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  "body": "\n## Summary\n\nThe eigenvalue method solves a [[homogeneous linear system with constant coefficients|0VJF99CZ0Z]] $x' = A x$ with the eigenvalues and eigenvectors of $A$.\nIf $A v = \\lambda v$ with $v \\ne 0$, then $x(t) = e^{\\lambda t} v$ is a solution.\nIf $A$ has $n$ linearly independent eigenvectors $v_1, \\dots, v_n$ with eigenvalues $\\lambda_1, \\dots, \\lambda_n$, then each solution is $x(t) = c_1 e^{\\lambda_1 t} v_1 + \\dots + c_n e^{\\lambda_n t} v_n$ for unique constants $c_k$.\nFor a complex eigenvalue $\\lambda$ of a real matrix with eigenvector $v$, the real and imaginary parts of $e^{\\lambda t} v$ are two real solutions.\n\n## Remarks\n\nExample: $A = \\begin{pmatrix} 1 \u0026 1 \\\\ 4 \u0026 1 \\end{pmatrix}$ has the eigenvalues $3$ and $-1$, with eigenvectors $(1, 2)$ and $(1, -2)$. So $x(t) = c_1 e^{3t} (1, 2) + c_2 e^{-t} (1, -2)$.\nIf $A$ has too few eigenvectors, generalized eigenvectors give solutions of the form $e^{\\lambda t}$ times a polynomial in $t$ with vector coefficients.\nThe eigenvalues decide the [[stability of the system|0SKVEMXJWF]].\n\n## Explanations\n\n### L1 human\n\nAlong an eigenvector, the matrix $A$ acts as multiplication by a number.\nSo on this line the system is the simple equation $y' = \\lambda y$, with solution $e^{\\lambda t}$.\nWrite the start vector in a basis of eigenvectors, and let each part grow or decay at its own rate.\n",
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