Why are hyperbolic toral automorphisms (e.g. Arnold's cat map) ergodic?

Let $\varphi : \mathbb{T}^2 \to \mathbb{T}^2$ be a hyperbolic automorphism of the torus, induced by a linear map $A : \mathbb{R}^2 \to \mathbb{R}^2$ of determinant $\pm 1$ with no eigenvalues of modulus 1. What is an easy way to prove that $\varphi$ is ergodic?

It's true that the stable and unstable manifolds at $(0,0) \in \mathbb{T}^2$ (projections of the eigenspaces of $A$ to the torus) are dense in the torus, which can be used to prove that such maps are topologically mixing. An example is Arnold's cat map.

Arnold and Avez show in Ergodic Problems in Classical Mechanics that Arnold's cat map is ergodic by proving that it has "Lebesgue spectrum", which implies that it is strong mixing, which implies that it is ergodic. Is there a more direct way to prove this?

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A few weeks ago, someone here asked for a proof that these maps are chaotic. I found one in a textbook by Elaydi on discrete dynamical systems. The hardest part was proving the map transitive. –  Gerry Myerson Dec 20 '12 at 4:57
@GerryMyerson I saw that thread. I found a simpler proof of a stronger statement (topologically mixing) in the book by Broer and Takens (proposition 2.15, page 111). But I don't see how topological transitivity or mixing can help me prove that these maps are ergodic. –  Ricardo Buring Dec 20 '12 at 11:25

$\varphi$ is ergodic if and only if every $f\in L^2(\mathbb{T}^2)$ such that $f\circ \varphi = \varphi$ is constant function.
Now let's use this criterion to prove $\varphi$ is ergodic. Suppose $f\in L^2$ with $f\circ \varphi = \varphi$. Decompose $f$ into its Fourier series $$f = \sum_{m,n\in \mathbb{Z}} = \alpha_{(m\,n)}e^{2\pi imx}e^{2\pi iny},$$ with coefficients $\alpha_{(m\,n)}\in \mathbb{C}$. Then, if $\varphi$ is given by the matrix $$A = \left(\begin{matrix} a & b\\c & d\end{matrix}\right),$$ it is easy to compute that $$f\circ \varphi = \sum_{m,n}\alpha_{(m\,n)}e^{2\pi i(ma+nc)x}e^{2\pi i(mb+nd)y}.$$ Since $f$ is invariant, the Fourier series for $f$ and $f\circ \varphi$ must agree, so $\alpha_{(m\,n)} = \alpha_{(ma+nc\,mb+nd)}$ for all $m,n\in \mathbb{Z}$. We can express this more simply as follows. If $v = (m\,\,n)\in \mathbb{Z}^2$, then $\alpha_v = \alpha_{vA}$. By iterating, $\alpha_v = \alpha_{vA^k}$ for each $k\in \mathbb{Z}$.
Suppose that $v\in \mathbb{Z}^2$. Either the sequence $vA^k$ of vectors with integer coordinates is periodic, or else $\|vA^k\|\to \infty$ as $k\to \infty$. Note that the first case cannot happen unless $v = 0$, since if $v = vA^k$ for some $k$, then $A^k$ would have $1$ as an eigenvalue, which contradicts the assumption of hyperbolicity. Thus either $v = 0$, or $\|vA^k\|\to \infty$. Suppose $v\neq 0$. Since $f\in L^2$, the coefficients $\alpha_{(m\,n)}\to 0$ as $\|(m\,\,n)\|\to \infty$, and thus $\alpha_v = \alpha_{vA^k}\to 0$, i.e., $\alpha_v = 0$. We have therefore shown that the only way $\alpha_v$ can be nonzero is if $v = 0$. The Fourier series for $f$ is then $f = \alpha_0$, so $f$ is a constant function.