Since P(X) _|_ x−P(X) , we have norm(x)^2=norm(p(x))^2+norm(x−P(x))^2. Can someone please explain why this is true?

Since , we have . thus $||x|{|}^{2}$=⟨x,x⟩ =$||P\left(x\right)|{|}^{2}$ +||x-$P\left(x\right)|{|}^{2}$
Therefore, $||P\left(x\right)|{|}^{2}\le ||x|{|}^{2}.since||P\left(x\right)||\ge 0$ and $||x||\ge 0,wehave||P\left(x\right)||\le ||x||$
We jump to the conclusion of $||P\left(x\right)|{|}^{2}\le ||x|{|}^{2}.$ because of $||P\left(x\right)||\ge 0$ and $||x||\ge 0.$
but where do we prove that $||P\left(x\right)||\ge 0$ and $||x||\ge 0$ in the first place?
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Frederick Espinoza
If $x\perp y$ then
$‖x+y{‖}^{2}=⟨\left(x+y\right),\left(x+y\right)⟩$
$=⟨x,x⟩+⟨x,y⟩+⟨y,x⟩+⟨y,y⟩=‖x{‖}^{2}+0+0+‖y{‖}^{2}.$
Replace x by $Px$ and y by x−$Px$ in this.
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solvarmedw
Let $u:=p\left(x\right)$ and $v:=x-p\left(x\right).$ Then u⊥v, therefore, by Pythagoras:
$||x|{|}^{2}=||u+v|{|}^{2}=||u|{|}^{2}+||v|{|}^{2}.$