1. ## Positive definite

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.

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 $A^{-1}\leqslant B^{-1}$.