Is the set of matrices that has an eigenvector orthogonal to a fixed vector Zariski closed? This is related to the question I asked.
Suppose $x \in \mathbb C^n$ is a fixed vector$.$ Define a set
$$\begin{align*}
\mathcal E = \{A \in M_n(\mathbb C): \exists \text{ an eigen-pair }(\lambda, v) \text{ of } A, \text{i.e., }Av = \lambda v \text{ such that } v \perp x \}.
\end{align*}$$
It is shown in the answer that $\mathcal E$ is closed in the Euclidean topology. I want to know whether this set is closed in Zariski topology.
 A: Yes.  We may assume $x=(0,0,\dots,1)$.  Then $\mathcal{E}$ is the set of $A$ such that there exists $\lambda\in\mathbb{C}$ such that the matrix formed by the first $n-1$ columns of $\lambda I-A$ has rank less than $n-1$.  This happens for a particular value of $\lambda$ iff all of the $(n-1)\times(n-1)$ minors of the first $n-1$ columns of $\lambda I-A$ vanish.  Each of these minors is a polynomial $p_k(A,\lambda)$ in the entries of $A$ and $\lambda$, and we are looking for the set of $A$ such that these polynomials have a common root in $\lambda$.  
Let $X_k\subseteq M_n(\mathbb{C})\times \mathbb{C}$ be the vanishing set of the $k$th minor $p_k(A,\lambda)$ (obviously a Zariski closed set).  Then $\mathcal{E}$ is the projection of $\bigcap X_k$ onto $M_n(\mathbb{C})$.  Let $Y_k$ be the Zariski closure of $X_k$ in $M_n(\mathbb{C})\times\mathbb{P}^1$, or equivalently the vanishing set of the polynomial $q_k(A,s,t)=t^dp_k(A,s/t)$ where $d$ is the degree of $p_k$ with respect to $\lambda$ (here $[s:t]$ are the homogeneous coordinates on $\mathbb{P}^1$).  Note that the minor $p_n$ formed by omitting the bottom row is monic in $\lambda$, and so $q_n$ does not vanish at $t=0$ and thus $X_n=Y_n$.  It follows that $\bigcap Y_k=\bigcap X_k$.  But the projection of $\bigcap Y_k$ to $M_n(\mathbb{C})$ is Zariski-closed, since the projection $M_n(\mathbb{C})\times\mathbb{P}^1\to M_n(\mathbb{C})$ is a closed map in the Zariski topology.  Since $\bigcap Y_k=\bigcap X_k$, this means $\mathcal{E}$ is Zariski-closed.
