TITLE: Matrix-convexity of inverse of the cofactor matrix
QUESTION [6 upvotes]: Consider the matrix-valued function $f(A) = \frac{A}{\det(A)}$ on the set of $3\times 3$ positive-definite matrices. Is this function matrix-convex ? (i.e., is $tf(A) + (1-t)f(B) - f(tA+(1-t)B)$ positive semi-definite $\forall \ t \in [0,1]$?)

REPLY [2 votes]: Well, this is not an answer. But I cannot resist to mention the following equivalent property. Let $A\mapsto \hat A$ denote the cofactor map, and $B\mapsto \check B$ its inverse. For positive definite symmetric matrices, $\hat A=(\det A)A^{-1}$ and $\check B=(\det B)^{\frac1{n-1}}B^{-1}$. Then the harmonic mean is less than or equal to the cofactor mean over the cone of positive definite matrices:
$$\langle A^{-1}\rangle^{-1}\le\widehat{\langle\hat A\rangle}.$$