Consider irrational rotation $T:S^1\to S^1, T(x) = x + \alpha$ where $\alpha\notin \mathbb{Q}$ (you may assume additional number theoretic properties of $\alpha$, say $\alpha = \sqrt{2}$ is already interesting for me). Assume that we have a very concrete smooth map $g:S^1 \to GL(n, \mathbb{R})$ (it is given by some rather unpleasant but finite formula). I want to understand the behavior of $f(m, x) = g(T^{m-1}x)g(T^{m-2}x)\ldots g(Tx)g(x)$ for big $m$. As far as I know, this is done by the multiplicative ergodic theorem which reads as follows:
There are $k\le n$ real numbers $\lambda_1 > \ldots > \lambda_k$ and for a.e. $x\in S^1$ there are $k$ subspaces $\mathbb{R}^n = V_1(x) \supset V_2(x) \supset \ldots \supset V_k(x)$ such that for $v\in V_l(x) \backslash V_{l+1}(x)$ we have $\frac{\log ||f(m, x)v||}{m}\to \lambda_l$.
I want two things:
Some rigorous algorithm to prove that none of $\lambda_l$ is equal to $1$ so that if I will do some finite computation with the function $g$ with enough precision I will prove it;
Quantitive and, more important, uniform in $x$ limit from above. Actually, I will be satisfied with the bound of the following form: for $v\in V_l(x)$ we have $$||f(m, x)v|| \le C_\varepsilon (\lambda_l + \varepsilon)^m ||v||,$$ where $\varepsilon > 0$ is arbitrary small and $C_\varepsilon$ does not depends on $x$.
Note that for $n = 1$ both of these tasks are doable: in that case $\lambda_1 = \int \log |g(x)|dx$ so we just have to calculate this integral with enough precision to deduce 1. For 2. we can subdivide $S^1$ into many small arcs and approximate the integral by the Riemann sum and then say that irrational rotation visits each arc approximately equal number of times (to quantify the first part we need some knowledge about modulus of continuity of $g$ (that is smoothness), while for the second part we need to know that $\alpha$ does not have any good rational approximations though with some abstract $C_\varepsilon$ the bound from 2. is true for any $g\in C(S^1)$ and any $\alpha \notin\mathbb{Q}$).