Thread: Self-adjoint operator

Hi,
Is there someone who can help me with this?
(a) Prove that a normal operator on a complex inner product space with real eigenvalue is self-adjoint.
(b) Let T: V->V be a self-adjoint operator. Is it true that T must have a cube root? (A cube root of T is an operator S:V->V such that S^3=T).
Thank you so much!

Originally Posted by mivanova

(a) Prove that a normal operator on a complex inner product space with real eigenvalue is self-adjoint.

Your answer to this will depend on how much you know about normal operators. With luck, you will know that a normal operator T is unitarily diagonalisable. So there exist a unitary operator U and a diagonal operator D such that . What's more, the diagonal elements of D are the eigenvalues of T. If these are real then D will be self-adjoint ... .

Originally Posted by mivanova

(b) Let T: V->V be a self-adjoint operator. Is it true that T must have a cube root? (A cube root of T is an operator S:V->V such that S^3=T).

Again, use the fact that T can be diagonalised to reduce the problem to the case of a diagonal operator. (If you can find a cube root C for D, then will be a cube root for T.)

Hint for finding C: see if you can take C also to be diagonal.