TITLE: loewner ordering of symetric positive definite matrices and their inverse
QUESTION [5 upvotes]: $M_1$ and $M_2$ are symetric positive definite matrix and $M_2>M_1$ in Loewner ordering ($M_2-M_1$ is positive definite). does this imply that $M_1^{-1}>M_2^{-1}$?

REPLY [7 votes]: 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 $M_1^{-1}>M_2^{-1}$.