# Math Help - Normal projectors, Hermitian projectors

1. ## Normal projectors, Hermitian projectors

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. Originally Posted by 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 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.
Hi, math8. All your questions show physicsforums.com logo. You need to check it on your school computer rather than your home computer. A stupid phisicsforum.com is asking copyrights for a student's textbook exercise.

Anyhow, it is not a good idea to post the exact same questions on different forums unless you get any response from either of them.

3. The missing formula is P=P^2.

I think as long as your browser doesn't have a referer, physicsforum will serve you the right image. So you can see them by copying the image location and opening it in a new tab. Once they're cached they will show up correctly, which might be why you're not noticing math8.

4. Originally Posted by 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 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.
If P is normal then P*P is a projection (because then $(P^*P)^2 = P^*PP^*P = P^{*2}P^2 = P^*P$). The kernel of $P^*P$ is the same as the kernel of P, and since P is normal this is also the kernel of P*. The range of P is the orthogonal complement of the kernel of P*. So P is a projection whose kernel and range are complementary subspaces. That means that P is an orthogonal projection, and therefore selfadjoint.