Let $k$ be an [algebraically closed field](/page/Algebraically%20Closed%20Field), let $V$ be a finite-dimensional [vector space](/page/Vector%20Space) over $k$ with $\dim_k V \geq 1$, and let $T: V \to V$ be a $k$-[linear map](/page/Linear%20Map). Then there exist $\lambda \in k$ and $v \in V \setminus \{0\}$ such that
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\begin{align*}
T(v)=\lambda v.
\end{align*}
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Equivalently, every $k$-linear endomorphism of a nonzero finite-dimensional vector space over an algebraically closed field has an eigenvector.