Exchange matrix and positive definiteness

randommacuser
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Homework Statement



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.

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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.
 
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randommacuser said:
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.
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.
 
Beautiful. Thanks for your help.
 
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