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  "body": "\n## Statement\n\nLet $F$ be an algebraically closed field, let $V$ be a finite-dimensional vector space over $F$ with $V \\ne 0$, and let $T \\colon V \\to V$ be a linear map.\nThen $T$ has at least one [[eigenvalue|0BQAA2BK3Q]] in $F$.\n\n## Remarks\n\nThe [[characteristic polynomial|0J5EMJRRFM]] of $T$ has degree $\\dim V \\ge 1$, so it has a root in $F$.\nThe statement applies to $F = \\C$, by the fundamental theorem of algebra.\nIt fails over $\\R$: the rotation of $\\R^2$ by $90$ degrees has no real eigenvalue.\nIt also fails in infinite dimensions: the map $p(x) \\mapsto x\\, p(x)$ on $\\C[x]$ has no eigenvalue.\n\n## Explanations\n\n### L1 human\n\nOver the complex numbers, every square matrix has at least one eigenvector.\nThe reason is that every nonconstant complex polynomial has a root.\nOver the real numbers, a rotation can move every line, so no real eigenvector exists.\n",
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