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Let A and B be two positive definite, real, symmetric matrices. The eigenvalues of A, B and AB, denoted by $\lambda(X)$, obey the relation (from Bhatia): $$ \lambda^\downarrow(A) \cdot \lambda^\uparrow(B) \prec \lambda(AB) \prec \lambda^\downarrow(A) \cdot \lambda^\downarrow(B) $$ where $\downarrow$ indicates decreasing order, $\uparrow$ increasing order, $x \cdot y := (x_1y_1,\ldots ,x_ny_n)$ for $x,y \in \mathbb{R}^n$ and $\prec$ is the majorization preorder.

My question is, for a given set of eigenvalues $\lambda(A),\, \lambda(B)$ and $\lambda(AB)$ which satisfies the above, does there necessarily exist an A and B such that they and AB have the desired eigenvalues?

I've found plenty of material on similar inequalities, but nothing that states if every solution of the inequalities can be realised.

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    $\begingroup$ How about $A$ is the identity matrix, and $B$ is the diagonal matrix with the given eigenvalues? Can you make your question a bit more precise and generally understandable? $\endgroup$
    – Steven Gubkin
    Commented Jan 8, 2015 at 19:21
  • $\begingroup$ Reworded the question to, hopefully, be clearer. $\endgroup$
    – ScienceSnake
    Commented Jan 8, 2015 at 19:34
  • $\begingroup$ If I understand everything correctly, just define $A$ and $B$ to be diagonal matrices, where the eigenvalues are placed in descending order. Can you confirm this is what you are asking about? If so it is really off topic for this site. $\endgroup$
    – Steven Gubkin
    Commented Jan 8, 2015 at 19:44
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    $\begingroup$ I guess I am maybe more confused than I thought. If $\lambda(A)$ is only a set (not an ordered set), what does $\prec$ mean here? $\endgroup$
    – Steven Gubkin
    Commented Jan 8, 2015 at 20:15
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    $\begingroup$ I've checked the example above: the eigenvalues of A, B and AB are definitely right. I do however agree with you Shamisen, something doesn't make sense. In fact, there is no reason in general that $\sum_i \lambda^\downarrow(A)_i \lambda^\uparrow(B)_i = \sum_i \lambda^\downarrow(A)_i \lambda^\downarrow(B)_i$ so it's not even necessarily true that $\lambda^\downarrow(A) \cdot \lambda^\uparrow(B)\prec \lambda^\downarrow(A) \cdot \lambda^\downarrow(B)$. Ive looked through the textbook that first states this (Bhatia) and the only explanation I can find is that he means $\prec_w$ instead of $\prec$. $\endgroup$
    – ScienceSnake
    Commented Jan 9, 2015 at 15:03

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