TITLE: Is inverse matrix convex?
QUESTION [12 upvotes]: I wonder a generalization of Jensen's inequality:
let $\mathbf{X,Y}$ be two positive definite matrices, can we obtain the following Jensen like inequality
$$(1-\lambda)\mathbf{X}^{-1}+\lambda\mathbf{Y}^{-1} \succeq((1-\lambda)\mathbf{X}+\lambda\mathbf{Y})^{-1}$$
for any $\lambda \in (0,1)$, where the notation $\mathbf{A}\succeq \mathbf{B}$ represents matrix $(\mathbf{A-B})$ positive semi-definite.

REPLY [7 votes]: Let $\mathbf{P}=\mathbf{X}^{-1/2}\mathbf{Y}\mathbf{X}^{-1/2}$. By left- and right- multiplying both sides of the equation by $\mathbf{X}^{1/2}$, the inequality is equivalent to
$(1-\lambda)I+\lambda\mathbf{P}^{-1} \succeq((1-\lambda)I+\lambda\mathbf{P})^{-1}$. Since $\mathbf P$ is positive definite, it can be unitarily diagonalised and hence we may assume that it is a diagonal matrix. So, the inequality reduces down to the scalar case $(1-\lambda)+\lambda p_{ii}^{-1} \ge ((1-\lambda)+\lambda p_{ii})^{-1}$, which is true because the function $f(t)=\frac1t$ is convex for $t>0$.