Timeline for Prime/undecomposable matrices
Current License: CC BY-SA 4.0
24 events
| when toggle format | what | by | license | comment | |
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| S May 3, 2023 at 8:36 | history | suggested | The Amplitwist | CC BY-SA 4.0 |
fixed broken links to springerlink.com and informaworld.com; replaced direct link to PDF on JSTOR with link to landing page; edited full citations using the citation helper; fixed list formatting
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| May 3, 2023 at 6:18 | review | Suggested edits | |||
| S May 3, 2023 at 8:36 | |||||
| Apr 15, 2023 at 12:01 | comment | added | The Amplitwist | Reposting a link mentioned in a previous comment so that it appears in the "Linked" questions list: Is the semigroup of nonnegative integer matrices with determinant 1 finitely generated? | |
| Aug 14, 2019 at 6:13 | answer | added | Wlod AA | timeline score: 1 | |
| Aug 13, 2019 at 15:42 | answer | added | Ali Taghavi | timeline score: 0 | |
| Jun 22, 2010 at 9:29 | vote | accept | Unknown | ||
| Jun 22, 2010 at 9:29 | history | bounty ended | Unknown | ||
| Jun 16, 2010 at 0:26 | history | edited | Unknown | CC BY-SA 2.5 |
Added question
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| Jun 15, 2010 at 22:19 | history | bounty started | Unknown | ||
| Jun 15, 2010 at 7:14 | comment | added | Unknown | In effect, I am looking for sources like the one mentioned by Amy Glen. | |
| Jun 15, 2010 at 5:16 | comment | added | Victor Protsak | See also mathoverflow.net/questions/24131/…? | |
| Jun 15, 2010 at 4:34 | history | edited | Timothy Chow |
Added "reference-request" tag
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| Jun 14, 2010 at 15:29 | history | edited | Unknown | CC BY-SA 2.5 |
added 8 characters in body
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| Jun 14, 2010 at 15:23 | history | edited | Unknown | CC BY-SA 2.5 |
revision
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| May 27, 2010 at 15:19 | answer | added | Amy Glen | timeline score: 6 | |
| May 27, 2010 at 12:26 | comment | added | Unknown | @Wadim. Thanks. I have found Course in Arithmetic to be helpful in this regard. But there it says S is [0,-1;1,0].... Never mind | |
| May 27, 2010 at 11:54 | vote | accept | Unknown | ||
| Jun 14, 2010 at 15:22 | |||||
| May 27, 2010 at 11:44 | history | edited | Pete L. Clark | CC BY-SA 2.5 |
added 274 characters in body; edited body
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| May 27, 2010 at 11:35 | comment | added | Homology | T and its transpose seem to work: you can "simplify" any matrix in M by multiplying it (left or right) by the inverse of T or its transpose. | |
| May 27, 2010 at 10:36 | comment | added | Wadim Zudilin | Victor, today you really puzzle me: if only nonnegative integers are allowed, then $S$ is missed, so what are the two prime matrices?! Well, if one considers only positive powers, then $T^{-1}$ has to be included as well. But having determinant 1, so staying inside the modular group has no analogy with the prime factorisation in $\mathbb Z$. I guess that the author is just curious... | |
| May 27, 2010 at 9:14 | comment | added | Victor Protsak | These are $\textit{group}$ generators, their inverses aren't in the set. Cf my comment to coudy's answer. | |
| May 27, 2010 at 9:05 | answer | added | coudy | timeline score: 5 | |
| May 27, 2010 at 8:40 | comment | added | Wadim Zudilin | But these (I mean thw two matrices) are the two generators, $T=[1,1;0,1]$ and $S=[0,1;-1,0]$ of the modular group! Look at Wikipedia article or better in Serre's "Course in arithmetic". | |
| May 27, 2010 at 8:23 | history | asked | Unknown | CC BY-SA 2.5 |