Timeline for Gordan's Theorem with $Ax=b$
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 2, 2011 at 17:13 | vote | accept | bandini | ||
| Jun 2, 2011 at 17:11 | history | edited | bandini | CC BY-SA 3.0 |
Clarification
|
| May 20, 2011 at 17:39 | answer | added | Robert Israel | timeline score: 4 | |
| May 20, 2011 at 17:32 | answer | added | Brian Borchers | timeline score: 2 | |
| May 20, 2011 at 16:26 | answer | added | Igor Rivin | timeline score: 0 | |
| May 20, 2011 at 16:23 | history | edited | bandini | CC BY-SA 3.0 |
Clarified in response to comments.
|
| May 20, 2011 at 16:20 | comment | added | bandini | I'm looking for a version on Gordan's Theorem (stated) in which $Ax=0$ is replaced by $Ax=b$. | |
| May 20, 2011 at 15:08 | comment | added | Federico Poloni | So $A$ is square and $m=n$? | |
| May 20, 2011 at 15:03 | comment | added | Igor Rivin | I am conjecturing that he wants $Ax = b$ or $A^t y > b^t...$ | |
| May 20, 2011 at 14:53 | comment | added | Federico Poloni | What is the role of $b$ in all this? Maybe there is a misprint? | |
| May 20, 2011 at 13:48 | history | asked | bandini | CC BY-SA 3.0 |