Prove: Normal Projector iff Self Adjoint

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To prove that a projector P is normal if and only if it is self-adjoint, one must first establish that if P is self-adjoint, then it is normal. This is straightforward since self-adjointness implies that P = P*, leading to the condition PP* = P*P. The challenge lies in proving that if P is normal, then it must also be self-adjoint. Key ideas include examining the relationship between the kernels of P and P*, and utilizing properties of orthogonal projections. The discussion emphasizes the need for a deeper understanding of the implications of normality on the structure of the projector.
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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.
 
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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*).
 
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.
 
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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?
 
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).
 
as VeeEight suggest, how about if you have the complex inner product defined, then
|Px|^2 = <Px,Px> = (Px)^*Px= x^*P^*Px =...
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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