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  "body": "\n## Summary\n\nLet $U \\subseteq \\mathbb{R}^n$ be open, let $f : U \\to \\mathbb{R}^n$ be continuously differentiable, and let $x^*$ be an [[equilibrium point|0G4X36AE0X]] of $x' = f(x)$.\nLet $A = Df(x^*)$ be the Jacobian matrix of $f$ at $x^*$. The linearisation of the system at $x^*$ is the linear system $y' = A y$, where $y$ stands for $x - x^*$.\nIt comes from the first-order Taylor expansion $f(x) = A (x - x^*) + o(\\|x - x^*\\|)$.\n\n## Remarks\n\nIf each eigenvalue of $A$ has negative real part, then $x^*$ is asymptotically stable for the nonlinear system. If one eigenvalue has positive real part, then $x^*$ is unstable.\nIf some eigenvalue has zero real part and none has positive real part, the linearisation does not decide stability. For $x' = -x^3$ and $x' = x^3$, the linearisation at $0$ is $y' = 0$ in both cases, but the first is stable and the second is not.\nNear a hyperbolic equilibrium, the [[Hartman–Grobman theorem|0S8WX8A4V9]] says that the linearisation describes the phase portrait up to a homeomorphism.\n\n## Explanations\n\n### L1 human\n\nNear an equilibrium, a smooth vector field looks like a linear one.\nWe replace the field by its best linear approximation and study the simpler linear system.\nThis works well unless the linear system is on the border between stable and unstable.\n",
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