Lyapunov exponent

Remarks

The value does not depend on the choice of norm, because any two norms on $\R^n$ are within constant factors of each other. A nonzero solution never vanishes, by uniqueness of solutions, so the logarithm is defined for all $t$. If $A$ is bounded on $I$, then every exponent is finite.

Example: for constant $A$ with an eigenvalue $\mu$ and eigenvector $v$, the solution $x(t) = e^{\mu t} v$ satisfies $\norm{x(t)} = e^{(\operatorname{Re} \mu) t} \norm{v}$. So $\frac{1}{t} \ln \norm{x(t)} = \operatorname{Re} \mu + \frac{1}{t} \ln \norm{v} \to \operatorname{Re} \mu$, and $\lambda(x) = \operatorname{Re} \mu$.

Non-example: the zero solution has no Lyapunov exponent in this sense, since $\ln \norm{0}$ is undefined; some authors set it to $-\infty$.

For a nonlinear system $x' = f(x)$ with $f$ of class $C^1$, the exponents along a trajectory are those of the variational equation $y' = Df(x(t))\, y$. The largest of them measures the rate of separation of nearby trajectories.