1. ## Positive definite

Question

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

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

What I have tried

I tried showing $\displaystyle B^{-1}-A^{-1}$ 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 $\displaystyle A-B>0$ , then $\displaystyle B^{-1}-A^{-1}>0$
The proof that I know relies on two facts. First, if $\displaystyle A\geqslant0$ and T is a matrix with Hermitian adjoint T*, then $\displaystyle T^*AT\geqslant0$. Second, if $\displaystyle T^*T\leqslant I$ then $\displaystyle TT^*\leqslant I$. I'll also need the fact that a positive matrix $\displaystyle A$ has a positive square root $\displaystyle A^{1/2}$.
Given positive definite (therefore invertible) matrices A and B with A – B > 0, it follows that $\displaystyle 0\leqslant A^{-1/2}(A-B)A^{-1/2} = I - A^{-1/2}BA^{-1/2}$. Therefore $\displaystyle (B^{1/2}A^{-1/2})^*(B^{1/2}A^{-1/2})\leqslant I$. Hence $\displaystyle B^{1/2}A^{-1}B^{-1}B^{1/2} = (B^{1/2}A^{-1/2})(B^{1/2}A^{-1/2})^*\leqslant I$, from which $\displaystyle A^{-1}\leqslant B^{-1}$.