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 $\mu>0?$


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 $A_X$).

  • $\begingroup$ sorry, can you elaborate a bit on the assumptions, also you say $\operatorname{SL}(n,\mathbb R)$, however, this matrix is complex-valued. $\endgroup$
    – Kung Yao
    Jun 22 at 16:37
  • $\begingroup$ I've added some details. The same results works for $\mathbb{R}$ or $\mathbb{C}$. $\endgroup$
    – FMB
    Jun 22 at 16:50
  • $\begingroup$ I am really sorry, but would you be able to elaborate on this subspace condition, what it actually means here? $\endgroup$
    – Kung Yao
    Jun 22 at 18:18
  • $\begingroup$ I've expanded the definition. $\endgroup$
    – FMB
    Jun 22 at 19:00
  • $\begingroup$ sorry, but how did you verify that condition? $\endgroup$
    – Kung Yao
    Jun 22 at 20:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.