Let $\Sigma_{m}:=\{(\ldots,x_{-1},x_{0},x_{1},\ldots):x_{i}\in\{0,1,\ldots,m-1\}\}$. We define $X\subset\Sigma_{2}$ to be the subshift of finite type that has $000$ and $111$ as its only forbidden words. I want to calculate the topological entropy of the left shift map $\sigma\colon X\to X$.
It can be shown that if $Y$ is a vertex shift (i.e. induced by a graph), then the topological entropy of the left shift map on $Y$ is equal to the logarithm of the largest eigenvalue of the adjency matrix $A$ associated to the graph $\Gamma$ (i.e. $a_{ij}=1$ iff there is an edge from $i$ to $j$ in $\Gamma$). Also, two conjugate dynamical systems have the same topological entropy. For a more detailed explanation, check Brin-Stuck, Chapter 3 (and 2).
So I tried to construct a conjugacy $c\colon X\to Y\subset\Sigma_{4}$ (i.e. a surjective continuous map $c$ such that $\sigma\circ c=c\circ\sigma$), where $Y$ is a vertext shift. Inspired by the proof of proposition 3.2.1 in Brin-Stuck, I defined $c(x)_{i}:=x_{i}x_{i+1}$ (where $x_{i}x_{i+1}$ is a word of length $2$). Note that $\{00,01,10,11\}$ are the allowed words that occur in $X$ of length $2$. So $Y$ is a vertex shift with the following graph $\Gamma$ (sorry for bad graphical representation):
- $00\to01$
- $01\to10$, $01\to11$
- $10\to00$, $10\to01$
- $11\to10$
Then the associated matrix is given by $$A:=\begin{pmatrix}0&1&0&0\\ 0&0&1&1\\ 1&1&0&0\\ 0&0&1&0\end{pmatrix}.$$ The characteristic polynomial of $A$ is given by $$\lambda^{4}-\lambda^{2}-2\lambda-1.$$ But I think it is way to hard to calculate the zeroes of this polynomial algebraicly (which should be doable, since it is an exercise from an old exam). Am I missing something? Any suggestions are greatly appreciated!