Positive definiteness of difference of inverse matrices

Question

Let $A$ and $B$ be two $n \times n$ symmetric and positive definite matrices.

If $A \prec B$, then is it true that $B^{-1} \prec A^{-1}$?

Here, $A \prec B$ means that $B-A$ is positive definite.

Answer 1

HINT:

Try proving that if $A$ is positive definite and $B-A$ is non-negative definite, then $\det(B-\lambda A)=0$ has all its roots $\lambda\ge 1$ and conversely if $\lambda\ge1$, then $B-A$ is non-negative definite. (it's an exercise from a book by the way)

Then, $\det(A^{-1}-\mu B^{-1})=0$

$\Rightarrow \det B\det(A^{-1}-\mu B^{-1})\det A\ge0$ $\quad($ as $A$ and $B$ are both p.d., $\det A,\det B>0)$

$\Rightarrow \det(B-\mu A)=0$

So the roots of $\det(A^{-1}-\mu B^{-1})=0$ are the same as the roots of $\det(B-\mu A)=0$.

As $B-A$ is n.n.d. we have $\mu\ge1$, which in turn makes $A^{-1}-B^{-1}$ n.n.d. (by the claim above)