Random sequence with positive Lyapunov exponent?

Question

Consider the following self-adjoint matrix

$A_X = \begin{pmatrix} 0 & -i \\ i & X \end{pmatrix},$ where $i$ is the imaginary unit and $X$ is a uniformly distributed random variable on some interval $[-\varepsilon,\varepsilon].$

Now, take the product

$$M_n = A_{X_n} \cdot...\cdot A_{X_1}$$

where $X_1,...,X_n$ are iid copies of $X.$

We observe that $\operatorname{det}(M_n)=(-1)^n.$ Thus, $M_n$ has an eigenvalue that is at least of modulus $1.$

We then consider the Lyapunov exponent $$ \mu = \lim_{n \to \infty}\frac{1}{n} \log \Vert M_n\Vert.$$

By the above reasoning we have that $\mu \ge 0.$

Is it true that $\mu>0?$

Answer 1

We have the classical result of Furstenberg: consider a random walk $\mu$ on $SL(n,\mathbb{C})$ with finite first moment such that the semigroup generated by the support of the measure is strongly irreducible and unbounded. Then its greater Lyapunov exponent is positive.

(You can find this statement e.g. in this survey of Alex Furman in the section "Furstenberg's condition for positive growth")

Finite first moment means that $\int \log N(g) d \mu(g) < \infty $ where $N(g)$ is any matrix norm.

A semigroup is strongly irreducible if it does not stabilize any finite union of proper subspaces: there is no family $V_1,\dots,V_k$ of subspaces, $V_i \subsetneq \mathbb{C}^n$ such that $\bigcup_i V_i$ is left globally invariant by the semigroup.

It can be checked that the random walk you consider satisfies those requirements (in your case, $\mu$ is the law of the matrix $A_X$).