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Timeline for Commuting matrices in GL(n,Z)

Current License: CC BY-SA 4.0

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S Apr 18 at 21:22 history suggested The Amplitwist CC BY-SA 4.0
fixed broken link to Wikipedia
Apr 18 at 20:39 review Suggested edits
S Apr 18 at 21:22
Jul 14, 2022 at 13:35 comment added David E Speyer All I need is some integers $(a,b,c)$ such that $u^a v^b w^c > 1$ and $v^a w^b u^c$ and $w^a u^b v^c < 1$. I have no memory of why I came up with the specific choice $(1,0,4)$, I assume I tried a few triples until one worked.
Jul 14, 2022 at 6:00 comment added ah-- Is there a hint what kind of "numerical experimentation" would be needed to find the $f(\lambda_1)$ (i.e. $uw^4$) in the example you gave?
S Oct 10, 2013 at 17:59 history suggested SashaKolpakov CC BY-SA 3.0
edited latex a bit
Oct 10, 2013 at 17:40 review Suggested edits
S Oct 10, 2013 at 17:59
Feb 16, 2011 at 19:03 comment added David E Speyer Right. Let $M$ have eigenvalues $(u,v,w) \approx (0.307979, 0.643104, 5.04892)$. There will be some other matrix $N$, corresponding to the unit $u w^4$, whose eigenvalues are $(u w^4, v u^4, w v^4) \approx (200.131, 0.0057858, 0.863622)$. So they are both hyperbolic.
Feb 16, 2011 at 18:48 vote accept Nikita Sidorov
Feb 16, 2011 at 18:47 comment added Nikita Sidorov David, thanks! It is clear to me now how these matrices are related to units. As for the second part of my question, I'm afraid I didn't make myself 100% clear... What I meant was whether there exists a hyperbolic $M'$ commuting with $M$ with the <b>same property as $M$</b> - that is, it has only one eigenvalue outside the unit disc. It seems that the basis of the centralizer are matrices from different "clusters" (cluster = $m$ eigenvalues inside the unit disc, $n-m$ outside). Does this mean that the answer to this question is actually no?
Feb 16, 2011 at 17:37 history answered David E Speyer CC BY-SA 2.5