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} $?
 A: Consider $B-A\geq 0$ using Schur complement, this is equivalent to
$$\begin{bmatrix}B&I\\I&A^{-1}\end{bmatrix}\geq 0, \quad B>0$$
Since $A^{-1}>0$, 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 $A^{-1}\geq B^{-1}$.
A: I will show that (for $A > 0$ and $B > 0$) we have that $A\le B$ is equivalent to $\sigma(A^{-1}B)\subset [1,\infty)$, where $\sigma(T)$ denotes the spectrum of a matrix $T$ (i.e., the set of eigenvalues). Then the claim immediately follows since $\sigma(BA^{-1}) = \sigma(A^{-1}BA^{-1}A) = \sigma(A^{-1}B)$.
For the equivalence, note that $T:=A^{-1}B - I$ is selfadjoint with respect to the (positive definite) inner product $\langle A\cdot,\cdot\rangle$. Now, $A\le B$ is equivalent to $T$ being positive semi-definite with respect to this inner product. But this is itself equivalent to $\sigma(T)\subset [0,\infty)$, i.e., $\sigma(A^{-1}B) = \sigma(T+I)\subset [1,\infty)$.
A: 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
 $$B^{-1} ≤ A^{-1}$$
