Since $~P(X) ~\bot~ x-P(X)~ $, we have $~||x||^2=||p(x)||^2+||x-P(x) ||^2~$

Question

"Since $~P(X) \ \bot ~\ x -P(X) ~$, we have $~||x||^2=||p(x)||^2+||x-P(x) ||^2~$ ". Can someone please explain why this is true?

Since $~P(X) \ \bot ~\ x -P(X) ~$, we have $~||x||^2=||p(x)||^2+||x-P(x) ||^2~$. thus $||x||^2$=$ \langle x,x \rangle $ =$||P(x)||^2$ +||x-$P(x)||^2$.

Therefore, $||P(x)||^2 \leq ||x||^2. since ||P(x)||\geq 0$ and $||x||\geq 0, we have ||P(x)||\leq ||x||$.

We jump to the conclusion of $||P(x)||^2 \leq ||x||^2.$ because of $ ||P(x)||\geq 0$ and $||x||\geq 0. $.

but where do we prove that $||P(x)||\geq 0$ and $||x||\geq 0$ in the first place?

Answer 1

If $x \perp y$ then $$\|x+y\|^{2}=\langle (x+y) , (x+y) \rangle$$ $$ =\langle x , x \rangle +\langle x , y \rangle+\langle y , x \rangle+\langle y , y \rangle=\|x\|^{2}+0+0+\|y\|^{2}.$$

Replace $x$ by $Px$ and $y$ by $x-Px$ in this.

Answer 2

Let $u:=p(x)$ and $v: =x-p(x).$ Then $u \perp v$, therefore, by Pythagoras:

$$||x||^2=||u+v||^2=||u||^2+||v||^2.$$