Concavity of $\det^{1/n}$ over $HPD_n$. - MathOverflow

Concavity of $\det^{1/n}$ over $HPD_n$.

2010-10-18T08:08:28Z

One of my beloved theorems in matrix analysis is the fact that the map $H\mapsto (\det H)^{1/n}$, defined over the convex cone $HPD_n$ of Hermitian positive definite matrices, is concave. This is accurate, if we think that this map is homogeneous of degree one, thus linear over rays.

it has important applications in many branches of mathematics,
it has many elegant proofs. I know at least three complety different ones.

I am interested to learn in both aspects. Which is your prefered proof of the concavity ? Is it useful in your own speciality ? In order to avoid influencing the answers, I decide not to give any example. But those who have visited my page may know my taste.

Answer by Deane Yang for Concavity of $\det^{1/n}$ over $HPD_n$.

2010-10-18T13:49:11Z

The concavity of $(\det A)^{1/n}$ for a positive definite symmetric matrix $A$, as well as its generalization known as the Brunn-Minkowski inequality, are absolutely fundamental and critical to differential and integral geometry, as well as geometric analysis (here, I mean functional inequalities like the Sobolev and Poincare inequalities). It is used, for example, in the proof of isoperimetric inequalities and something known as the Bishop-Gromov inequality on a Riemannian manifold.

The first proof I learned is simply differentiating $(\det A(t))^{1/n}$ twice, where $A(t) = A_0 + A_1t$.

Answer by Alekk for Concavity of $\det^{1/n}$ over $HPD_n$.

2010-10-19T09:33:12Z

An easy reduction shows that one can suppose that one of the matrices is the identity and the other diagonal: the inequality then reduced to the convexity of $f(x)=\ln(1+e^x)$.

Answer by Denis Serre for Concavity of $\det^{1/n}$ over $HPD_n$.

2013-04-04T07:03:33Z

Here is a interesting calculus proof. Let $f:A\mapsto(\det A)^{1/n}$, defined over $SPD_n$. Differentiating twice, we find the Hessian $${\rm D}^2f_A(X,X)=\frac1{n^2}f(A)\left(({\rm Tr} M)^2-n{\rm Tr}(M^2)\right),$$ where $M=A^{-1}X$. This matrix, being the product of two symmetric matrices with one of them positive definite, is diagonalisable with real eigenvalues $m_1,\ldots,m_n$. The parenthesis above is now $$\left(\sum_jm_j\right)^2-n\sum_jm_j^2,$$ a non-positive quantity, according to Cauchy-Schwarz. We infer that ${\rm D}^2f_A\le0$ and that $f$ is concave.