Timeline for equations over (some) lie groups
Current License: CC BY-SA 3.0
16 events
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May 7, 2012 at 11:54 | history | edited | user6976 | CC BY-SA 3.0 |
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May 7, 2012 at 11:44 | comment | added | user6976 | No, in the case $n=3$, every matrix of the form $x(x^{-1})^t$ has one of the eigenvalues 1. For non-unipotent matrices, it looks like the only restriction. In general, I think that if $a$ is an eigenvalue of that matrix, then $1/a$ is also an eigenvalue, and that is the only restriction for non-unipotent matrices. It can be rigorously proved, I am sure. | |
May 7, 2012 at 10:40 | comment | added | Igor Rivin | That's very interesting: I did not see whether you checked whether all non-unipotent matrices are representable, or is that still a conjecture only... | |
May 6, 2012 at 21:59 | history | edited | user6976 | CC BY-SA 3.0 |
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May 6, 2012 at 18:02 | history | edited | user6976 | CC BY-SA 3.0 |
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May 6, 2012 at 11:52 | comment | added | Igor Rivin | Ah, OK, I am too jet lagged to fix typos :) | |
May 6, 2012 at 11:51 | comment | added | Igor Rivin | Though since every invertible matrix is a square, as you say, my first question is equivalent to, for every $b,$ the solvability of $x = y b y.$ | |
May 6, 2012 at 11:50 | comment | added | user6976 | A misprint: you need to substitute $z=ya$ (I fixed that). Then $yaya^{-1}=zza^{-1}a^{-1}=z^2a^{-2}$. | |
May 6, 2012 at 11:49 | history | edited | user6976 | CC BY-SA 3.0 |
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May 6, 2012 at 11:48 | comment | added | Igor Rivin | I might be more jet lagged than I thought, but it seems to me that substituting $y=za$ gives you $x=z a a z a a^{-1} = z a^2 z \neq z^2 a^{-2}.$ | |
May 6, 2012 at 9:53 | history | edited | user6976 | CC BY-SA 3.0 |
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May 6, 2012 at 9:46 | history | edited | user6976 | CC BY-SA 3.0 |
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May 6, 2012 at 9:34 | history | edited | user6976 | CC BY-SA 3.0 |
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May 6, 2012 at 9:24 | history | undeleted | user6976 | ||
May 6, 2012 at 9:06 | history | deleted | user6976 | ||
May 6, 2012 at 8:51 | history | answered | user6976 | CC BY-SA 3.0 |