How Can Determinant Inequalities for Positive Semi-Definite Matrices Be Proven?

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SUMMARY

The discussion focuses on proving the determinant inequality for two n x n positive semi-definite matrices A and B, specifically that det(A+B)^(1/n) ≥ det(A)^(1/n) + det(B)^(1/n). The Brunn-Minkowski Inequality is referenced as a potential tool for the proof, suggesting that the determinants represent volumes of n-dimensional compact bodies. The conversation also highlights the relationship between positive semi-definite matrices and Gram matrices, indicating that the determinants can be interpreted in terms of the volumes spanned by corresponding vectors.

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  • Understanding of positive semi-definite matrices
  • Familiarity with the Brunn-Minkowski Inequality
  • Knowledge of Gram matrices and their properties
  • Basic concepts of convexity in linear algebra
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  • Study the properties of Gram matrices and their applications in geometry
  • Explore the concept of convex cones in the context of positive operators
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Mathematicians, students studying linear algebra, researchers in optimization, and anyone interested in the properties of positive semi-definite matrices and their applications in geometry.

Combinatorics
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Homework Statement



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}

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 !
 
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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!
 
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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