# Normal Projector

1. Apr 10, 2010

### math8

How do you prove that a projector is normal if and only if it is self adjoint?

I know a matrix P is a projector if $$P=P^{2}$$ and P is normal if PP* = P*P and P is self adjoint (or hermitian) if P= P*.

I think I know how to prove that if the projector P is self adjoint then P is normal.

But I am not sure how to proceed to prove that if the projector P is normal, then it is Self adjoint.

2. Apr 11, 2010

### Staff: Mentor

I'm bumping this to see if anyone has any ideas on this I've put in close to a couple of hours but haven't made much headway.

We have a matrix P that is a projector (so P2 = P), and normal (so PP* = P*P), and need to show that P is Hermitian (hence P = P*).

3. Apr 11, 2010

### VeeEight

Perhaps we can try showing that under the hypothesis, it is an orthogonal projection and thus self adjoint. And also using the fact that if P is normal, then ker(P) = ker(P*). This will involve some messy orthogonal subspaces stuff, however.

Last edited: Apr 11, 2010
4. Apr 12, 2010

### math8

Thanks, I get the idea.

How do you prove that if if P is normal, then ker(P) = ker(P*) ? I know it involves the fact that Ker (P)= Ker (P*P), but how do we prove this?

5. Apr 12, 2010

### VeeEight

I think to show this, try showing that |Px| = |P*x|, using the fact that P*P - PP* = 0 (I do not know the exact method, however, off the top of my head).

6. Apr 12, 2010

### lanedance

as VeeEight suggest, how about if you have the complex inner product defined, then
$$|Px|^2 = <Px,Px> = (Px)^*Px= x^*P^*Px =...$$