Is there a matrix C so that the trace of C^n is dense in R?

Question

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 $z$.

Answer 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 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 $\mathbb{R}$.

Answer 2

In answer to the question as to whether $\text{tr}(C^n)$ is dense in $(-2,2)$: Choose $z=\exp(2 \pi i \theta)$ where $\theta$ is irrational, and let $C$ be the diagonal matrix with $z$ and $\overline{z}$ on the diagonal. By Weyl's criterion , the fractional parts of $n \theta$ are equidistributed modulo 1, and thus $\{z^n\}$ is dense in the unit circle. From this it follows easily that $\text{Re}(z^n)$ is dense in $(-1,1)$.

Answer 3

Bjorn has already answered this question in the affirmative, and shown that such matrices do exist. I'd like to add a further comment here though - 'almost no' matrices satisfy the required property. That is, the collection of 2x2 matrices such that Tr(C^n) is dense in R has zero Lebesgue measure.

We know that Tr(C^n) = a^n + b^n where a,b are the roots of the characteristic polynomial of C. If a and b are both real then it is not possible for C to have the required property. The only possibility is where they are complex conjugates, a = r exp(iθ), b = r exp(-iθ) for r >1. Then, Tr(Cⁿ)=2rcos(nθ). Suppose that θ is uniformly distributed over [-π,π], so that exp(inθ) is uniformly distributed on the unit circle for each n. For any positive K, |Tr(C^n)|<K is equivalent to |cos(nθ)|<r^-nK/2. The set of values of exp(inθ) for which this holds forms a pair of arcs of length r ^-nK (to leading order). So,

$$\mathbb{P}(\vert{\rm Tr}(C^n)\vert\lt K)\approx r^{-n}K/\pi$$

to leading order. Summing over n, this is finite. Then, the Borel-Cantelli lemma says that, with probability one, |Tr(Cⁿ)|<K only finitely often. So, with probability 1, |Tr(Cⁿ)| diverges to infinity.

Answer 4

(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.