Concavity Of The Determinant

  • #1

Homework Statement



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

[itex] det(A+B)^\frac{1}{n} \geq det(A) ^\frac{1}{n} + det(B)^\frac{1}{n} [/itex]

Homework Equations



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

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 !
 

Answers and Replies

  • #2
9
0
Maybe you can use that because [itex]A[/itex] and [itex]B[/itex] are positive semi-definite, they can be written as the Gram-matrix of certain vectors [itex]a_1, a_2, ..., a_n[/itex] and [itex]b_1, b_2, ..., b_n[/itex] respectively, and the determinant of a Gram-matrix is the content (or volume) squared of the [itex]n[/itex]-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
chiro
Science Advisor
4,797
133
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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