matrices - 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 $I_0=[0,2pi]$. 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 $2pi/n_j$ of a point inside this union and within $2pi/n_j$ 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 $mathbb{R}$.