Timeline for Expressing $-\operatorname{adj}(A)$ as a polynomial in $A$?
Current License: CC BY-SA 3.0
25 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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| Aug 13, 2015 at 18:54 | history | edited | darij grinberg | CC BY-SA 3.0 |
attempt @ fixing the latex
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| Dec 11, 2010 at 11:19 | history | edited | Pierre-Yves Gaillard | CC BY-SA 2.5 |
added 329 characters in body
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| Oct 1, 2010 at 23:01 | comment | added | Bill Dubuque | @Pierre-Yves Gaillard: +1, thanks for the interesting post. | |
| Aug 31, 2010 at 6:31 | history | edited | Pierre-Yves Gaillard | CC BY-SA 2.5 |
Edit clearly indicated.
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| Aug 31, 2010 at 5:16 | history | edited | Pierre-Yves Gaillard | CC BY-SA 2.5 |
Minor changes.
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| Aug 30, 2010 at 14:01 | comment | added | Pierre-Yves Gaillard | Thanks a lot!!! I've just added Atiyah-MacDonald's formulation, which is extremely concise, and doesn't even use any indeterminate. If you read the two pages I link to, you'll see that I just rephrased (hopefully without introducing mistakes) what they wrote. | |
| Aug 30, 2010 at 13:48 | history | edited | Pierre-Yves Gaillard | CC BY-SA 2.5 |
Added "Atiyah-MacDonald's formulation".
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| Aug 30, 2010 at 13:06 | comment | added | darij grinberg | Nice coordinate version. Up to now, the most readable of all. | |
| Aug 30, 2010 at 12:30 | history | edited | Pierre-Yves Gaillard | CC BY-SA 2.5 |
Added reference to Atiyah-MacDonald.
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| Aug 30, 2010 at 9:43 | history | edited | Pierre-Yves Gaillard | CC BY-SA 2.5 |
EDIT clearly indicated.
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| Aug 1, 2010 at 8:36 | history | edited | Pierre-Yves Gaillard | CC BY-SA 2.5 |
EDIT clearly indicated
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| Jul 26, 2010 at 19:37 | comment | added | darij grinberg | Don't know. I think I used to prove it by graph theory when I first learnt about it, but this is far from a proof I would like to write up. | |
| Jul 20, 2010 at 11:13 | comment | added | Pierre-Yves Gaillard | Dear darij grinberg: Thank you for your interest! How would YOU prove the theorem? | |
| Jul 20, 2010 at 11:00 | comment | added | darij grinberg | Okay, I think I've got it. No, the books that I know do not present this in a clearer way. | |
| Jul 20, 2010 at 10:34 | comment | added | Pierre-Yves Gaillard | Dear darij grinberg: For the meaning of $e\chi$: $H$ is a right $M_n(K[X])$-module and $\chi$ is in $K[X]\subset M_n(K[X])$. I think the argument is a pain to understand because I'm not explaining it well. Clearly presented, it would look very easy, I believe. | |
| Jul 20, 2010 at 10:25 | comment | added | darij grinberg | (For some reason, it seems to me that every proof of Cayley-Hamilton is either a pain to write - like the Zariski-based and the graph-theoretical ones - or a pain to understand - like yours and any other of the short arguments...) | |
| Jul 20, 2010 at 10:23 | comment | added | darij grinberg | Okay, next: what does $e\chi $ mean? There is a type mismatch in here. | |
| Jul 20, 2010 at 10:19 | comment | added | Pierre-Yves Gaillard | Dear darij grinberg: How would you call it? | |
| Jul 20, 2010 at 10:17 | comment | added | darij grinberg | Ah, so that' what you mean by evaluation at $A$! | |
| Jul 20, 2010 at 10:15 | comment | added | Pierre-Yves Gaillard | Dear darij grinberg: It maps $X^i v$ in $K[X]^n=K^n[X]$ to $A^i v$ in $K^n$. (Here $v$ is in $K^n$.) | |
| Jul 20, 2010 at 10:09 | comment | added | darij grinberg | Why does it map $K[X]^n$ to $K^n$ ? Shouldn't it map $K[X]^n$ to $\left(\mathrm M_n\left(K\right)\right)^n$ ? | |
| Jul 20, 2010 at 10:09 | comment | added | Pierre-Yves Gaillard | Dear darij grinberg: Because it maps $K[X]^n$ to $K^n$ and is $K[X]$-linear. | |
| Jul 20, 2010 at 9:58 | comment | added | darij grinberg | Why is your $e$ in $H$? | |
| Jul 20, 2010 at 9:35 | history | answered | Pierre-Yves Gaillard | CC BY-SA 2.5 |