Timeline for How much redundancy resides in an $n \times n$ orthogonal matrix?
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 12, 2015 at 20:41 | answer | added | Denis Serre | timeline score: 2 | |
| Jun 9, 2015 at 17:23 | answer | added | loup blanc | timeline score: 4 | |
| Jun 8, 2015 at 1:37 | answer | added | Richard Stanley | timeline score: 15 | |
| Jun 6, 2015 at 13:47 | vote | accept | Joseph O'Rourke | ||
| Jun 6, 2015 at 3:53 | comment | added | Manfred Weis | if the sign of the determinant is given, then a higher number of entries can be recovered, specifically $2$ for $n=2$ | |
| Jun 6, 2015 at 0:51 | answer | added | Robert Israel | timeline score: 11 | |
| Jun 6, 2015 at 0:51 | comment | added | Joseph O'Rourke | @NoamD.Elkies: Thank you for the clarity of your observations, which have uncovered the essence of the situation explored in my question. | |
| Jun 6, 2015 at 0:18 | comment | added | Noam D. Elkies | And in the other direction, if an orthogonal matrix has $\pm 1$'s on the diagonal then every other entry is zero because each row has norm $1$; so in this sense $k$ can be as large as $n^2-n$, and that's clearly maximal because any more would leave an entire row $r$ undetermined and $-r$ works as well. | |
| Jun 6, 2015 at 0:15 | comment | added | Noam D. Elkies | What's the quantification over matrices and locations? It's not always true that even a single diagonal entry can be recovered uniquely: what's $x$ such that ${\rm diag}(1,x)$ is orthogonal? | |
| Jun 6, 2015 at 0:04 | history | asked | Joseph O'Rourke | CC BY-SA 3.0 |