# Homework Help: Orthogonal Projections

1. Sep 10, 2009

### buzzmath

1. The problem statement, all variables and given/known data

P is mxm complex matrix, nonzero, and a projector (P^2=P). Show 2-norm ||P|| >= 1
with equality if and only if P is an orthogonal projector (P=P*)

2. Relevant equations
Let ||.|| be the 2-norm

3. The attempt at a solution

a. show ||P|| >= 1
let v be in the range of P. Pv = v. so ||Pv|| = ||v||. By definition of matrix norm ||Pv|| <= ||P||*||v||. thus ||P|| >= 1. otherwise, ||Pv|| > ||P||*||v|| which is a contradiction.

b. show equality if and only if P = P*
first assume P =P*
||Px|| ^2 = <Px,Px> = <x,P*Px>=<x,Px> <= ||Px|| * ||x|| by cauchy-schwarz inequality
so we have ||Px||^2 <= ||Px|| * ||x||
we know ||P|| = max ||Px|| where ||x|| = 1
so let ||x||=1
we have ||Px||^2 <= ||Px|| so dividing by ||Px|| givex
||Px|| <= 1 thus Thus 1<=||P||<=1 so ||P||=1

This is the part of the problem that I'm having trouble with
assume ||P|| = 1 show P = P*
<Px,Px> = 1
<Px,Px> = <Px,PPx> = <P*Px,Px> = 1
<Px,Px> - <P*Px,Px> = <Px-P*Px,Px> = 0
P-P*P = 0 so P = P*P
does this imply that P = P*?
is so why? I'm stuck at this point. Thanks for any help pointing me in the right direction

2. Sep 11, 2009

### gabbagabbahey

P also equals P2=PP, so P-P*P = 0 implies PP-P*P = (P-P*)P=0 and P is non-zero so ______

3. Sep 11, 2009

### buzzmath

So I understand now that (P-P*)P=0 implies P=P* since P is nonzero. I'm questioning my method for getting there now. The reason is I don't use the fact that ||P|| = 1
I only use the fact that <Px,Px>=<P*Px,Px>
For example
What if ||P|| = 2 then <Px,Px> = <Px,PPx> = <P*Px,Px> = 2 and
<Px,Px> - <P*Px,Px> = <Px-P*Px,Px> = 0 still getting me to the same place.
then I would have P = P* and from part 1 ||P|| = 1 a contradiction.
Does this make sense?
I'm not sure how I can use the ||P|| = 1

Last edited: Sep 11, 2009
4. Sep 11, 2009

### gabbagabbahey

Hmmm...yes, I just noticed that you go directly from $\langle Px-P^*Px\vert Px\rangle=0$ to $P-P^*P=0$....how do you justify that?

Also, your notation looks pretty suspect to me....is it supposed to be Dirac notation, or something else?

Last edited: Sep 11, 2009
5. Sep 11, 2009

### buzzmath

Yes, I see the problem now. I was thinking I could use ||.|| = 0 only is . = 0
but <Px-P*Px,Px> = 0 doesn't mean that the first part is zero.
I've been playing with this and can't find how to use the fact that ||P|| = 1
any pointers?
Thanks

6. Sep 11, 2009

### gabbagabbahey

Hmmm... how about using $$\langle(P-P^*P)x,Px\rangle=\langle(P-P^*P)x,(P-P^*P+P^*P)x\rangle=\langle(P-P^*P)x,(P-P^*P)x\rangle+\langle(P-P^*P)x,P^*Px\rangle$$

7. Sep 11, 2009

### buzzmath

so I have <(P-P*P)x,Px> = <(P-P*P)x,(P-P*P)x>+<(P-P*P)x,P*PX>
I know <Px,Px> = 1 = x*P*Px
now I write <(P-P*P)x,Px>=x*P*Px-x*PP*Px=1-x*PP*Px

and <(P-P*P)x,(P-P*P)x> = ||(P-P*P)x||

and <(P-P*P)x,P*PX> = x*P*P*Px-x*PP*P*Px = 1-x*PP*Px
Then <(P-P*P)x,Px> = 1-x*PP*Px = <(P-P*P)x,(P-P*P)x>+<(P-P*P)x,P*PX>
= ||(P-P*P)x|| + 1-x*PP*Px

subtract 1-x*PP*Px from each sde

then we have
||(P-P*P)x|| = 0 and thus P = P*P so (P-P*)P =0 P isn't 0 so thus P=P*

This seems right but I have one concern
If ||P|| = 2 for example then <(P-P*P)x,P*PX> = 2-x*PP*Px = ||(P-P*P)x|| + 2-x*PP*Px which gives me the same thing of P = P* but from the first part of the proof P = P* implies ||P|| = 1 a contradiction

8. Sep 12, 2009

### buzzmath

I'm still confused because we're not using the fact that ||P|| = 1 just the fact that
<(P-P*P)x,Px> = <(P-P*P)x,P*Px>
and that <(P-P*P)x,Px> =||(P-P*P)x|| - <(P-P*P)x,P*Px>
so ||(P-P*P)x|| = 0 and thus P-P*P = 0 thus P = P*
but we never use ||P|| = 1 so wouldn't this work for and finite value of ||P||?