# Orthogonal Projections Proof

• Dec 12th 2013, 07:36 PM
turbozz
Orthogonal Projections Proof
Let x Rn and let p be the orthogonal projection of x onto W where W is a subspace of Rn. Provethat for all y W ,
||x(p+y)||2 =||xp||2 +||y||2.

Now expanding out using the defn of norm (dot product) doesn't get me very far, any hints?
• Dec 12th 2013, 08:36 PM
romsek
Re: Orthogonal Projections Proof
$\displaystyle \|(x-p)+y)\|^2= <(x-p)+y,(x-p)+y>\text{ } =$

$\displaystyle <x-p,x-p> -2<x-p,y>+<y,y> \text{ }=$

$\displaystyle \|x-p\|^2+\|y\|^2-2<x-p,y>$

now y is in W, x-p is orthogonal to W so

$\displaystyle <x-p,y>=0$
• Dec 12th 2013, 10:12 PM
romsek
Re: Orthogonal Projections Proof
[deleted]
• Dec 12th 2013, 10:18 PM
romsek
Re: Orthogonal Projections Proof
Quote:

Originally Posted by romsek
$\displaystyle \|(x-p)+y)\|^2= <(x-p)+y,(x-p)+y>\text{ } =$

$\displaystyle <x-p,x-p> -2<x-p,y>+<y,y> \text{ }=$

$\displaystyle \|x-p\|^2+\|y\|^2-2<x-p,y>$

now y is in W, x-p is orthogonal to W so

$\displaystyle <x-p,y>=0$

bah this is all wrong. It should be

$\displaystyle \|(x-p)-y)\|^2= <(x-p)-y,(x-p)-y>\text{ } =$

$\displaystyle <x-p,x-p> -2<x-p,y>+<y,y> \text{ }=$

$\displaystyle \|x-p\|^2+\|y\|^2-2<x-p,y>$

and the same result.