Janessa Benson

2022-10-06

Since $\text{}P(X)\text{}\mathrm{\perp}\text{}\text{}x-P(X)\text{}$ , we have $\text{}||x|{|}^{2}=||p(x)|{|}^{2}+||x-P(x)|{|}^{2}\text{}$ . thus $||x|{|}^{2}$=⟨x,x⟩ =$||P(x)|{|}^{2}$ +||x-$P(x)|{|}^{2}$

Therefore, $||P(x)|{|}^{2}\le ||x|{|}^{2}.since||P(x)||\ge 0$ and $||x||\ge 0,wehave||P(x)||\le ||x||$

We jump to the conclusion of $||P(x)|{|}^{2}\le ||x|{|}^{2}.$ because of $||P(x)||\ge 0$ and $||x||\ge 0.$

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

Therefore, $||P(x)|{|}^{2}\le ||x|{|}^{2}.since||P(x)||\ge 0$ and $||x||\ge 0,wehave||P(x)||\le ||x||$

We jump to the conclusion of $||P(x)|{|}^{2}\le ||x|{|}^{2}.$ because of $||P(x)||\ge 0$ and $||x||\ge 0.$

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

Frederick Espinoza

Beginner2022-10-07Added 7 answers

If $x\perp y$ then

$\Vert x+y{\Vert}^{2}=\u27e8(x+y),(x+y)\u27e9$

$=\u27e8x,x\u27e9+\u27e8x,y\u27e9+\u27e8y,x\u27e9+\u27e8y,y\u27e9=\Vert x{\Vert}^{2}+0+0+\Vert y{\Vert}^{2}.$

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

$\Vert x+y{\Vert}^{2}=\u27e8(x+y),(x+y)\u27e9$

$=\u27e8x,x\u27e9+\u27e8x,y\u27e9+\u27e8y,x\u27e9+\u27e8y,y\u27e9=\Vert x{\Vert}^{2}+0+0+\Vert y{\Vert}^{2}.$

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

solvarmedw

Beginner2022-10-08Added 3 answers

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

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

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

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