Timeline for How many commuting nilpotent matrices are there?
Current License: CC BY-SA 2.5
15 events
| when toggle format | what | by | license | comment | |
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| Apr 19, 2014 at 13:03 | answer | added | NN guest | timeline score: 1 | |
| Jan 5, 2011 at 17:00 | comment | added | David E Speyer | In what way is this not answered by the theorem of Schur, discussed in mathoverflow.net/questions/19591 ? | |
| Jan 5, 2011 at 13:36 | answer | added | Pasha Zusmanovich | timeline score: 4 | |
| Sep 19, 2010 at 1:27 | answer | added | Richard Borcherds | timeline score: 4 | |
| Sep 19, 2010 at 0:30 | comment | added | Jonas Meyer | By Schur's result 1 + the floor of n^2/4 is an upper bound, and I guess (for what it's worth) that the floor of n^2/4 is actually the maximum, but I don't know how to prove it. | |
| Sep 19, 2010 at 0:28 | comment | added | Jonas Meyer | Related questions: mathoverflow.net/questions/19591/…, mathoverflow.net/questions/19755/…. It came up there that you can achieve the maximal dimension for a commutative subalgebra, 1 plus the floor of n^2/4, by taking scalars plus 2-by-2 block strictly upper triangular matrices. (Apparently the proof of maximality is due to Schur.) Now you can throw out the multiples of the identity to obtain an example of commuting nilpotents having dimension the floor of n^2/4. | |
| Sep 19, 2010 at 0:20 | comment | added | Yuhao Huang | Thank you for changing the title, I feel it is confusing but couldn't figure out the correct wording. | |
| Sep 18, 2010 at 23:09 | comment | added | Yemon Choi | Have changed title, as per Charles' comment | |
| Sep 18, 2010 at 23:09 | history | edited | Yemon Choi | CC BY-SA 2.5 |
fixed title
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| Sep 18, 2010 at 22:45 | comment | added | darij grinberg | Oh right. But at least we now know that we are looking for maximal commuting subspaces of $\mathrm{u}_n$. Is it the same (dimensionwise) as asking for maximal tori of $\mathrm{U}_n$ ? | |
| Sep 18, 2010 at 22:35 | comment | added | Charles Staats | I think the substitution of "commuting" for "commutative" would address Qiaochu's concern. | |
| Sep 18, 2010 at 22:28 | comment | added | Qiaochu Yuan | "Commutative" is not a property of a matrix, but of a set of matrices, so I would rather you state the property you want more precisely. | |
| Sep 18, 2010 at 22:25 | comment | added | Yuhao Huang | I get this point, but upper triangular matrices don't necessarily commute with each other, right? | |
| Sep 18, 2010 at 22:12 | comment | added | darij grinberg | Commuting matrices can be simultaneously trigonalized over the algebraic closure. If the matrices are nilpotent, their trigonalizations will be strictly upper triangular (because any nonzero entries on the diagonal would survive taking powers, contradicting the nilpotency). So the maximal dimension is $\frac{n\left(n-1\right)}{2}$. | |
| Sep 18, 2010 at 22:04 | history | asked | Yuhao Huang | CC BY-SA 2.5 |