11
$\begingroup$

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.

$\endgroup$
  • 2
    $\begingroup$ Community wiki, seeing as there is no single "best answer"? $\endgroup$ – Yemon Choi Oct 18 '10 at 8:28
  • $\begingroup$ I think that, as far as elementary solutions are concerned, it's hard to beat the proof in ex.219. $\endgroup$ – Gjergji Zaimi Oct 18 '10 at 10:42
12
$\begingroup$

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

$\endgroup$
7
$\begingroup$

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 $A(t) = A_0 + A_1t$.

$\endgroup$
  • 1
    $\begingroup$ Why Brunn-Minkowski for bodies is called generalization of matrix determinant inequality? Does BM imply matrix determinant? $\endgroup$ – Paata Ivanishvili Oct 8 '16 at 16:09
  • 2
    $\begingroup$ 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. $\endgroup$ – Deane Yang Oct 9 '16 at 1:35
  • $\begingroup$ @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. $\endgroup$ – stewbasic Jan 4 '18 at 3:20
5
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for?Browse other questions tagged or ask your own question.