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Eigen question

  1. May 23, 2007 #1
    1. The problem statement, all variables and given/known data

    For which real numbers c and d does the matrix have real eigenvalues and three orthogonal eigenvectors?

    120
    2dc
    053

    2. Relevant equations

    im having trouble getting started on this one.

    Ive tried using solving for the eigenvalues pretending that c and d are constants but that doesnt seem to help any. Can anyone nudge me in the correct direction.



    3. The attempt at a solution
     
  2. jcsd
  3. May 23, 2007 #2

    Dick

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    If the matrix has real eigenvalues and three orthogonal eigenvectors, then the corresponding linear transformation is self-adjoint. What does that mean in terms of its matrix representation?
     
    Last edited: May 23, 2007
  4. May 23, 2007 #3
    so... would I find the co-factor matrix of

    123
    2dc
    053

    take the transpose to find the adj.

    which I did actually.

    new values aprear in the d c spots.... it is its own adjoint so those would be the fill in values?

    I think im off track
     
  5. May 23, 2007 #4

    Dick

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    Not the adjugate, the adjoint. The conjugate transpose. You are making this much harder than it should be.
     
  6. May 24, 2007 #5
    I dont know why this one is so hard for me...
    does this have to do with the real spectral theorem?
     
  7. May 24, 2007 #6

    Dick

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    It's the converse of the spectral theorem. The spectral theorem says that if an operator has a certain property then its eigenspace has a certain property. This problem says if the eigenspace has a certain property then the operator has a certain property. If you write the linear transformation corresponding to the matrix as M, then self adjoint means (Mx).y=x.(My). Can you show thats true in this case? What might that have to do with a certain matrix being hermitian?
     
  8. May 24, 2007 #7
    I dont think that I've seen the operations the way that showing them...
    I do see something now though.

    Wont the matrix

    120
    2dc
    053


    have to be

    120
    2d5
    053

    It will have to symetric?

    I dont see how to find d besides by trying different d until the found eigenvectors are orthogonal. Any real number choice for d should give real eigenvalues by the spectral theorem?
     
  9. May 24, 2007 #8

    Dick

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    Correct. Any real choice for d will work.
     
  10. May 24, 2007 #9
    That easy?

    120
    205
    053

    120
    215
    053

    120
    225
    053

    120
    235
    053

    will all have orthogonal eigenvectors?

    I'll try a few to see if it works.
     
  11. May 24, 2007 #10

    Dick

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    You don't trust the spectral theorem? That's healthy skepticism! Check away.
     
  12. May 24, 2007 #11
    One more question... is it possible to choose d so that one of eigenvalues is repeated makeing it so there are not actually three othogonal eigenvectors. Because I want to say right now that any choice of d will work.
     
  13. May 24, 2007 #12
    I appreciate the help on this one... thanks'a bunch
     
  14. May 24, 2007 #13

    Dick

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    A repeated eigenvalue does not mean you don't have three orthogonal eigenvectors. The zero matrix has them. Take (1,0,0), (0,1,0) and (0,0,1). They are orthogonal and all have eigenvalue 0. You're welcome!
     
  15. May 24, 2007 #14
    nice.....

    even if it wasnt so, gram-schmidt could make them into orthonormal set I suppose.
     
  16. May 24, 2007 #15

    Dick

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    Yes. The problem comes with matrices like [[1,1],[0,1]]. 1 is a double eigenvalue - but there is only one linearly independent eigenvector.
     
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