Let $A$ be an $n\times n$ positive-definite matrix. Let $0<\lambda \_1 \leq \lambda\_2 \leq \lambda \_3 \ldots \leq \lambda \_n$ be the eigenvalues of $A$. Let $n\geq k\geq 1$. Is the function $f(A) = \frac{1}{\lambda \_{1} \lambda \_{2} \ldots \lambda \_{k}}$ convex in $A$ ? 



Convexity of a function of matrices
It is sufficient to prove that $A\mapsto\lambda\_1\cdots\lambda\_k$ is concave. This is true and is the consequence of the stronger property:



> 
> $A\mapsto(\lambda\_1\cdots\lambda\_k)^{1/k}$ is concave. 
> 
> 
> 


The latter property follows from two facts:


* $\lambda\_1\cdots\lambda\_k=\min\_{\dim F=k}\det(A|\_F),$
* the map $B\mapsto(\det B)^{1/k}$ is concave over the cone of $k\times k$ positive definite symmetric matrices.


