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Matrix/eigenvalue proof

  1. Nov 12, 2008 #1
    1. The problem statement, all variables and given/known data
    Given a matrix with eigenvalues [tex]\lambda_{i}[/tex], show that if the inverse of the matrix exists, its eigenvalues are [tex]\frac{1}{\lambda}[/tex].


    2. Relevant equations



    3. The attempt at a solution This shouldn't be so hard. I've come up with a few trivial examples, but I would like to get a general proof in 3 (or more) dimensions. I've tried solving for the eigenvalues, getting the eigenvectors and trying a unitary transformation. I've tried substituting the product of the matrix with its inverse in the characteristic equation and playing around with that before actually calculating the determionant. I've tried other symbolic brute force attempts, but they get very complicated very quickly. I've researched a bunch of determinant identities to no avail. I feel that there must be some key relation that I'm just missing. Any help would be appreciated.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Nov 12, 2008 #2
    Just write down the definition of eigenvalue and eigenvector, and then apply the inverse matrix to that equation. Nothing complicated involving determinants and/or unitary transformations are necessary.
     
  4. Nov 12, 2008 #3
    Is this the idea?
    [tex]$\begin{array}{l}
    \Lambda \left| \psi \right\rangle = \lambda \left| \psi \right\rangle \\
    \Lambda ^{ - 1} \left( {\Lambda \left| \psi \right\rangle } \right) = \Lambda ^{ - 1} \left( {\lambda \left| \psi \right\rangle } \right) \\
    \left( {\Lambda ^{ - 1} \Lambda } \right)\left| \psi \right\rangle = \left( {\Lambda ^{ - 1} \lambda } \right)\left| \psi \right\rangle \\
    I\left| \psi \right\rangle = \lambda \Lambda ^{ - 1} \left| \psi \right\rangle \\
    \left| \psi \right\rangle = \lambda \Lambda ^{ - 1} \left| \psi \right\rangle \\
    I = \lambda \Lambda ^{ - 1} \\
    \end{array}$[/tex]
     
  5. Nov 12, 2008 #4

    Avodyne

    User Avatar
    Science Advisor

    Fine except for the last line; you can't drop the state because it's not a generic state; it's an eigenstate of Lambda. You can, however, multiply both sides of the next-to-last line by 1/lambda.
     
  6. Nov 12, 2008 #5
    Except for the very last line, you're there!
     
  7. Nov 12, 2008 #6
    Yes, I see. Thanks very much!
     
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