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  "body": "\n## Definition\n\nLet $I = [t_0, \\infty)$ and let $A \\colon I \\to \\R^{n \\times n}$ be continuous.\nConsider the [[linear system|037ZNY7NAG]] $x' = A(t) x$, and fix a [[norm|0V2WP9B1MF]] $\\norm{\\cdot}$ on $\\R^n$.\nLet $x$ be a nonzero [[solution|07VX34QXAV]] of this system on $I$.\nThe *Lyapunov exponent* (or characteristic exponent) of $x$ is\n$$\\lambda(x) = \\limsup_{t \\to \\infty} \\frac{1}{t} \\ln \\norm{x(t)} \\in [-\\infty, \\infty],$$\nwhere $\\limsup$ is the [[limit superior|0T3RZQXDYD]].\n\n## Remarks\n\nThe value does not depend on the choice of norm, because any two norms on $\\R^n$ are within constant factors of each other.\nA nonzero solution never vanishes, by uniqueness of solutions, so the logarithm is defined for all $t$.\nIf $A$ is bounded on $I$, then every exponent is finite.\n\nExample: for constant $A$ with an eigenvalue $\\mu$ and eigenvector $v$, the solution $x(t) = e^{\\mu t} v$ satisfies $\\norm{x(t)} = e^{(\\operatorname{Re} \\mu) t} \\norm{v}$.\nSo $\\frac{1}{t} \\ln \\norm{x(t)} = \\operatorname{Re} \\mu + \\frac{1}{t} \\ln \\norm{v} \\to \\operatorname{Re} \\mu$, and $\\lambda(x) = \\operatorname{Re} \\mu$.\n\nNon-example: the zero solution has no Lyapunov exponent in this sense, since $\\ln \\norm{0}$ is undefined; some authors set it to $-\\infty$.\n\nFor a nonlinear system $x' = f(x)$ with $f$ of class $C^1$, the exponents along a trajectory are those of the variational equation $y' = Df(x(t))\\, y$.\nThe largest of them measures the rate of separation of nearby trajectories.\n",
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