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 $$ A_{n}(x)=A\bigl(x+(n1) \alpha\bigr)\cdot A\bigl(x+(n2) \alpha\bigr) \cdots A(x+\alpha)\cdot A(x). $$ 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 $$\operatorname{LE}(\lambda) > 0? $$
1 Answer
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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.

1$\begingroup$ The Micheal Herman method can only get the Lyapunov exponent here to be greater than or equal to zero, but what I need here is greater than zero. $\endgroup$– xia xuCommented Apr 26, 2021 at 1:03