= Solution
For $n\geq1$, an <eigenvalue> is a number $\lambda\in\mathbb C$ for which there is a nonzero <vector> $v$ satisfying $Av=\lambda v$. Equivalently, $\ker(A-\lambda I)\ne\{0\}$, or $\det(A-\lambda I)=0$. The <characteristic polynomial> $\det(tI-A)$ has degree $n$ and leading coefficient one. By the <fundamental theorem of algebra> it has a complex root $\lambda$. At that root $A-\lambda I$ is a <singular matrix>, so it has a nonzero <kernel> <vector>, proving that $A$ has an <eigenvalue>.
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