Timeline for Simultaneous "Monomialization" of a set of operators.
Current License: CC BY-SA 3.0
3 events
| when toggle format | what | by | license | comment | |
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| Nov 21, 2012 at 16:05 | comment | added | Piotr Migdal | @Eric Thanks. (I see, I had in mind a different, stronger condition.) | |
| Nov 20, 2012 at 17:28 | comment | added | Eric Rowell | In "$\Rightarrow$" you assume there is a basis with respect to which $X$ is diagonal and $Y$ is a monomial matrix. This is stronger than "simultaneously monomializable." Indeed, take $X,Y$ to be the $3\times 3$ permutation matrices corresponding to $(1\/2)$ and $(1\/2\/3)$. Then they are monomial in the standard basis, but $YXY^-1$ corresponds to $(2\/3)$, which does not commute with $(1\/2)$. | |
| Nov 19, 2012 at 20:00 | history | answered | Piotr Migdal | CC BY-SA 3.0 |