# Normal operators with real eigenvalues are self-adjoint

1. Oct 11, 2009

### Doom of Doom

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.

2. Oct 11, 2009

### Hurkyl

Staff Emeritus
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.

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