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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.

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Community wiki, seeing as there is no single "best answer"? –  Yemon Choi Oct 18 '10 at 8:28
    
I think that, as far as elementary solutions are concerned, it's hard to beat the proof in ex.219. –  Gjergji Zaimi Oct 18 '10 at 10:42
    
Yes that's a great proof. It also shows that it is a special case of Brunn-Minkowski, although I do not know if this counts as an elementary proof... –  Piero D'Ancona Oct 18 '10 at 10:55

3 Answers 3

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 $f(x)=\ln(1+e^x)$.

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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 $A(t) = A_0 + A_1t$.

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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.

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