Loewner ordering of symetric positive definite matrices and their inverse

Question

Let $M_1$ and $M_2$ be symmetric positive definite matrices and $M_2 > M_1$ in the Loewner ordering, i.e., $M_2 - M_1$ is positive definite. Does this imply that $M_1^{-1} > M_2^{-1}$?

Answer 1

The answer is yes. Two facts first:

(1) The statement $M_2>M_1$ is equivalent to $x^TM_2x>x^TM_1x$ for any $x\neq 0$;

(2) For any symmetric positive definite matrix $M$, there exist a positive definition matrix $L$ such that $M=L^2$ (called the square root of $M$).

We can show it is true when $M_1$ is the identity matrix $I$: for $M_2=L_2^2$, $$ x^TM_2^{-1}x=x^TL_2^{-T}L_2^{-1}x=(L_2^{-1}x)^T(L_2^{-1}x) \leq (L_2^{-T}x)^TM_2(L_2^{-T}x)=x^Tx. $$

In the general case for $M_1=L_1^2$, the condition $M_2>M_1$ is equivalent to $L_1^{-1}M_2L_1^{-1}>I$, which implies that $ I>(L_1^{-1}M_2L_1^{-1})^{-1}=L_1M_2^{-1}L_1 $ or $M_1^{-1}>M_2^{-1}$.