Timeline for A query about modular arithmetic on a matrix
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
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| Oct 24, 2022 at 14:57 | comment | added | xyz | thank you. much clearer.. | |
| Oct 24, 2022 at 14:49 | comment | added | Max Alekseyev | @xyz: A small non-trivial example is given by $M$ with a solution over integers. For example, $M$ containing vectors $(0,1,1)$, $(1,1,0)$, and $(0,1,0)$ has a solution $(1,1,p-1)$ for any prime $p$. | |
| Oct 24, 2022 at 14:22 | comment | added | xyz | thanks again. Is there a small non trivial example (soluble modulo all primes) to make things clearer that satisfies these conditions, namely: (1) subset of rows don't add up to 1 vector (2) Ranks of $M$ and $M'$ are equal (3) Radicals of of $kth$ determinant divisors of $M$ and $M′$ are the same for each $k$? | |
| Oct 24, 2022 at 13:25 | comment | added | Max Alekseyev | @xyz: Full-rank matrices are just an example, I've corrected the answer to avoid confusion. As for scenario when the equation soluble modulo all primes, it takes place when the ranks over reals of $M$ and $M'$ are equal, and the radicals of $k$th determinant divisors of $M$ and $M'$ are the same for each $k$. | |
| Oct 24, 2022 at 13:22 | history | edited | Max Alekseyev | CC BY-SA 4.0 |
corrected
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| Oct 24, 2022 at 2:11 | history | edited | Max Alekseyev | CC BY-SA 4.0 |
deleted 46 characters in body
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| Oct 23, 2022 at 19:41 | comment | added | xyz | And a follow up query: Is there a scenario where: Given a matrix M such that there are no subset of rows which add up to all 1 rows. But, $xM=ιK,$ can be solved for all primes $X$? Can we have a test or property that characterizes such matrices completely? | |
| Oct 23, 2022 at 19:30 | vote | accept | xyz | ||
| Oct 23, 2022 at 19:16 | comment | added | xyz | Thank you very much. Few clarifications: As I understand the R-C theorem provides a test whenever the system will have a solution. I am unclear about the second "Let us assume.." part: Does this test of insolubility works only when $M$ has a full rank and fails otherwise when $M$ does not have a full rank? Or the insoluble scenario arises iff $M$ has a full rank and thus this test is self-sufficient for all scenarios? | |
| Oct 23, 2022 at 15:06 | history | edited | Max Alekseyev | CC BY-SA 4.0 |
added 13 characters in body; deleted 27 characters in body
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| Oct 23, 2022 at 14:59 | history | answered | Max Alekseyev | CC BY-SA 4.0 |