Matrix inequality after taking inverse Let A and B be Positive definite Matrices with $ A\leq B$ in the sense that $B-A$ is positive definite. Is it true that $A^{-1} \geq B^{-1} $?
Consider $B-A\geq 0$ using Schur complement, this is equivalent to
$$\begin{bmatrix}B&amp;I\\I&amp;A^{-1}\end{bmatrix}\geq 0, \quad B&gt;0$$
Since $A^{-1}&gt;0$, now apply the Schur complement one more time to obtain
$$A^{-1}-I(B)^{-1}I=A^{-1}-B^{-1}\geq  0$$
therefore we have $A^{-1}\geq B^{-1}$.
It is true. If you have matrices with this conditions ,then you can start from 
$A≤B$. 

Now multiply with $A^{-1}$  on the left
$$A^{-1} A ≤ A^{-1} B$$
Now you have the identity Matrix on the left. Multiply with $B^{-1}$ on the right and you will get
 $$B^{-1} ≤ A^{-1}$$