is the trace of inverse of positive, positive definite matrix decreasing? Let $A, B$ be non-negative, and symmetric positive definite matrices.
If $A\le B$, i.e., all the entries of $B-A$ are non-negative, is it true that $\mbox{trace}(A^{-1}) \ge \mbox{trace}(B^{-1})$?
No. Let $A=I\le B=\pmatrix{1+\frac1t&amp;1\\ 1&amp;1}$ with $t&gt;0$. Then $B^{-1}=\pmatrix{t&amp;-t\\ -t&amp;1+t}$ and hence $\operatorname{tr}(A^{-1})=2&lt;1+2t=\operatorname{tr}(B^{-1})$ when $t$ is sufficiently large.
By $A \le B $, taking $a_{ij}\le b_{ij}$, 
we have $A=PDP^{-1}$ and $B=QMQ^{-1}$, $ Tr(A)=Tr(D)$ and $Tr(B)=Tr(D)$, $Tr(A^{-1})=Tr(D^{-1})$, so $Tr(D^{-1})$ is sum of reciprocals of the diagonal entries of D. If the diagonal elements of D are all greater than or equal to one then $Tr(D)\ge Tr(D^{-1})$

$A\le B$

$Tr(A) \le Tr(B)$

$Tr(A^{-1}) \ge Tr(B^{-1})$