Normal operators with real eigenvalues are self-adjoint

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
1 reply · 4K views
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.