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?

Janessa Benson 2022-10-06 Answered
Since   P ( X )       x P ( X )   , we have   | | x | | 2 = | | p ( x ) | | 2 + | | x P ( x ) | | 2   . thus | | x | | 2 =⟨x,x⟩ = | | P ( x ) | | 2 +||x- P ( x ) | | 2
Therefore, | | P ( x ) | | 2 | | x | | 2 . s i n c e | | P ( x ) | | 0 and | | x | | 0 , w e h a v e | | P ( x ) | | | | x | |
We jump to the conclusion of | | P ( x ) | | 2 | | x | | 2 . because of | | P ( x ) | | 0 and | | x | | 0.
but where do we prove that | | P ( x ) | | 0 and | | x | | 0 in the first place?
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Answers (2)

Frederick Espinoza
Answered 2022-10-07 Author has 7 answers
If x y then
x + y 2 = ( x + y ) , ( x + y )
= x , x + x , y + y , x + y , y = x 2 + 0 + 0 + y 2 .
Replace x by P x and y by x− P x in this.
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solvarmedw
Answered 2022-10-08 Author has 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 .
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