Timeline for Almost Hadamard matrices
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
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| Jun 30, 2023 at 8:32 | vote | accept | Roland Bacher | ||
| Nov 9, 2018 at 12:15 | answer | added | Wolfgang | timeline score: 8 | |
| Nov 1, 2018 at 6:57 | comment | added | Gottfried Helms |
The matrix, in compact form, is 0222000020220022020222202022220220020202000000022000202000220020022002220202200202202002200220200222000222202000000022222220222200220000020222220222020222220220022022202 where you extract the digits rowwise into a $13\times13$-matrix and subtract $1$
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| Nov 1, 2018 at 6:49 | comment | added | Gottfried Helms | @joro - just found a solution for n=13 (using random rotation-matrices as basis and then using that "varimin"-rotation discussed elsewhere). | |
| Feb 5, 2014 at 15:58 | comment | added | The Masked Avenger | Ehlich, Barba, and Wojtas did the work on matrix determinants for (some permutation of) orders 1,2, and 3 mod 4. Some recent accounts (e.g. Osborn or Orrick) of the Hadamard problem may talk enough about Gram matrices to say why you can't do optimal for 1 mod 4. | |
| Feb 5, 2014 at 15:46 | answer | added | Yuichiro Fujiwara | timeline score: 7 | |
| Feb 5, 2014 at 14:11 | history | edited | Roland Bacher | CC BY-SA 3.0 |
added 1 characters in body
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| Feb 5, 2014 at 14:03 | comment | added | Roland Bacher | Indeed, a solution for $n=5$ is given by considering a symmetric square matrix of size $5$ with $1$ on the diagonal and $-1$ everywhere else. | |
| Feb 5, 2014 at 13:50 | comment | added | joro | According to the computer there are no solutions for n=9 and n=13. It found solution for n=5. Might be wrong. | |
| Feb 5, 2014 at 9:29 | history | asked | Roland Bacher | CC BY-SA 3.0 |