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The set of pairs $(A,v)\in Mat_{n\times n}(k)\times k$ such that $v$ is an eigenvector of $A$. I saw that this set is an affine algebraic set since the condition is equivalent to ${\rm rank}(A,v) \leq 2$. But I don't know the meaning of ${\rm rank}(A,v)$.

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    Then that is false? I saw that http://www.math.purdue.edu/~dvb/preprints/algeom.pdf p.42011-10-05

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If $A$ is a $n \times n$ matrix and $v$ is a vector, then $v$ is an eigenvector of $A$ if and only if there exists $\lambda \in k$ such that $Av = \lambda v$ if and only if $\mathrm{rank} (Av \mid v) \leq 1$, where $(Av \mid v) \in Mat_{n,2}(k)$ is the matrix whose columns are $Av$ and $v$. It is clear that the condition $\mathrm{rank} (Av \mid v) \leq 1$ gives polynomial relations among entries of $A$ and coordinates of $v$, so the set $ X = \{ (A,v) \in Mat_{n,n}(k) \times k^n \mid \mathrm{rank} (Av \mid v) \leq 1 \} $ is an algebraic subset of $Mat_{n,n}(k) \times k^n \cong k^{n^2 + n}$.