# Lyapunov exponent of singular values and operator norm

Discussion in 'Mathematics' started by Michal, Oct 8, 2018.

1. ### MichalGuest

Consider a product of i.i.d. $3\times 3$ random matrices $A_{i}$ (with $\mathbb{E}\log\left\Vert A_{i}\right\Vert <\infty$) acting on a non-zero vecor $V \in \mathbb{R^3}$, i.e. $$A_{n}\cdots A_{1}X.$$

According Oseledets theorem,Lyapunov exponents describe the exponential growth properties of $$\left\Vert A_{n}\cdots A_{1}X\right\Vert .$$ Thus we define the the Lyapunov exponent as $$\lambda\left(X\right):=\lim_{n\to\infty}\frac{1}{n}\log\left\Vert A_{n}\cdots A_{1}X\right\Vert \;\;(1).$$

Let us assume for every points we have three Lyapunov exponent $$\lambda_{3}\leq \lambda_{2}\leq\lambda_{1}.$$

Now, let $\sigma_{i,n}$ be the $i$th singular value of the matrix product $A_{n}\cdots A_{1}$ such that $$\sigma_{3,n}\leq\sigma_{2,n}\leq\sigma_{1,n}.$$

then According to the Muliplicative Ergodic Theorem $$\lim_{n\to\infty}\frac{1}{n}\log\sigma_{i,n}=\lambda_{i}.\;\;(2)$$

Question what is related between (1) and (2) for second singular value?

I mean we know that $\sigma_{1}=sup_{v}\frac{||Av||}{||V||}$ and $\sigma_{3}=inf_{v}\frac{||Av||}{||v||}$, thus relation between (1) and (2) are clear.Does one know can we have explicitly formula for $\sigma_{2}$ like $\sigma_{1}$and $\sigma_{3}$?.