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I came across this question and was wondering how it could be proven.

Find all pairs $(m,n)$ of positive integers for which there exists an $m\times n$ matrix $A$ and an $n\times m$ matrix $B$, both with real entries, such that all diagonal entries of $AB$ are positive and all off-diagonal entries are negative.

I have no idea how to solve it.

Anthony Deber
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  • If $n=m$ then you can take $A = I_n$, making life easy. – Alex G. Aug 22 '15 at 20:13
  • Also note that for $n > m$ you can write $A$ and $B$ where $A$ has a zero block and a $m \times m$ square identity block, and $B$ has a zero block and whatever $m \times m$ square matrix you want as one of its blocks. Then you simply get the $m \times m$ square block of $B$ back for the product, so that case is easy too. – user2566092 Aug 22 '15 at 20:26

1 Answers1

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Answer:

(1) $m=1$, $n\geq 1$;

(2) $m\in\{2,3\}$, $n\geq 2$;

(3) $m\geq 4$, $n\geq 3$.

Hint: For $m=1$ and $m=2$, everything is trivial. For $m=3$, the matrix $$\begin{pmatrix}2&-1&-1\\-1&2&-1\\-1&-1&2\end{pmatrix}$$ has rank $2$, and we proceed for all $n\geq2$ as in the comment by @user2566092. For $m\geq4$, define $$u_{ij}=\left(2+\frac{1}{2m^2}\right)ij-i^2-j^2.$$ The matrix $U$ has rank three, positive entries on the diagonal and negative entries everywhere else. When we checked this stuff, it remains to prove that $n=2$ is impossible for $m\geq 4$. Actually, it suffices to check that $(m,n)=(4,2)$ is impossible, which is a routine task.

user2097
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