Timeline for Encoding primes via ranks of sign matrices
Current License: CC BY-SA 4.0
7 events
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| Mar 23, 2023 at 21:32 | comment | added | math54321 | I tried playing with prime $n$, and it seems like your suggestion of $e_k - e_{n-k}$, $1 \le k < n/2$ being in the row (= column) space checks out - although the coefficients I found for the linear combinations (wrt rows of $A_n$) are not easy to predict, they do seem to be only supported in the latter half of rows (with a single exception for $k = 1$). Also I believe the additional 2 vectors needed in this case can be chosen to be $e_0$ and $\sum_{k=1}^{\lfloor n/2 \rfloor} e_k$ (both of which only involve 2 rows of $A_n$) | |
| Mar 23, 2023 at 19:47 | comment | added | Peter Mueller | @math54321 Not really. I guess one again has to write out a sufficiently large set of linearly independent vectors of the row space. The vectors $e_k-e_{n-k}$, $1\le k<n/2$, seem to be in this space. Probably the additional vectors coming from the divisors of $n$ have a similarly simple shape. But at a first glance I did not see a pattern even in how to linearly combine $e_k-e_{n-k}$ them from the rows of $A$. Even if $n$ is prime it might be difficult. | |
| Mar 23, 2023 at 18:11 | comment | added | math54321 | In fact I did use 0-based indexing in my own calculations (now edited). This is a nice construction of the correct number of linearly independent vectors in $\ker(A_n)$, so it remains to show that they span the kernel - any ideas? | |
| Mar 23, 2023 at 13:14 | history | edited | Peter Mueller | CC BY-SA 4.0 |
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| Mar 23, 2023 at 12:42 | history | edited | Peter Mueller | CC BY-SA 4.0 |
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| Mar 22, 2023 at 23:27 | history | edited | Peter Mueller | CC BY-SA 4.0 |
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| Mar 22, 2023 at 22:04 | history | answered | Peter Mueller | CC BY-SA 4.0 |