Timeline for Proving "almost all matrices over C are diagonalizable".
Current License: CC BY-SA 2.5
16 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 2, 2010 at 23:37 | vote | accept | Anweshi | ||
| Feb 2, 2010 at 23:37 | comment | added | Anweshi | @Mariano. It follows from Sard's theorem, that the set has measure zero. | |
| Jan 23, 2010 at 9:51 | comment | added | Kevin Buzzard | All this fuss about "the analytic part"---just use the Zariski topology :-) | |
| Jan 22, 2010 at 22:16 | comment | added | Anweshi | @Qiaochu. Nothing important. I think you are right. | |
| Jan 22, 2010 at 20:39 | comment | added | Qiaochu Yuan | Anweshi, I don't understand what statement of mine you're trying to correct. | |
| Jan 22, 2010 at 19:14 | comment | added | Anweshi | @Mariano. I wish I could accept your answer. However, Emerton gave the same answer with only a few seconds delay, and I was put in a dilemma. See meta here. tea.mathoverflow.net/discussion/178 | |
| Jan 22, 2010 at 18:26 | comment | added | Anweshi | @Qiaochu. The point is that the set Ryan mentions is closed. Dense and closed would of course mean that it is the whole set. I do not think you can really throw out the use of measure in the context of these type of question. | |
| Jan 22, 2010 at 17:43 | comment | added | Qiaochu Yuan | I wasn't making a claim about measure: I was making a claim about what one could do without measure! Density alone is enough for the application Ryan Budney mentions above, for example. It's also not hard to see that zero sets are nowhere dense (although again I am aware that this does not imply measure zero). | |
| Jan 22, 2010 at 17:33 | comment | added | Mariano Suárez-Álvarez | @Anweshi: see en.wikipedia.org/wiki/Smith%E2%80%93Volterra%E2%80%93Cantor_set for an example of a closed set which contains no open and of positive measure. | |
| Jan 22, 2010 at 17:27 | comment | added | Anweshi | @Tom. This set is closed, and it does not contain any open sets. So it has measure zero. That part was easy enough. | |
| Jan 22, 2010 at 17:23 | comment | added | Tom LaGatta | @Anweshi: The analytic part enters when Mariano waves his hands---"Now the set where a non-zero polynomial vanishes is very, very thin"---so there is a little more work to be done. | |
| Jan 22, 2010 at 17:22 | comment | added | Anweshi | @Qiaochu. Dense sets can be of measure zero. For examples the rationals. Even a condition such as "closed" won't help. For example the Cantor set has measure zero. | |
| Jan 22, 2010 at 17:20 | comment | added | Qiaochu Yuan | In particular, even if you don't want to do any measure theory, it's not hard to see that the complement of the set where a non-zero polynomial vanishes is dense. | |
| Jan 22, 2010 at 16:56 | comment | added | Anweshi | Thanks. This use of the discriminant provided the rigor I was looking for. So the missing part was not analytic, it was still algebraic.. | |
| Jan 22, 2010 at 16:55 | vote | accept | Anweshi | ||
| Jan 22, 2010 at 17:01 | |||||
| Jan 22, 2010 at 16:52 | history | answered | Mariano Suárez-Álvarez | CC BY-SA 2.5 |