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Trace of a matrix and an expansion of eigenvectors

  1. Sep 16, 2009 #1
    I'm trying to derive the Kullback-Leibler divergence between two multi-variate gaussian distributions, and I need the following property. Is there a simple way to understand this?

    Prove that:
    Given that E has orthonormal eigenvectors [tex]u_{i}[/tex] and eigenvalues [tex]\lambda_{i}[/tex]

    [tex]trace(A*E) = \sum_{i}u^{T}_{i}*A*u_{i}*\lambda_{i}[/tex]

    I'm not quite sure how to start. I suspected that it can be proven by looking at block matrices but I didn't get anywhere with that. Thanks a lot for your help.
  2. jcsd
  3. Sep 17, 2009 #2


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    Hi Patrick! :smile:

    (have a sigma: ∑ and a lambda: λ and try using the X2 tag just above the Reply box :wink:)

    If the ui are orthonormal, then they can be used as a basis, and in that basis tr(AE) = ∑ λiAii

    Then transform back to a general basis, and you get the result given.

    (there's probably a more direct way of doing it, also! :wink:)
  4. Sep 17, 2009 #3
    Thanks Tiny_Tim. It took me a while to figure out the details but I finally got it. Your advice worked perfectly!

    PS: And thanks for the generous greek letters. =)
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