Timeline for Matrices that are Hadamard products of $X$ and $X^{-T}$
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
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| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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| May 22, 2013 at 8:45 | vote | accept | Federico Poloni | ||
| Apr 26, 2011 at 14:01 | comment | added | Denis Serre | The question is whether every matrix $A$ such $A\vec e=A^T\vec e=\vec e$ (with $\vec e=(1,\ldots,1)^T$) can be written as $X\odot X^{-T}$. When $A$ has non-negative entries, this is a question about bistochastic matrices. A special case is that of ortho-stochastic matrices, for which such an $X$ exists, a unitary one. But it is known that ortho-stochastic matrices form a small part of bistochastic ones. In addition, we don't want to restrict to non-negative entries. | |
| Apr 26, 2011 at 13:44 | comment | added | Denis Serre | yes, $a+b=1$. Sorry for the misprint. | |
| Apr 26, 2011 at 13:25 | history | edited | Federico Poloni | CC BY-SA 3.0 |
added 52 characters in body; added 104 characters in body
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| Apr 26, 2011 at 13:16 | comment | added | Federico Poloni | I think you mean $a+b=1$ --- anyway good point, I omitted it in my computations because it didn't matter for the problem and then forgot about it. | |
| Apr 26, 2011 at 12:42 | comment | added | Denis Serre | In the $2\times2$ case, there is a restriction on the output: $a+b=0$ (maybe you forgot to divide by the determinant of $X$, which occurs from $X^{-T}$). This makes an affine set that is stable under $\times$, but is not a multiplicative group, because of the matrix with $a=b=\frac12$. | |
| Apr 26, 2011 at 12:30 | answer | added | greg coxson | timeline score: 3 | |
| Apr 26, 2011 at 11:59 | history | asked | Federico Poloni | CC BY-SA 3.0 |