# Existence of a matrix product from its eigenvalues

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.

• 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? Jan 8, 2015 at 19:21
• Reworded the question to, hopefully, be clearer. Jan 8, 2015 at 19:34
• 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. Jan 8, 2015 at 19:44
• 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? Jan 8, 2015 at 20:15
• 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 $\prec$. Jan 9, 2015 at 15:03