A question on Lyapunov exponents associated with a fixed point of a vector field.

Question

The Lyapunov exponent $\chi(x_{0}, e)$ in a direction $e \in \mathbb{R}^{n}$ along the trajectory $x(t,x_{0})$ of a vector field $\dot{x} = f(x)$ through a point $x_{0}$ is defined to be

$$\chi(x_{0},e) = \lim_{t \to \infty}\frac{1}{t} \log \frac{||X(t;x(t,x_{0}))e||}{||e||}$$ where $X(t;x(t,x_{0}))$ is the fundamental solution matrix of the linearisation of the vector field orbit about $x(t,x_{0})$. Here $x \in \mathbb{R}^{n}$.

Let us consider the case of the the orbit being a fixed point. It is known that then $\chi$ is just the real parts of the eigenvalues associated with the matrix of the vector field linearised about the fixed point. How does one prove this? Why do the imaginary parts of e-values do not contribute to $\chi$?

Answer 1

When $x_0$ is a critical point we have $X(t;x(t,x_{0}))=e^{At}$ with $A=d_{x_0}f$.

So what matters are the vectors in each generalized eigenspace.

For example, if $Av=\lambda v$ with $v\ne0$, then $e^{At}v=e^{\lambda t}v$ and so $\chi(x_{0},v)={\rm Re\,}\lambda$. This gives what you want (by taking sums of generalized eigenvectors).