1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    e=m^2

    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

    Matterwave

    User Avatar
    Science Advisor
    Gold Member

    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

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Identifying Self-Adjoint Operators
  1. Self Adjoint Operator (Replies: 9)

  2. Self-adjoint operator (Replies: 1)

Loading...