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  • $\begingroup$ Thank you for this proof which is also quite elegant. I like it a great deal. I like David's answer slightly more, so I am accepting his response, but it was a hard call to make. Again, thank you. $\endgroup$ Commented Apr 18, 2011 at 15:48
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    $\begingroup$ I like David's proof (which was posted as I was writing mine) as well---it's really a simpler way to see this particular fact, I agree with your choice. BTW, there's another interesting MO question asking a related but deeper question, mathoverflow.net/questions/24131/… $\endgroup$ Commented Apr 18, 2011 at 18:05
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    $\begingroup$ This proof (as well as the proof I propose below) gives a slightly stronger result: If $A$ is invertible and both matrices $A$ and $A^{-1}$ have only non-negative real coefficients, then $A$ is a permutation matrix times a diagonal matrix with strictly positive diagonal coefficients. $\endgroup$ Commented Apr 19, 2011 at 6:42