz inequality: $$|<y,y>|^{2} = |<x,y>|^{2} \; \leqslant \;<x,x>\cdot<y,y>$$ Since $<y, y> \;\geqslant 0$.
$$<y,y> \; \leqslant \; <x,x> $$ Q.E.D.