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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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Very nice! $\,\,\!$ –  Berci Feb 25 '13 at 21:24
Don't you want to divide, instead of subtracting? Nice answer, +1. –  1015 Feb 25 '13 at 21:26
Great, we maybe to reach the perfection must add that the matrix is not the zero matrix so it's rank not equal $0$. –  Sami Ben Romdhane Feb 25 '13 at 21:38
@julien No, I want to subtract, so that I get, apart from a lot of common factors, the determinant of a $2\times2$ submatrix. –  Andreas Blass Mar 4 '13 at 2:15
Ah, yes, sorry. Dividing works, but subtracting is the way. –  1015 Mar 4 '13 at 2:40
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