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  "body": "\n## Statement\n\nLet $F$ be a field, let $V$ be a finite-dimensional vector space over $F$, and let $T \\colon V \\to V$ be a linear map.\nLet $p_T$ be the [[characteristic polynomial|0J5EMJRRFM]] of $T$, and let $\\lambda \\in F$.\nThen $\\lambda$ is an [[eigenvalue|0BQAA2BK3Q]] of $T$ if and only if $p_T(\\lambda) = 0$.\n\n## Remarks\n\nThe proof: $\\lambda$ is an eigenvalue if and only if $\\lambda \\id_V - T$ is not injective, if and only if $\\det(\\lambda \\id_V - T) = 0$.\nSo $T$ has at most $\\dim V$ eigenvalues.\nFor $A = \\begin{pmatrix} 2 \u0026 1 \\\\ 1 \u0026 2 \\end{pmatrix}$, $p_A(t) = t^2 - 4t + 3 = (t - 1)(t - 3)$, so the eigenvalues are $1$ and $3$.\n\n## Explanations\n\n### L1 human\n\nTo find the eigenvalues of a matrix, compute its characteristic polynomial and find its roots.\nEach root is an eigenvalue, and each eigenvalue is a root.\nThen, for each eigenvalue $\\lambda$, solve $(A - \\lambda I)x = 0$ to find the eigenvectors.\n",
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