Matrix inequalities - Positive matrices

Question

Given the inequality $A - B > 0$, we can consider that $A > B$ [1]. Is it also true that, for $A \, B > I$, $det(A) \neq 0$, then B > inv(A)?

[1] Bellman, R. (1997). Introduction to matrix analysis (Vol. 19). Siam.

Answer 1

An easy counterexample is $A = -I$, $B = -2I$.

EDIT: What is true is this. Suppose $A > 0$, $B$ is hermitian and $AB > I$. Then $A^{1/2} B A^{1/2} > I$, where $A^{1/2}$ is the unique positive definite square root of $A$, and so $B > A^{-1}$.

In fact, since $AB$ is typically not hermitian, the hypothesis can be weakened to: $A > 0$, $B$ is hermitian, and all eigenvalues of $AB$ are greater than $1$.

Answer 2

The answer is no. We might have, for example, $\det(A)<0$.

Here is an example (with A and B scalars!) Let $A=-1, B= -2.$

$A-B>0$ and $A>B$ and $AB>I$ but inv(A)>B !