Normal operators with real eigenvalues are self-adjoint

Doom of Doom
Messages
85
Reaction score
0
Prove that a normal operator with real eigenvalues is self-adjoint


Seems like a simple proof, but I can't seem to get it.

My attempt: We know that a normal operator can be diagonalized, and has a complete orthonormal set of eigenvectors.

Let A be normal. Then A= UDU* for some diagonal matrix D and unitary U. Also, A*=U*D*U

Since D is the diagonal matrix of the eigen values of A, D is real, and thus D=D*.

Thus D=U*AU = UA*U*.

Then, I just get stuck on A=U²A*(U*)².
 
Physics news on Phys.org
You seem to be using the substitution
(ST)* --> S*T*,​
but the left hand side is not equal to the righthand side; making this substitution is not a valid deduction.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top