Does $A \geq B$ imply $I \geq BA^{-1}$?

Question

Using the Loewner order, if $A \geq B$ and $A$ is invertible, then is $I \geq BA^{-1}$?

Assume all matrix dimensions are $n \times n$.

Answer 1

Note that the Loewner order only applies to Hermitian matrices matrices (or matrices whose difference is Hermitian). Given that $A,B$ are Hermitian, $BA^{-1}$ will be Hermitian if and only if $A$ and $B$ commute.

On the other hand: if $A$ is positive definite, then we can indeed say that $$ A \geq B \implies \operatorname{Id} \geq A^{-1/2}BA^{-1/2}. $$