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This is mostly a reference request, as this must be well known!

Let $A$ and $B$ be two real symmetric matrices, one of which is positive definite. Then it is easy to see that the product $AB$ (or $BA$, which has the same eigenvalues) is similar to a symmetric matrix, so has real eigenvalues. Now take the vectors of eigenvalues of $A$ and of $B$, sorted in decreasing order, and let their componentwise product be $ab$. Then the question is: what is known about the relationship (for instance, inequalities) between $ab$ and the vector of eigenvalues of the product $AB$ (also taken in decreasing order). Some experimentation gives the conjecture that there is a majorization order between them, for instance. This must be well-known!

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Not exactly what you're asking for, but I assume you're familiar with von Neumann's trace inequality, Richter's corresponding lower bound and L. Mirsky's elementary proofs of these. –  cardinal Sep 2 '12 at 19:01
    
Somewhat related: math.stackexchange.com/a/47830/7003 –  cardinal Sep 2 '12 at 19:03

1 Answer 1

up vote 18 down vote accepted

Here are the results that you are probably looking for.

The first one is for positive definite matrices only (the theorem cited below fixes a typo in the original, in that the correct version uses $\prec_w$ instead of $\prec$).

Theorem (Prob.III.6.14; Matrix Analysis, Bhatia 1997). Let $A$ and $B$ be Hermitian positive definite. Let $\lambda^\downarrow(X)$ denote the vector of eigenvalues of $X$ in decreasing order; define $\lambda^\uparrow(X)$ likewise. Then, \begin{equation*} \lambda^\downarrow(A) \cdot \lambda^\uparrow(B) \prec_w \lambda(AB) \prec_w \lambda^\downarrow(A) \cdot \lambda^\downarrow(B), \end{equation*}

where $x \cdot y := (x_1y_1,\ldots ,x_ny_n)$ for $x,y \in \mathbb{R}^n$ and $\prec_w$ is the weak majorization preorder.

However, when dealing with matrix products, it is more natural to consider singular values rather than eigenvalues.

Therefore, the relation that you might be looking for is the log-majorization \begin{equation*} \log \sigma^\downarrow(A) + \log\sigma^\uparrow(B) \prec \log\sigma(AB) \prec \log\sigma^\downarrow(A) + \log\sigma^\downarrow(B), \end{equation*} where $A$ and $B$ are arbitrary matrices, and $\sigma(\cdot)$ denotes the singular value map.

Reference

  1. R. Bhatia. Matrix Analysis. Springer, GTM 169. 1997.
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What is $\lambda(AB)$? You only defined $\lambda^\uparrow$ and $\lambda^\downarrow$. –  David E. Roberson Jan 2 at 10:53
    
@DavidE.Roberson $\lambda(AB)$ is the vector whose elements are the eigenvalues of $AB$. I edited the question and added the definitions of $\cdot$ and $\prec$ so that the inequality should be more clearer now. –  Shamisen Jan 9 at 2:37
    
@shamisen but eigenvalues in what order? –  Michał Masny Jan 9 at 7:22
1  
@MichałMasny It does not matter what order because the majorization preorder does not depend on order. By the way, does anyone know a direct proof of the first result? In Bhatia's book it proves the second result which is stronger, but I would like to know if there is a somewhat easier direct proof of the first result. –  David E. Roberson Jan 9 at 10:22

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