Timeline for The largest eigenvalue of a "hyperbolic" matrix
Current License: CC BY-SA 2.5
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 23, 2010 at 21:14 | vote | accept | Seva | ||
| Oct 23, 2010 at 14:33 | answer | added | Qiaochu Yuan | timeline score: 1 | |
| Oct 23, 2010 at 12:56 | answer | added | Fedor Petrov | timeline score: 6 | |
| Oct 23, 2010 at 12:35 | comment | added | Suvrit | Some simple calculations show that the trace of the matrix is $\lfloor \sqrt{n} \rfloor$, and the rank is $\lfloor \sqrt{4n+1}\rfloor-1$---but this does not seem to be immediately useful... | |
| Oct 23, 2010 at 10:49 | answer | added | KexiangXu | timeline score: 0 | |
| Oct 23, 2010 at 9:59 | history | edited | Seva | CC BY-SA 2.5 |
added 10 characters in body; deleted 3 characters in body
|
| Oct 23, 2010 at 9:48 | history | edited | Seva | CC BY-SA 2.5 |
added 145 characters in body; deleted 147 characters in body
|
| Oct 23, 2010 at 9:39 | history | edited | Seva | CC BY-SA 2.5 |
added 397 characters in body; added 3 characters in body; added 11 characters in body; added 2 characters in body
|
| Oct 23, 2010 at 6:31 | comment | added | Denis Serre | You're right. I wrongly supposed that so many rows are identical. I'll think again. | |
| Oct 22, 2010 at 18:17 | comment | added | Seva | It is motivated by a problem in algebraic graph theory. | |
| Oct 22, 2010 at 16:51 | comment | added | Nikita Sidorov | What's your motivation if I may ask? | |
| Oct 22, 2010 at 11:57 | history | asked | Seva | CC BY-SA 2.5 |