Topological entropy of non-hyperbolic toral automorphism I have a matrix $ A = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$ and its associated toral automorphism $f_A$. That is;
$\begin{equation} {f}^{n}_A := ( x + ny \text{ mod } 1 , y ) \end{equation}$

I am trying to show that its topological entropy, $h_{\text{top}}(f_A) = 0$. My thought process would be to construct its $(n,\epsilon)$ - spanning sets and then show that their exponential growth rate is 0, and thus the topological entropy is equal to 0. However I am struggling to construct them, so any tips would be appreciated.
I cannot use that if an toral automorphism has a repeated eigenvalue of 1 then $h_{\text{top}}(f_A) = \log(1) = 0$.
 A: Since $f$ is a continuous map on the compact metric space $\mathbb{T}^2$, one form of the variational principle says that
$$
h_{top}(f) = \sup_{\mu \in \mathscr{M}^f_e(\mathbb{T}^2)} h_\mu(f)
$$
where $\mathscr{M}^f_e(\mathbb{T}^2)$ is the set of all ergodic $f$-invariant probability measures on $\mathbb{T}^2$.  So it is enough to show that $h_\mu(f) = 0$ for any ergodic $f$-invariant measure $\mu$.
A key observation is the following.  If $L_y = \{(x,y): x \in \mathbb{T}\}$ is the horizontal line in $\mathbb{T}^2$ at height $y$, then each $L_y$ is an $f$-invariant set.  Therefore if $\mu$ is any ergodic measure, there is a unique $y$ such that $\mu$ is supported on $L_y$.  Now when restricted to the line $L_y$, the action of $f$ is clearly isomorphic to the circle rotation $x \mapsto x+y$ on $\mathbb{T}$.  Since $\mu$ is supported on a line, it is naturally identified with a measure on $\mathbb{T}$, and so our system $(\mathbb{T}^2, f, \mu)$ is naturally isomorphic to the circle rotation $(\mathbb{T}, x \mapsto x+y, \mu)$.
It's well known that 1-dimensional circle rotations have zero topological entropy, so in particular any invariant measure for a circle rotation has zero measure entropy as well.  Therefore $h_\mu(f) = 0$.  Since this holds for any ergodic $f$-invariant measure $\mu$ we are done.
