In the 2-by-2 case, the answer is no. (Something like this argument should go through in general).

After rescaling, we can assume the matrix has determinant 1. If C is elliptic (real trace between -2 and 2), then all powers are elliptic, so that's no good. If it's parabolic (trace equal to -2 or 2), then all powers all parabolic, again no good. If it's loxodromic, the traces of the powers have real part going to infinity with n, and so they can't be dense.

(Oops, the rescaling part is bogus in the below. So this only works for C with determinant 1.)

In the 2-by-2 case, the answer is no. (Something like this argument should go through in general).

After rescaling, we can assume the matrix has determinant 1. If C is elliptic (real trace between -2 and 2), then all powers are elliptic, so that's no good. If it's parabolic (trace equal to -2 or 2), then all powers all parabolic, again no good. If it's loxodromic, the traces of the powers have real part going to infinity with n, and so they can't be dense.

In the 2-by-2 case, the answer is no. (Something like this argument should go through in general).

If C is elliptic (real trace between -2 and 2), then all powers are elliptic, so that's no good. If it's parabolic (trace equal to -2 or 2), then all powers all parabolic, again no good. If it's loxodromic, the traces of the powers have real part going to infinity with n, and so they can't be dense.

In the 2-by-2 case, the answer is no. (Something like this argument should go through in general).

After rescaling, we can assume the matrix has determinant 1. If C is elliptic (real trace between -2 and 2), then all powers are elliptic, so that's no good. If it's parabolic (trace equal to -2 or 2), then all powers all parabolic, again no good. If it's loxodromic, the traces of the powers have real part going to infinity with n, and so they can't be dense.