Timeline for $E_n(\ell^\infty)=SL_n(\ell^\infty)$?
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
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| Mar 15, 2017 at 18:02 | comment | added | YCor | @MatthewDaws Also depends on what you call elementary row operator. In coherence with this convention (which is well-admitted in algebraic K-theory), elementary row/column operators mean $C_i\leftarrow C_i+tC_j$ for $i\neq j$. | |
| Mar 14, 2017 at 19:55 | comment | added | Matthew Daws | @YCor I had thought that it was at least reasonably standard to say that an "elementary matrix" is any matrix you can get from the identity by applying an elementary row operator. So "of determinant 1" was my, probably clumsy, way of singling out those "of the first type" (as Artin seems to say). Thank you very much for the link! | |
| Mar 14, 2017 at 18:49 | history | edited | YCor |
edited tags
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| Mar 14, 2017 at 18:49 | comment | added | YCor | @MatthewDaws Elementary matrices have determinant 1 by definition. And this fact is basic linear algebra. References are given in math.stackexchange.com/questions/654206 | |
| Mar 14, 2017 at 18:46 | comment | added | Matthew Daws | Do you have a nice reference that elementary matrices of determinant one generate $SL_n(\mathbb C)$? | |
| Mar 14, 2017 at 17:48 | answer | added | Luc Guyot | timeline score: 5 | |
| S Mar 14, 2017 at 17:48 | history | suggested | Eilin | CC BY-SA 3.0 |
Added the condition $i\neq j$
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| Mar 14, 2017 at 16:58 | review | Suggested edits | |||
| S Mar 14, 2017 at 17:48 | |||||
| Mar 14, 2017 at 15:17 | review | First posts | |||
| Mar 14, 2017 at 15:18 | |||||
| Mar 14, 2017 at 15:15 | history | asked | KevinC | CC BY-SA 3.0 |