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

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.

• I think that, as far as elementary solutions are concerned, it's hard to beat the proof in ex.219. Oct 18, 2010 at 10:42

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 $f(x)=\ln(1+e^x)$.
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 $A(t) = A_0 + A_1t$.
• The classical Brunn-Minkowski inequality implies the $L^2$ Brunn-Minkowski inequality (see Theorem 4 in sciencedirect.com/science/article/pii/S0196885811001126). The $L^2$ Brunn-Minkowski inequality for two ellipsoids centered at the origin is equivalent to the matrix determinant inequality. Oct 9, 2016 at 1:35
• @DeaneYang could you elaborate on how to derive the matrix inequality from the $L^2$ Brunn-Minkowski inequality? My naive thought is to apply the latter to $XB$ and $YB$, where $B$ is the unit ball and $X,Y\in HPD_n$. However I don't think $XB+_2YB=(X+Y)B$ holds in general. Jan 4, 2018 at 3:20
Here is an 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.