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Lyapunov exponent of singular values and operator norm

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

  1. Michal

    Michal Guest

    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 $\sigma_{3}$?.

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