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Eigenvalue of product of matrices

  1. Jul 10, 2013 #1
    I have two real symmetric matrices [itex]A[/itex] and [itex]B[/itex] with the following additional properties. I would like to know how the eigenvalues of the product [itex]AB[/itex], is related to those of [itex]A[/itex] and [itex]B[/itex]? In particular what is [itex]\mathrm{trace}(AB)[/itex]?

    [itex]A[/itex] contains only 0s on its diagonal. Off diagonal terms are either 0 or 1.
    [itex]B[/itex] also contains only 0s on its diagonal. Its off diagonal terms are positive real numbers.

    If equalities don't exist, some bounds would also be helpful.

    Thanks.
     
  2. jcsd
  3. Jul 10, 2013 #2
    Let ##a_{i,j}## be the element of matrix A from row i and column j, and let ##b_{i,j}## be the element of matrix A from row i and column j. Then,
    $$\operatorname{tr}(\textbf{AB})=\sum_{i}\sum_{j}a_{j,i}b_{i,j}.$$
    Thus, from the fact that the non-diagonal terms of A are either 0 or 1, we obtain the bounds that
    $$0 \leq \operatorname{tr}(\textbf{AB}) \leq \sum_{i}\sum_{j}b_{i,j}.$$
     
  4. Jul 10, 2013 #3
    Thanks.
    I forgot to mention that A is sparse, i.e., most of its off diagonal terms are zero. So the above bound would be pretty bad.
     
  5. Jul 10, 2013 #4
    Do you have bounds on the sparsity of the matrix? I could probably do a little better with an idea of how dense the matrix is. Also, is there any idea as to the size of the matrices?

    I have nothing to do and I want something to work on. :tongue:
     
  6. Jul 10, 2013 #5
    A is an n x n matrix. m = constant * n of the off diagonal terms are 1. n is large.
     
  7. Jul 10, 2013 #6

    Office_Shredder

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    So you have a bound like the trace is smaller than
    [tex] m* max_{i,j}\left{ b_{ij} \right} [/tex]
     
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