Timeline for Is the transpose of an infinite Hadamard matrix also Hadamard?
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 13 at 10:53 | vote | accept | Dominic van der Zypen | ||
| Sep 13 at 9:54 | answer | added | Sean Eberhard | timeline score: 3 | |
| Sep 13 at 6:54 | comment | added | HenrikRüping | Then we could repeat each column once. Then the bound doubles,but after transposing all pairs $(2k,2k+1)$ would violate this. What if we relax the condition further so that for a given $k$ there can only be finitely many violatings with that $k$ ? | |
| Sep 13 at 6:42 | comment | added | Fedor Petrov | @HenrikRüping What if we relax the definition to "all but finitely many rows are almost orthogonal"? | |
| Sep 13 at 6:27 | history | edited | Dominic van der Zypen | CC BY-SA 4.0 |
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| Sep 6 at 9:56 | comment | added | HenrikRüping | Ah yes. I got confused with rows and columns. The stated definition here is that rows are almost orthogonal. So lets add two columns of ones add the left instead. Then the bound $C_0$ increases by two, but the two columns are not almost orthogonal, so transposing doesn't work. | |
| Sep 6 at 9:53 | comment | added | The Amplitwist | @HenrikRüping But then $\left\lvert \sum_{k=0}^n \bigl( M[0,k] \cdot M[1,k] \bigr) \right\rvert = n+1$, which is not bounded by any positive integer $C_0$, so the first two rows aren't almost orthogonal, right? | |
| Sep 6 at 6:46 | comment | added | HenrikRüping | No. just add two rows of ones at the top. | |
| Sep 6 at 6:38 | history | asked | Dominic van der Zypen | CC BY-SA 4.0 |