Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Normal Projector

  1. Apr 10, 2010 #1
    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 [tex]P=P^{2}[/tex] 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. jcsd
  3. Apr 11, 2010 #2


    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*).
  4. Apr 11, 2010 #3
    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
  5. Apr 12, 2010 #4
    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?
  6. Apr 12, 2010 #5
    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).
  7. Apr 12, 2010 #6


    User Avatar
    Homework Helper

    as VeeEight suggest, how about if you have the complex inner product defined, then
    [tex] |Px|^2 = <Px,Px> = (Px)^*Px= x^*P^*Px =...[/tex]
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook