# Orthogonal Projections Proof

• December 12th 2013, 08:36 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?
• December 12th 2013, 09:36 PM
romsek
Re: Orthogonal Projections Proof
$\|(x-p)+y)\|^2= <(x-p)+y,(x-p)+y>\text{ } =$

$ -2+ \text{ }=$

$\|x-p\|^2+\|y\|^2-2$

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

$=0$
• December 12th 2013, 11:12 PM
romsek
Re: Orthogonal Projections Proof
[deleted]
• December 12th 2013, 11:18 PM
romsek
Re: Orthogonal Projections Proof
Quote:

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

$ -2+ \text{ }=$

$\|x-p\|^2+\|y\|^2-2$

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

$=0$

bah this is all wrong. It should be

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

$ -2+ \text{ }=$

$\|x-p\|^2+\|y\|^2-2$

and the same result.