Convexity of a function of matrices

Question

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 $A$ ?

Answer 1

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.

Answer 2

The canonical result to answer this sort of question is Chandler Davis' Theorem (the linked to paper gives a reference to the original proof, which is one page long).