Timeline for $\omega\times\omega$-Hadamard matrices
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 30 at 14:33 | comment | added | Dominic van der Zypen | OK I see thanks. | |
| Aug 30 at 13:54 | comment | added | Oscar Lanzi | @DominicvanderZypen in the comment to the referenced question the dot product of $f_k$ and $f_{2k}$ gets as big absolutely as $k$ for any whole number $k$ by truncating after any odd multiple of $k$ terms. You are still pigeonholed. | |
| Aug 30 at 11:17 | comment | added | Dominic van der Zypen | @OscarLanzi I am not sure your argument works - see the comment here : mathoverflow.net/questions/477848/… | |
| Aug 30 at 9:35 | comment | added | Oscar Lanzi | You nay wnt to pur absolute values aroubd the lim inf argument. | |
| Aug 30 at 6:12 | comment | added | HenrikRüping | For example, if we compute the "inner product" of columns $1,2$ we get the following partial sums: $1,0,-1,0,... $ (and then it repeats). Thus the $\liminf$ would be $-1$. | |
| Aug 29 at 19:26 | comment | added | Oscar Lanzi | It would not work for that more restrictive rule, nor could any IMO. You are pigeonholed into having indefinitely long parts of rows or columns all $1$ or all $-1$. But the fraction of such row or column pairs diminishes for larger truncations of this matrix. | |
| Aug 29 at 19:25 | vote | accept | Dominic van der Zypen | ||
| Aug 29 at 19:23 | comment | added | Dominic van der Zypen | Thanks a lot Oscar! Actually it occurred to me that the most natural definition for approximately orthogonal would be: there is a global constant $C_0\in\omega$ such that $\big|\sum_{k=0}^n \big(f(n)g(n)\big)\big| < C_0$ for all $n\in\omega$. I think your example works with this as well | |
| Aug 29 at 19:02 | history | answered | Oscar Lanzi | CC BY-SA 4.0 |