Timeline for Are the trace relations among matrices generated by cyclic permutations?
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
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| Jan 8 at 14:20 | history | edited | Nikita Safonkin | CC BY-SA 4.0 |
added 7 characters in body
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| Jan 6 at 7:26 | comment | added | Nikita Safonkin | @LSpice Thank you! I've added a clarification about the field. When I was writing the post yesterday, I couldn't recall the command for the appropriate brackets. I've corrected this too. Thanks again. | |
| S Jan 6 at 7:18 | review | First answers | |||
| Jan 6 at 12:34 | |||||
| S Jan 6 at 7:18 | history | edited | Nikita Safonkin | CC BY-SA 4.0 |
added 48 characters in body
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| Jan 5 at 23:24 | comment | added | LSpice |
Your proof never says it explicitly, but I guess that $\Bbbk$ is meant to be any field of characteristic $0$? \\ TeX note: $\Bbbk\left<x_1, \dotsc, x_d\right>$ \Bbbk\left<x_1, \dotsc, x_d\right> produces a space between $\Bbbk$ and $\left<\right.$ that's larger than might be intended. You can switch it off with an extra pair of braces, as $\Bbbk{\left<x_1, \dotsc, x_d\right>}$ \Bbbk{\left<x_1, \dotsc, x_d\right>}, but, unless you really want the sizing, $\Bbbk\langle x_1, \dotsc, x_d\rangle$ \Bbbk\langle x_1, \dotsc, x_d\rangle may be better.
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| S Jan 5 at 23:22 | review | First answers | |||
| Jan 6 at 1:33 | |||||
| S Jan 5 at 23:22 | history | edited | LSpice | CC BY-SA 4.0 |
Link to @JosephVanName's proof
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| S Jan 5 at 19:52 | review | First answers | |||
| Jan 5 at 21:11 | |||||
| S Jan 5 at 19:52 | history | answered | Nikita Safonkin | CC BY-SA 4.0 |