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Exchange matrix and positive definiteness

  1. Aug 29, 2008 #1
    1. The problem statement, all variables and given/known data

    Let E be the exchange matrix (ones on the anti-diagonal, zeroes elsewhere). Suppose A is symmetric and positive definite. Show that B = EAE is positive definite.

    2. Relevant equations

    3. The attempt at a solution

    I've tried showing directly that for any conformable vector h, h'Bh > 0 whenever h'Ah > 0. This looks like a dead end. I suspect the easiest way to get the result is to show all the eigenvalues of B are positive, using the fact that all the eigenvalues of A are positive. However, I don't know how to show this.
  2. jcsd
  3. Aug 29, 2008 #2


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    That's a good idea. Suppose k is an eigenvalue of B. Then EAEx=kx for some nonzero x. What happens if you multiply both sides by E?

    Also note that for this to be a valid proof, you have to first show that B is symmetric, but this is trivial.
  4. Aug 29, 2008 #3
    Beautiful. Thanks for your help.
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