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Math Help - Positive definite

  1. #1
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    Positive definite

    Hi guys, can you please help me out??

    Question

    A and B are nonsingular matrices and positive definite. Show that:

    If A-B>0 , then B^{-1}-A^{-1}>0


    What I have tried

    I tried showing B^{-1}-A^{-1} is a quadratic form but firstly I'm guessing that doesn't even show it's positive definite, only positive semi-definite. And secondly, I couldn't get it to work anyway.

    Please help!!

    Thanks!
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  2. #2
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    Quote Originally Posted by WWTL@WHL View Post
    A and B are nonsingular matrices and positive definite. Show that:

    If A-B>0 , then B^{-1}-A^{-1}>0
    The proof that I know relies on two facts. First, if A\geqslant0 and T is a matrix with Hermitian adjoint T*, then T^*AT\geqslant0. Second, if T^*T\leqslant I then TT^*\leqslant I. I'll also need the fact that a positive matrix A has a positive square root A^{1/2}.

    Given positive definite (therefore invertible) matrices A and B with A B > 0, it follows that 0\leqslant A^{-1/2}(A-B)A^{-1/2} = I - A^{-1/2}BA^{-1/2}. Therefore (B^{1/2}A^{-1/2})^*(B^{1/2}A^{-1/2})\leqslant I. Hence B^{1/2}A^{-1}B^{-1}B^{1/2} = (B^{1/2}A^{-1/2})(B^{1/2}A^{-1/2})^*\leqslant I, from which A^{-1}\leqslant B^{-1}.
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