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Math Help - Self-adjoint operator

  1. #1
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    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!
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  2. #2
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    Quote Originally Posted by mivanova View Post
    (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 T=U^*DU. What's more, the diagonal elements of D are the eigenvalues of T. If these are real then D will be self-adjoint ... .

    Quote Originally Posted by mivanova View Post
    (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 U^*CU will be a cube root for T.)

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