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 the farthest distance between points in neighboring intervals that every real number is less than within $2\pi/n_j$, 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$\mathbb{R}$. 1 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 the farthest distance between points in neighboring intervals is less than$2\pi/n_j$, so if$n_j$is chosen large enough, one such interval 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}\$.