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Feb 12, 2010 at 1:12 history edited Autumn Kent CC BY-SA 2.5
rescaling is bogus.; added 46 characters in body
Feb 12, 2010 at 0:56 comment added Victor Miller @GL: You have a point, I'll have to think about it.
Feb 12, 2010 at 0:32 comment added George Lowther Yes, cos(nt) is dense in [-1,1] when theta is irrational. However, on its own that doesn't imply that 2 r^n cos(nt) either is or isn't dense in the reals. If theta is algebraic then you can use Roth's theorem to deduce that cos(nt) can't be less than (1/n^2) infinitely often, so 2r^ncos(nt)-> infinity. However, there is still the transcendental theta case.
Feb 12, 2010 at 0:23 comment added Victor Miller I think yes. Write $t=2 \pi \theta$. There are two cases: 1) $\theta$ is rational -- then $\cos(nt)$ takes on only a finite set of values, and unless $\theta =$ an odd integer$/4$, they are non-zero. 2) $\theta$ is irrational. Then by Weyl's criterion, $\cos(nt)$ is dense in $(-1,1)$
Feb 12, 2010 at 0:17 comment added George Lowther Victor, if it has complex eigenvalues and r>1 then the traces are 2r^n cos(nt). Does this have to grow with n?
Feb 12, 2010 at 0:12 comment added Victor Miller Here's a way around it: Let $r$ be the absolute value of the largest eigenvalue. If $r > 1$ then the traces are a sequence growing like $r^$ (perhaps times a polynomial in $n$). It's not hard to see that this sequence can't be dense in $\mathbb{R}^n$. If $r < 1$ it's also clear that the sequence can't be. Similarly for $r=1$.
Feb 12, 2010 at 0:03 comment added David E Speyer How do you reduce to the determinant 1 case?
Feb 11, 2010 at 23:23 comment added Tom LaGatta Is there an elliptic C such that tr(C^n) is dense in (-2,2)?
Feb 11, 2010 at 22:34 history edited Autumn Kent CC BY-SA 2.5
rescaled matrix.
Feb 11, 2010 at 22:26 history answered Autumn Kent CC BY-SA 2.5