I am looking for a matrix $C$ so that the sequence $tr(C^n)$ is dense in the set of real numbers. Equivalently (in the $2 \times 2$ case), find a complex number $z$ so that the sequence $z^n+w^n$ is dense in $\mathbb{R}$ where $w$ is the conjugate of $z$. 



Is there a matrix C so that the trace of C^n is dense in R?
The answer is yes, even in the $2 \times 2$ case. Let $q\_1,q\_2,\ldots$ be an enumeration of the rational numbers. Let $Q\_j$ be the closed interval $[q\_j-1/j,q\_j+1/j]$. Let $I\_0=[0,2\pi]$. Let $z=2e^{i \theta}$ for a $\theta \in I\_0$ to be determined.


By induction, we construct positive integers $n\_1 < n\_2 < \ldots$ and closed intervals $I\_0 \supseteq I\_1 \supseteq \cdots$ such that for each $j$, the trace $z^{n\_j} + \bar{z}^{n\_j}$ is in $Q\_j$ whenever $\theta \in I\_j$. Namely, if $n\_1,\ldots,n\_{j-1},I\_1,\ldots,I\_{j-1}$ have been determined already, then for any sufficiently large $n\_j$, the set of $\theta$ such that $z^{n\_j} + \bar{z}^{n\_j}$ is in $Q\_j$ is a union of closed intervals such that every real number is within $2\pi/n\_j$ of a point inside this union and within $2\pi/n\_j$ of a point outside this union, so if $n\_j$ is chosen large enough, one such interval in this union will be completely contained in $I\_{j-1}$ and we name it $I\_j$.


The intersection of a descending chain of closed intervals is nonempty, so we can choose $\theta$ such that $\theta \in I\_j$ for all $j$. Then $\lbrace z^n+\bar{z}^n : n \ge 1 \rbrace$ contains an element of $Q\_j$ for each $j$, so it is dense in $\mathbb{R}$.


