In Convex Optimization, Boyd & Vandenberghe 3.6 "Convexity with respect to generalized inequalities", example 3.46 says that : $\textbf{tr}(X^{-1})$ is matrix decreasing on $S^n_{++}$.
In the book, page 109 and 110
Which means if $X,Y$ are positive definite matrices, and $Y-X$ is positive semidefinite matrix, if $Y\neq X$, then $\textbf{tr}(X^{-1}) > \textbf{tr}(Y^{-1})$.
How to prove this? Thanks!
Form Convex optimization we know that $f(X)=\textbf{tr}(X^{-1})$ is convex on $S^n_{++}$.
According to First-order condition: $f(X)\geq f(Y) + \nabla f(Y) \cdot (X-Y), \forall X,Y \in S_{++}^n$.
Which implies: $\textbf{tr}(X^{-1}) \geq \textbf{tr}(Y^{-1}) + \dfrac{\partial \textbf{tr}(Y^{-1})}{ \partial Y} \cdot (X-Y) = \textbf{tr}(Y^{-1}) + (-Y^{-2T}) \cdot (X-Y) = \textbf{tr}(Y^{-1}) + (Y^{-2T}) \cdot (Y-X) = \textbf{tr}(Y^{-1}) + \textbf{tr}(Y^{-2} \cdot (Y-X)) $.
$\textbf{tr}(Y^{-2} \cdot (Y-X)) \geq 0 $, since both $Y^{-2}$ and $(Y-X)$ are positive semidifinite matrices.
So $\textbf{tr}(X^{-1}) \geq \textbf{tr}(Y^{-1}) $, if $X,Y$ are positive definite and $Y-X$ is positive semidefinite.