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I'm working out of Devaney's Introduction to Chaotic Systems, and one of the problems I'm working on is to construct a subshift of finite type in $\Sigma_3$ with no fixed or period two points, but with points of period 3.

I don't know how to approach this problem, but it seems to me that if the number of periodic points of a subshift is given by the trace of its transition matrix, then this problem has no solution, since it's asking me to find a matrix $A$ such that $trace$(A)$=$$trace$($A^2$)$=0$, but have $trace(A^3)\neq\=0$, but this is impossible for a $3x3$ matrix.

What am I misunderstanding here? Thanks.

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    $\begingroup$ Such $3\times 3$ matrix exists. Let $A=(a_{i,j})$ be a $3\times 3$ matrix, with $a_{1,2}=a_{2,3}=a_{3,1}=1$ and other place zero. The trace of $A$ and $A^2$ is zero but the trace of $A^3$ is 3, which is not zero. $\endgroup$
    – Siming Tu
    Apr 17, 2014 at 4:09
  • $\begingroup$ How silly of me to miss that! Thanks for the help! $\endgroup$ Apr 17, 2014 at 4:30

1 Answer 1

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Echoing Siming Tu's comment. Such a matrix exists. Let $A$ be a square matrix with $a_{1,2}=a_{2,3}=a_{3,1} = 1,$ else 0. This matrix shifts the top row to the bottom under matrix multiplication. So that the trace of $A$ and $A^2$ are $0$ as everything is off the diagonal. However, $A^3$ is the identity, so it has trace $1$.

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