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Identifying Self-Adjoint Operators

  1. Jan 14, 2010 #1
    1. The problem statement, all variables and given/known data

    If A has eigenvalues 0 and 1, corresponding to the eigenvectors (1,2) and (2, -1), how can one tell in advance that A is self-adjoint and real.

    2. Relevant equations


    3. The attempt at a solution

    I can show that A is real: it has real orthogonal eigenvectors and real eigenvalues which form a basis of R2, so any real vector is transformed by A to another vector in R2. It follows that all of the components of A are real.

    Showing that the matrix is self-adjoint, however, is trickier for me. I know that eigenvectors corresponding to distinct eigenvalues of a self-adjoint matrix are orthogonal, and clearly these eigenvectors are orthogonal. However, I don't think is is a sufficient condition. If a matrix has two orthogonal eigenvectors, surely that doesn't mean it is self-adjoint, right?

    So how can you know at a glance (that is, without reconstructing A by finding the unitary operators formed by taking its eigenvectors as columns), that A is self-adjoint?
  2. jcsd
  3. Jan 14, 2010 #2
    What can you say about the diagonal matrix of eigenvalues?
    What can you say about any conjugate of a self-adjoint matrix?
  4. Jan 14, 2010 #3


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    Perhaps it'd be easier if you noted that a self-adjoint matrix which is real, is simply a symmetric matrix.
  5. Jan 14, 2010 #4
    I'm not quite getting why these are relevant. The conjugate of a self-adjoint matrix A is just A transposed. Not sure what is so interesting about the diagonal matrix of eigenvalues...
  6. Jan 14, 2010 #5
    Write down the diagonal matrix of eigenvalues in this situation. Write down its adjoint. Can you see a relationship?
  7. Jan 15, 2010 #6


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    Think about how the original matrix A and the diagonal matrix of eigenvalues are related.

    Edit: Oh, I missed you had written this:
    Construct the matrix with the eigenvectors as columns. Do you notice anything about it?
  8. Jan 15, 2010 #7
    Bam!, I think I have it.

    A is diagonalizable, so...

    [tex]A = U^\ast \Lambda U[/tex]

    And so:

    [tex]A^\ast = U^\ast \Lambda ^\ast U[/tex]

    But the eigenvalues are all real, so big lambda is self-adjoint, and we can say that:

    [tex] A^\ast = U^\ast \Lambda U = A[/tex]

    I think that does it. Cool beans.
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