Timeline for how to find all the solutions to $I+A+\cdots+A^n=0.$ [closed]
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 13, 2014 at 8:30 | comment | added | Marc van Leeuwen | @user108005: I have no judgement about cross-posting, but it is something people care to know about. By the way I just posted an answer on MSE. | |
| Jan 12, 2014 at 1:52 | comment | added | user108005 | @MarcvanLeeuwen I'm a young man.I was so eager to solve a problem and needed this lemma.I didn't realize it is unmoral to ask the queation simutaneously on MO and math.stackexchange.I feel ashamed now and will be cautious.Please forgive me. | |
| Jan 11, 2014 at 23:28 | comment | added | user108005 | @GerryMyerson I'm sorry,Gerry Myerson.I wll be cautious from now on.I also appologize to everyone for my levity. | |
| Jan 11, 2014 at 19:33 | comment | added | Benoît Kloeckner | @AlexDegtyarev: you're right, of course - sorry for being sloppy. | |
| Jan 11, 2014 at 16:05 | comment | added | Alex Degtyarev | @Benoît Kloeckner: not quite true: instead of $A^n\ne I$ one should require that $A$ has no invariant vectors. | |
| Jan 11, 2014 at 16:03 | comment | added | Alex Degtyarev | I'm not sure that it's that simple as @lennon310 suggests and that the question does not qualify. As far as I know, integral representations of finite cyclic groups are still an open problem. The only hope is the small dimension. I would suggest (but not sure) that there are finitely many conjugacy classes. It is relatively easy to describe them in $GL_3(\mathbb{Q}[i])$, but the passage to $\mathbb{Z}[i]$ may be a bigger problem. | |
| Jan 11, 2014 at 15:32 | history | closed |
Dima Pasechnik Gerald Edgar Andrey Rekalo Stefan Kohl♦ Daniel Moskovich |
Not suitable for this site | |
| Jan 11, 2014 at 14:27 | review | Close votes | |||
| Jan 11, 2014 at 15:32 | |||||
| Jan 11, 2014 at 14:16 | comment | added | Marc van Leeuwen | Cross-posted to math.stackexchange | |
| Jan 11, 2014 at 13:40 | comment | added | Gerry Myerson | This website is for math research, and I'm not certain your question qualifies. A good start can be made, anyway, by multiplying your equation by $I-A$. Finding matrices with $A^m=I$ has a literature. | |
| Jan 11, 2014 at 13:40 | comment | added | Benoît Kloeckner | Your equation is equivalent to $A^{n+1}=I$ and $A\neq I$. | |
| Jan 11, 2014 at 13:06 | history | asked | user108005 | CC BY-SA 3.0 |