I don't find a way to prove this: given $A$, $B$, symmetric and positive definite:
$$A>B \Rightarrow A^{-1} < B^{-1},$$ where $A>B$ means that $A-B$ is positive definite.
I don't find a way to prove this: given $A$, $B$, symmetric and positive definite:
$$A>B \Rightarrow A^{-1} < B^{-1},$$ where $A>B$ means that $A-B$ is positive definite.