Timeline for Integer matrices with no integer eigenvalues
Current License: CC BY-SA 3.0
19 events
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| S Sep 18, 2016 at 13:41 | history | suggested | Rodrigo de Azevedo | CC BY-SA 3.0 |
Minor edits
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| Sep 18, 2016 at 13:15 | review | Suggested edits | |||
| S Sep 18, 2016 at 13:41 | |||||
| Nov 12, 2010 at 2:24 | answer | added | user6976 | timeline score: 6 | |
| May 16, 2010 at 21:14 | comment | added | Hej | Andrey, for example if $A=[3,1;0,1]$ and $B=[1,0;1,4]$, then $AB$ has integer eigenvalues. I think (3,2) is special in a mysterious way relating to the Collatz problem. | |
| May 16, 2010 at 21:02 | comment | added | Hej | JSE, I used Mathematica to support the conjecture; besides that I really don't have a good reason for it. Your heuristic argument might suggest that there is a finite number of elements with integer eigenvalues but I don't see how it would suggest that there are none. | |
| May 16, 2010 at 13:42 | comment | added | JSE | Hej, why do you make this conjecture? My dopey heuristic argument suggests it shouldn't be so uncommon for A and B to have this property; you've thought about this more, so I'm curious where the opposite intuition comes from. | |
| May 16, 2010 at 5:17 | comment | added | Andrey Rekalo | @Hej: So I would guess you can prove that the property doesn't hold for some specific matrices $A$ and $B$, can't you? I'm just curious what mechanism should make the pair of matrices in your question different from a generic pair. | |
| May 16, 2010 at 5:01 | comment | added | Hej | KConrad, here is another way one can think about the problem. Let $f(x)=3x+1$ and $g(x)=x/(x+2)$ show that there is no composition of $f$'s and $g$'s that has a rational fixed point except $f^n$ and $g^n$. | |
| May 16, 2010 at 4:42 | comment | added | Hej | Andrey, if $A=[a,1;0,1]$ and $B=[1,0;1,b]$ for integers $a,b>1$, then my conjecture is that the only cases that the semigroup generated by $A$ and $B$ have this property are the cases where $(a,b)=(2,3), (3,2)$. So such semigroups are rather special. | |
| May 16, 2010 at 4:34 | answer | added | Bjorn Poonen | timeline score: 30 | |
| May 16, 2010 at 2:08 | comment | added | JSE | Heuristically, the 2^n words of length n in your semigroup are going to have determinant around 6^n, so I guess the "probability" of the discriminant being a square is something like 6^{-n/2}; so I guess it seems reasonable to me that you'd expect only finitely many integer eigenvalues, and none if you don't find any early on. (Yes, I know there are the A^m and B^n, but hey, heuristics are heuristics...) It's much less clear to me what to expect when the matrices are in SL_2(Z). | |
| May 16, 2010 at 0:38 | comment | added | Wadim Zudilin | Well, this definitely means a certain explicit knowledge about the members of the semigroup. I have in mind a similar looking problem for matrices $A=[1,1;1,0]$ and $B=[2,1;1,0]$. Is it possible to describe the set $S$ of $(1,1)$-entries of matrices sitting in the semigroup generated by $A$ and $B$? The so-called continuant problem is about the density of the set $$S$ in $\mathbb N$: for example, is it true that between $n^2+c_1n$ and $n^2+c_2n$ (for some $c_1< c_2$) there is always an element of the set $S$? This does not help in the original problem but shows that it is hard(if not obvious). | |
| May 16, 2010 at 0:19 | comment | added | KConrad | Two comments. 1) Can you tell us why you are interested in this? Maybe someone can see another approach to take when it is seen what the problem's setting is. 2) Can you tell us what ways you have thought of? What you are asking is equivalent to asking if the discriminants of the characteristic polynomials of the matrices in that semigroup are never squares except for powers of A and B. (The characteristic polynomial of a product of matrices is unchanged by cyclic shifts in the order of multiplication, so you could assume the first matrix in your product is A, say.) | |
| May 16, 2010 at 0:06 | comment | added | Andrey Rekalo | @Hej: Do you know any two integer matrices $A$ and $B$ which generate the semigroup that doesn't have this property? | |
| May 15, 2010 at 23:07 | comment | added | Xandi Tuni | It could, in principle, be undecidable, but I very much doubt that. Without any further arguments, it sounds rather like defeatism to me at that point. | |
| May 15, 2010 at 22:59 | history | edited | S. Carnahan♦ | CC BY-SA 2.5 |
repair
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| May 15, 2010 at 22:53 | history | edited | Hej | CC BY-SA 2.5 |
deleted 100 characters in body; added 2 characters in body
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| May 15, 2010 at 22:53 | history | edited | fedja | CC BY-SA 2.5 |
fixed matrices
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| May 15, 2010 at 22:45 | history | asked | Hej | CC BY-SA 2.5 |