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  "body": "\n## Proof\n\nWe find a root of the characteristic polynomial.\n\n1. Let $n = \\dim V$. Since $V \\ne 0$, $n \\ge 1$. Fix a basis $B$ of $V$ and let $A = [T]_B$.\n2. By the definition of the [[characteristic polynomial of $T$|0BZ81AJNWF]], $p_T(t) = p_A(t) = \\det(t I_n - A)$.\n3. By the rule for the [[trace as a coefficient|0FRYPZ6D0X]], $p_A(t) = t^n - (\\tr A)\\, t^{n-1} + \\cdots$. So $p_T$ has leading coefficient $1$ and [[degree|0DPF6WRZW6]] $n \\ge 1$.\n4. The field $F$ is [[algebraically closed|02SJG1HC32]], so $p_T$ has a [[root|00EHX6FWC0]] $\\lambda \\in F$. That is, $p_T(\\lambda) = 0$.\n5. By the theorem that [[eigenvalues are the roots of the characteristic polynomial|053QVHF6N7]], $\\lambda$ is an [[eigenvalue|0ZND7NQ970]] of $T$.\n\n## Idea\n\nThe characteristic polynomial of $T$ has degree $\\dim V$, which is at least $1$. Over an algebraically closed field such a polynomial has a root, and each root is an eigenvalue.\n",
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