Assume that $(T, A)$ is a linear cocycle such that $T:X\rightarrow X$ is a homemorphism on compact metric space $X$ and $A:X\rightarrow SL(2, \mathbb{R})$ is a continuous function.


We say that an invariant probability measure $\mu$ is aperiodic for $T$ if set of periodic points of has zero measure.


There is a famous result due to Bochi-Mane that said :


Given any invertible aperiodic ergodic system $(T,\mu)$ on a compact Hausdorff space, every continuous cocycle $A:X→SL(2, \mathbf{R})$ which is not uniformly hyperbolic can be approximated in the $C^{0}$ topology by another whose Lyapunov exponents vanish at $\mu$-almost every point.


Consider the following examples:
$$
A\_1=diag(2, \frac{1}{2}) , A\_{2}=diag(1, 1)
 $$
$$
A\_1=diag(2, \frac{1}{2}) , A\_{2}=antidiag(1, 1)$$
Would one show me how to approximate the above linear cocycles such that it has zero Lyapunov exponent?



Example of zero Lyapunov exponentes
I’m guessing you mean the base dynamical system to be a Bernoulli shift with the unperturbed cocycle being $A(x)=A\_{x\_0}$? For the second one, no perturbation is necessary. The Lyapunov exponents are already 0. 


For the first system, pick an $N$. Now if $x\_{-k-1}\ldots x\_{N-k}=1222\ldots 21$ for some $0\le k\le N-1$, replace $I$ by a rotation by $\pi/(2N)$, so that the product over the block of 2’s is $\text{antidiag}(1,-1)$. This has zero Lyapunov exponents just as above. 


