Consider $\mathbb{T}^2=\mathbb{R}^2/\mathbb{Z}^2$ and the following unimodular matrix \begin{equation} A=\begin{bmatrix} 2& 1\\ 1& 1 \end{bmatrix}. \end{equation} We know $F\colon\mathbb{R}\rightarrow\mathbb{R}$, $F(\mathbf{v})=A\mathbf{v}$ induces a well-defined map $f\colon\mathbb{T}^2\rightarrow\mathbb{T}^2$, $f([\mathbf{v}])=[A\mathbf{v}]$ since $F(\mathbb{Z})=\mathbb{Z}$. Now I want to show $f$ is topological transitive, i.e., for any $U,V\subset \mathbb{T}^2$ open, there exists $n\in\mathbb{Z}$ such that $f^{(n)}(U)\cap V\not=\varnothing$ (since $A^{-1}$ is also unimodular, so $f$ is invertible).
Here is my attempt: I try to show $f$ has a dense orbit $\{f^{n}(p)\colon n\in\mathbb{Z}\}$ in $\mathbb{T}^2$. If so, then take any $x\in U$, $y\in V$, there exists subsequence $f^{(m_j)}(p)\rightarrow x$, $f^{(n_k)}(p)\rightarrow y$. So there exists $N_1,N_2\in\mathbb{Z}$ such that $f^{(N_1)}(p)\in U$, $f^{(N_2)}(p)\in V$. Then $p\in f^{(-N_1)}(U)$ and $f^{(N_2-N_1)}(U)\cap V\not=\varnothing$.
But the question is, I can't make sure such dense orbit exists. $A$ has two irrational eigenvalues \begin{equation} \lambda_1=\frac{3+\sqrt{5}}{2},\,\mathbf{v}_1=\begin{bmatrix} 1\\ \frac{-1+\sqrt{5}}{2} \end{bmatrix},\,\,\,\lambda_2=\frac{3-\sqrt{5}}{2},\,\mathbf{v}_2=\begin{bmatrix} 1\\ \frac{-1-\sqrt{5}}{2} \end{bmatrix}. \end{equation} I guess that $\left\{[\lambda_1^n\mathbf{v}_1]\colon n\in\mathbb{Z}\right\}$, $\left\{[\lambda_2^n\mathbf{v}_2]\colon n\in\mathbb{Z}\right\}$ are two dense orbits. We already know $\{[A^nq]\colon n\in\mathbb{Z}\}$ is a periodic orbit for every $q\in\mathbb{Q}^2$. So I am guessing the irrationality of $\lambda_1$, $\lambda_2$ produces density.
Could you provide me with any hints?
