Let $A=(a_{ij})_n$ a symmetric matrix with positive coefficients. We suppose that there is $\alpha>0$ such that, for all permutation $\sigma$ of $\{1,\ldots,n\}$, we have $$a_{1\sigma(1)}a_{2\sigma(2)}\cdots a_{n\sigma(n)}=\alpha.$$ How to prove that the rank of $A$ is $1$? Please any suggestions?
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Suppose $\tau$ is another permutation that is the same as $\sigma$ except at two places, say $j$ and $k$, where $\tau(j)=\sigma(k)$ and $\tau(k)=\sigma(j)$. Subtract the equation you wrote about $\sigma$ from the corresponding equation for $\tau$, and, in the result, cancel all the common factors. What remains will tell you that a certain $2\times2$ submatrix has determinant $0$. Repeat for enough $\sigma$'s and $\tau$'s to find that all $2\times2$ submatrices ave determinant $0$. |
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