determinantal variety

For $A\in M_{n}(k)$ and $v\in M_{n \times 1}(k)$, $X:=\{(A,v) \in \mathbb{A}^{n^{2}} \times \mathbb{A}^n \mid {\rm rank}(Av \mid v) \leq 1\}$ is an affine algebraic set in $\mathbb{A}^{n^2}\times \mathbb{A}^n$. Can we say that $X$ is a determinantal variety? If $X$ a determinantal variety, then can we say that $X$ is irreducible since the maximal rank of $(Av \mid v)$ is $2$?

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I don't think $X$ is a determinantal variety. Think about the case $n=2$. Denote $A=(a_{ij})$ and $v=(v_{1}, v_{2})^{t}$. In this case, $X$ is the zeros of the polynomial $(a_{11}v_{1} + a_{12}v_{2})v_{2} - (a_{21}v_{1} + a_{22}v_{2})v_{1}$, which is not equal to the zeros of $\det(A), \det \left( \begin{array}{cc} a_{11} & v_{1} \\ a_{21} & v_{2} \end{array} \right), \det \left( \begin{array}{cc} a_{12} & v_{1} \\ a_{22} & v_{2} \end{array} \right)$.

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