Positiveness of the largest Lyapunov exponent

Question

Let $\alpha\in \mathbb{R} / \mathbb{Q}$, let $A(x)$ be the $2$-by-$2$ matrix $$ A(x)=\begin{pmatrix} \dfrac{1}{{\lambda}^2}-2 \cos 2\pi x -1& 2\lambda \cos 2\pi x-\dfrac{1}{{\lambda}} \\ \dfrac{1}{{\lambda}} & -1 \end{pmatrix}, $$ and for $n\ge1$, let $$ A_{n}(x)=A\bigl(x+(n-1) \alpha\bigr)\cdot A\bigl(x+(n-2) \alpha\bigr) \cdots A(x+\alpha)\cdot A(x). $$ The Lyapunov exponent (LE) of $(\alpha, A)$ at $\lambda\in\mathbb R_{>0}$ is given by \begin{equation} \operatorname{LE}(\lambda)=\lim _{n \rightarrow \infty} \frac{1}{n} \int_{\mathbb{R} / \mathbb{Z}} \ln \bigl\|A_{n}(x)\bigr\| d x. \end{equation} How can one prove that when $\lambda>0$ is large enough, the Lyapunov exponent satisfies $$\operatorname{LE}(\lambda) > 0? $$

Answer 1

I think you can use the main result of the paper of Michael Herman "Une méthode pour minorer..." https://link.springer.com/article/10.1007%2FBF02564647 in order to show that LE is positive when $\lambda$ is sufficiently large.