Concavity of $\det^{1/n}$ over $HPD_n$. - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T23:40:00Z http://mathoverflow.net/feeds/question/42594 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/42594/concavity-of-det1-n-over-hpd-n Concavity of $\det^{1/n}$ over $HPD_n$. Denis Serre 2010-10-18T08:08:28Z 2013-04-04T07:03:33Z <p>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.</p> <ul> <li>it has important applications in many branches of mathematics,</li> <li>it has many elegant proofs. I know at least three complety different ones.</li> </ul> <p>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 <a href="http://www.umpa.ens-lyon.fr/~serre/DPF/exobis.pdf" rel="nofollow">my page</a> may know my taste.</p> http://mathoverflow.net/questions/42594/concavity-of-det1-n-over-hpd-n/42635#42635 Answer by Deane Yang for Concavity of $\det^{1/n}$ over $HPD_n$. Deane Yang 2010-10-18T13:49:11Z 2010-10-18T13:49:11Z <p>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.</p> <p>The first proof I learned is simply differentiating $(\det A(t))^{1/n}$ twice, where $A(t) = A_0 + A_1t$.</p> http://mathoverflow.net/questions/42594/concavity-of-det1-n-over-hpd-n/42760#42760 Answer by Alekk for Concavity of $\det^{1/n}$ over $HPD_n$. Alekk 2010-10-19T09:33:12Z 2010-10-19T09:33:12Z <p>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 $f(x)=\ln(1+e^x)$.</p> http://mathoverflow.net/questions/42594/concavity-of-det1-n-over-hpd-n/126485#126485 Answer by Denis Serre for Concavity of $\det^{1/n}$ over $HPD_n$. Denis Serre 2013-04-04T07:03:33Z 2013-04-04T07:03:33Z <p>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.</p>