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  "body": "\n## Statement\n\nLet $A$ be a real $n \\times n$ matrix and consider the [[linear system with constant coefficients|0VJF99CZ0Z]] $x' = A x$ on $\\mathbb{R}^n$, with the equilibrium point $0$.\n(1) The point $0$ is [[asymptotically stable|05Y1JAM3Y5]] if and only if each eigenvalue $\\lambda$ of $A$ has $\\operatorname{Re} \\lambda \u003c 0$.\n(2) If some eigenvalue $\\lambda$ of $A$ has $\\operatorname{Re} \\lambda \u003e 0$, then $0$ is unstable.\n(3) The point $0$ is [[Lyapunov stable|092S6VBECQ]] if and only if two conditions hold. Each eigenvalue has $\\operatorname{Re} \\lambda \\le 0$. Each eigenvalue with $\\operatorname{Re} \\lambda = 0$ has equal algebraic and geometric multiplicities.\n\n## Remarks\n\nThe proof uses the [[matrix exponential|0YC8JT513J]]. The entries of $e^{tA}$ are sums of terms $t^k e^{\\lambda t}$. Here $\\lambda$ is an eigenvalue and $k$ is less than the size of its largest Jordan block.\nIn case (1), there are constants $C, \\alpha \u003e 0$ with $\\|e^{tA}\\| \\le C e^{-\\alpha t}$ for $t \\ge 0$. So the origin is globally and exponentially stable.\nFor $A = \\begin{pmatrix} 0 \u0026 1 \\\\ 0 \u0026 0 \\end{pmatrix}$, both eigenvalues are $0$ but the solution $(t, 1)$ is unbounded. So $0$ is unstable, in accordance with (3).\nThe Routh–Hurwitz criterion tests the condition of (1) from the coefficients of the characteristic polynomial, without computing the roots.\n\n## Explanations\n\n### L1 human\n\nFor a linear system, the eigenvalues decide everything.\nEigenvalues with negative real part give decaying solutions, and an eigenvalue with positive real part gives a growing solution.\n",
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