Is there a matrix C so that the trace of C^n is dense in R? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T10:18:51Z http://mathoverflow.net/feeds/question/15041 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/15041/is-there-a-matrix-c-so-that-the-trace-of-cn-is-dense-in-r Is there a matrix C so that the trace of C^n is dense in R? Hej 2010-02-11T22:16:46Z 2010-02-12T08:45:56Z <p>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 by 2 case), find a complex number z so that the sequence z^n+w^n is dense in R where w is the conjugate of z. </p> http://mathoverflow.net/questions/15041/is-there-a-matrix-c-so-that-the-trace-of-cn-is-dense-in-r/15042#15042 Answer by Richard Kent for Is there a matrix C so that the trace of C^n is dense in R? Richard Kent 2010-02-11T22:26:47Z 2010-02-12T01:12:17Z <p>(Oops, the rescaling part is bogus in the below. So this only works for C with determinant 1.)</p> <p>In the 2-by-2 case, the answer is no. (Something like this argument should go through in general).</p> <p>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.</p> http://mathoverflow.net/questions/15041/is-there-a-matrix-c-so-that-the-trace-of-cn-is-dense-in-r/15052#15052 Answer by Victor Miller for Is there a matrix C so that the trace of C^n is dense in R? Victor Miller 2010-02-11T23:37:02Z 2010-02-11T23:54:33Z <p>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 <code>$\{z^n\}$</code> is dense in the unit circle. From this it follows easily that $\text{Re}(z^n)$ is dense in $(-1,1)$.</p> http://mathoverflow.net/questions/15041/is-there-a-matrix-c-so-that-the-trace-of-cn-is-dense-in-r/15056#15056 Answer by Bjorn Poonen for Is there a matrix C so that the trace of C^n is dense in R? Bjorn Poonen 2010-02-12T00:39:58Z 2010-02-12T00:45:26Z <p>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.</p> <p>By induction, we construct positive integers $n_1 &lt; n_2 &lt; \ldots$ 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 $I_j$.</p> <p>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 $\mathbb{R}$.</p> http://mathoverflow.net/questions/15041/is-there-a-matrix-c-so-that-the-trace-of-cn-is-dense-in-r/15061#15061 Answer by George Lowther for Is there a matrix C so that the trace of C^n is dense in R? George Lowther 2010-02-12T01:36:46Z 2010-02-12T01:50:37Z <p>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.</p> <p>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&nbsp;=&nbsp;r&nbsp;exp(i&theta;), b&nbsp;=&nbsp;r&nbsp;exp(-i&theta;) for r&nbsp;&gt;1. Then, Tr(C<sup>n</sup>)=2rcos(n&theta;). Suppose that &theta; is uniformly distributed over [-&pi;,&pi;], so that exp(in&theta;) is uniformly distributed on the unit circle for each n. For any positive K, |Tr(C^n)|&lt;K is equivalent to |cos(n&theta;)|&lt;r<sup>-n</sup>K/2. The set of values of exp(in&theta;) for which this holds forms a pair of arcs of length r<sup>&nbsp;-n</sup>K (to leading order). So,</p> <p>$$\mathbb{P}(\vert{\rm Tr}(C^n)\vert\lt K)\approx r^{-n}K/\pi$$</p> <p>to leading order. Summing over n, this is finite. Then, the <a href="http://en.wikipedia.org/wiki/Borel%E2%80%93Cantelli%5Flemma" rel="nofollow">Borel-Cantelli lemma</a> says that, with probability one, |Tr(C<sup>n</sup>)|&lt;K only finitely often. So, with probability 1, |Tr(C<sup>n</sup>)| diverges to infinity.</p>