# Orthogonal Projections

1. Sep 15, 2009

### buzzmath

1. The problem statement, all variables and given/known data
Let P be a projection. The definition used is P is a projection if P = PP. Show that ||P|| >=1 with equality if and only if P is orthogonal.

Let ||.|| be the 2-norm

2. Relevant equations
P = PP. P is orthogonal if and only if P =P*

3. The attempt at a solution

I've proved the first part of ||P|| >= 1 and the first part of the equality portion. Assume P is orthogonal prove ||P|| = 1. However, I'm having a lot of trouble with the second part: Assume ||P|| = 1 show P is orthogonal.

Can someone point me in the right direction or suggest any ideas on how to use ||P|| = 1 to show that P = P*?

I've been playing around with inner products to try to solve this part but haven't gotten anything good out of it that uses the fact that ||P|| = 1

2. Sep 15, 2009

### buzzmath

I know that the square roots of the nonzero eigenvalues of P*P or PP* are the singular values of P and thus also P*. The ||P|| = max{singular value} thus P and P* have the same 2-norm. ||P||=||P*||=1.
I write <(PP*-P*P)x,x> = <P*x,P*x>-<Px,Px>=||P*x||^2-||Px||^2
||P|| = sup (||Px||) where ||x|| = 1.
Does this necessarily mean that ||Px||=||P*x||? I'm thinking the max values of both norms are equal but does that mean that they are equal for all other x values?

If that is true then <(PP*-P*P)x,x> = 0 thus PP*=P*P so P is normal

and a normal matrix is self adjoint so this would complete the last part of the proof.

Does this make sense? I'm worried because I don't really use the fact that ||P|| = 1

I was thinking that maybe since the norm is 1 that could maybe imply ||P*x||=||PX|| and then I would follow as above.

I'm not sure if this is the right direction to go in or if I'm way off.

Thanks for any help