I have the equation:

$$

dot{x}_i = F_i(x)

tag{1}

$$

with $xin mathbb{R}^n$. To deal with the Lyapunov exponents, we write the equation for small displacements $delta x_i$:

$$

dot{delta x}_i = sum_j frac{partial}{partial x_j} F_i(x) delta x_j

tag{2}

$$

The rate of increase of the vectors is related to the Lyapunov exponent. Here I assume that the system is Lyapunov regular.

The definition of “Lyapunov vector” that I saw is the following. First, a matrix $Y_{i,j}(t)$ is considered, with equation:

$$

dot{Y_{i,j}}= sum_k frac{partial}{partial x_k} F_i Y_{k,j}

$$

Then a matrix $M$ is defined as:

$$

M = lim_{tto +infty} frac{log Y Y^T}{t}

tag{3}

$$

According to this definition, the Lyapunov exponents and vectors are the eigenvalues and eigenvectors of $M$.

I tried to investigate how the Lyapunov vectors depend on the starting point $x$, taking two points $x_A$ and $x_B$ along a trajectory: $x_A=x(t=0)$ and $x_B=x(t=tau)$.

I calculate $M$ in the two points:

$$

M(x_A) = lim_{tto +infty} frac{log Y(x_A,t) Y^T(x_A,t)}{t} tag{4}

$$

and:

$$

M(x_B) = lim_{tto +infty} frac{log Y(x_B,t) Y^T(x_B,t)}{t} tag{5}

$$

Since $Y$ is a cocycle:

$$

Y(x_A,t) = Y(x_B, t-tau) Y(x_A, tau)

$$

Then:

$$

M(x_A) = lim_{tto +infty} frac{log Y(x_B, t-tau) Y(x_A, tau) Y^T(x_A, tau) Y^T(x_B, t-tau)}{t} tag{6}

$$

If the $Y$s commuted, we would write the logarithm of the products as the sum of logarithms of the factors, and thus get $M(x_A)=M(x_B)$ (Eq. 6 would give the same limit as Eq. 5, since $tau$ is constant), i.e. $M$ would be constant along a trajectory. However, they do not commute, so maybe $M$ changes along the trajectory.

**My question is:** Is this correct? Actually, according to a previous answer I got on MO, **it is believed that $M$ changes if we evaluate it starting from $x_A$ or $x_B$ along the same trajectory.** Moreover, it seems that it is believed that the eigenvectors of $M$ evolve along a trajectory according to Eq. (2). Is this correct? If so, how can we see it from Eq. (6)?