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Trace Inequality Tr(ABAB)

  1. Apr 22, 2014 #1
    Here's the claim: Assume that [itex]A[/itex] and [itex]B[/itex] are both symmetric matrices of the same size. Also assume that at least other one of them does not have negative eigenvalues. Then

    [tex]
    \textrm{Tr}(ABAB)\geq 0
    [/tex]

    I don't know how to prove this!
     
  2. jcsd
  3. Apr 22, 2014 #2

    jbunniii

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    If ##\lambda## is an eigenvalue of ##AB## then ##\lambda^2## is an eigenvalue of ##ABAB##. Since the trace is the sum of the eigenvalues, it suffices to show that the eigenvalues of ##AB## are all real. This would be true if ##AB## were symmetric, but the product of two symmetric matrices need not be symmetric. But can you show that ##AB## is similar to a symmetric matrix?
     
  4. Apr 22, 2014 #3
    Never mind, I succeeded in proving this now. After months of wondering the proof appears in few minutes after posting to PF as usual :wink:

    My hint for this is that you should first prove that you can assume [itex]A[/itex] to be diagonal with non-negative entries.
     
  5. Apr 22, 2014 #4

    AlephZero

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  6. Apr 22, 2014 #5

    jbunniii

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    Only positive semidefinite in this case, so that proof needs to be modified (##K## may not be invertible). I scratched something together this morning but upon reviewing what I wrote, I see that it contains a mistake. Will take another look this evening if I get a chance.
     
  7. Apr 22, 2014 #6
    According to my formulas this is true too: If [itex]A[/itex] is positive definite, and [itex]B[/itex] is non-zero (and the other assumptions), then

    [tex]
    \textrm{Tr}(ABAB) > 0
    [/tex]

    It is not clear which would be the most desired form for this problem.
     
  8. Apr 22, 2014 #7

    AlephZero

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    My bad. I tripped over the meaning of
    Too many negatives!
     
  9. Apr 22, 2014 #8

    AlephZero

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    Clearly if ##A## can be semi-definite, it can be zero, and ##ABAB = 0##.

    The basic idea of my thinking is to show that ##AB## has a full set of real eigenvalues, by simultaneously diagonalizing ##A## and ##B##. The eigenvalues of ##(AB)(AB)## are then all positive.

    If the diagonalization fails because ##A## is singular, replace ##A## with the nonsingular matrix ##A + eI## where ##e > 0##, and take the limit as ##e \to 0##.
     
  10. Apr 22, 2014 #9

    jbunniii

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    I was thinking of doing something similar to how one can "compress" the singular value decomposition when some of the singular values are zero. Here's a sketch:

    Since ##A## is symmetric, it can be orthogonally diagonalized: ##A = QDQ^T## where ##D## is diagonal. Since ##A## is positive semidefinite, ##D## has nonnegative diagonal elements: namely, the eigenvalues of ##A##. We may remove any zero eigenvalues as follows. For each ##i## for which ##[D]_{i,i} = 0##, delete the ##i##'th row and column from ##D##. Call the result ##D_0##. Also for each such ##i##, delete the ##i##'th column from ##Q##. Call the result ##Q_0##.

    Thus if there is a zero eigenvalue with algebraic multiplicity ##k## and the dimension of ##A## is ##n\times n##, then ##D_0## is ##(n-k)\times(n-k)## and ##Q_0## is ##n \times (n-k)##.

    We have ##A = Q_0 D_0 Q_0^T## but now ##D_0## is invertible whereas ##D## was not.

    Caveat for any subsequent calculations: ##Q_0## is now rectangular, and we still have ##Q_0^T Q_0 = I## but ##Q_0 Q_0^T## will not be ##I##.

    I haven't checked whether this delivers the goods.
     
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