Let $\alpha\in \mathbb{R} / \mathbb{Q}$, \begin{equation} A(x)=\left(\begin{array}{ll} \frac{1}{{\lambda}^2}2 \cos 2\pi x 1& 2\lambda \cos 2\pi x\frac{1}{{\lambda}} \\ \frac{1}{{\lambda}} & 1 \end{array}\right) \end{equation} The Lyapunov exponent (LE) of $(\alpha, A)$ is given by \begin{equation} LE(\lambda)=\lim _{n \rightarrow \infty} \frac{1}{n} \int_{\mathbb{R} / \mathbb{Z}} \ln \left\A_{n}(x)\right\ d x \end{equation} where $$ A_{n}(x)=A(x+(n1) \alpha) A(x+(n2) \alpha) \cdots A(x). $$ How to prove there is a large positive number $\lambda_0>0$ so that when $\lambda>\lambda_0,$ the Lyapunov exponent is $LE>0$ ?
I think you can use the main result of the paper of Micheal 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 xuApr 26 at 1:03