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Let $A, B$ be given $n\times n$ matrices with positive entries. Does the matrix equation $XAY=B$ always has a solution $X, Y$ with nonnegative entries?

I tried to use Kronecker product to transform the equation as $(Y^\top\otimes X)\mathrm{vec}\,A=\mathrm{vec}\,B$... how to proceed then? Thanks.

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up vote 4 down vote accepted

No, because it may be the case that $B$ has greater rank than $A$, but $\mbox{rank}(XAY) \le \mbox{rank}(A)$. If you exclude this case, the answer is still no. For example, let $A = \left [ \begin{matrix} 1 & a \\ a & 1 \end{matrix} \right ]$ and $B = \left [ \begin{matrix} b & 1 \\ 1 & b \end{matrix} \right ]$ where $ 1 < a < b$, then one can check it's impossible for both $X$ and $Y=A^{-1} X^{-1} B$ to be nonnegative.

I believe the problem is hard to solve in general, since it's similar to checking if a matrix is completely positive.

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Thanks, now I am looking for a counterexample when $A$ has full rank. – Sunni Nov 13 '11 at 14:06
@Sunni Added a counterexample, let me know if it doesn't make sense. – p.s. Nov 13 '11 at 21:47

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