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

### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)

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.

-

2
Community wiki, seeing as there is no single "best answer"? – Yemon Choi Oct 18 2010 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 2010 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 2010 at 10:55

## 3 Answers

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.

-

### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.

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 $A(t) = A_0 + A_1t$.

-
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 15, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9503922462463379, "perplexity_flag": "head"}