# Concavity Of The Determinant

1. Jul 11, 2012

### Combinatorics

1. The problem statement, all variables and given/known data

How can I prove that given two $nXn$ positive semi-definite matrices $A,B$, then the following inequality holds:

$det(A+B)^\frac{1}{n} \geq det(A) ^\frac{1}{n} + det(B)^\frac{1}{n}$

2. Relevant equations

Brunn-Minkowski Inequality:
http://en.wikipedia.org/wiki/Brunn–Minkowski_theorem

3. The attempt at a solution
I've tried proving that these kind of determinants represent volumes of n-dimensional compact bodies, but without any success.. Is there any algebraic/computational way of doing it?

Thanks in advance !

2. Jul 11, 2012

### Latrace

Maybe you can use that because $A$ and $B$ are positive semi-definite, they can be written as the Gram-matrix of certain vectors $a_1, a_2, ..., a_n$ and $b_1, b_2, ..., b_n$ respectively, and the determinant of a Gram-matrix is the content (or volume) squared of the $n$-dimensional box spanned by the vectors of which it is composed? Then maybe the Brunn-Minkowski inequality? This seems intuitively right to me, but it's not a rigorous argument!

3. Jul 12, 2012

### chiro

Hey Combinatorics.

On this page:

http://en.wikipedia.org/wiki/Positive-definite_matrix#Further_properties

It says on 11. the property of convexity for the matrices themselves. For the actual convexity condition, looking at this says that the set of all elements of the C*-algebra form a convex cone:

http://en.wikipedia.org/wiki/Positive_operator.

It looks like anything that is well-ordered in a sense is what is called a positive cone and my guess is that we establish the well-ordered property for the positive-definite matrices and then go from there to establish convexity.

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