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Normal operators with real eigenvalues are self-adjoint

  1. Oct 11, 2009 #1
    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. jcsd
  3. Oct 11, 2009 #2


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    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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