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Let \begin{align} \Omega=\begin{bmatrix} L_1 & \cdots & L_{n-1} \\ M_1 & \cdots & M_{n-1} \end{bmatrix}\end{align} be a matrix of linear forms on $\mathbb{P}^n$, i.e. homogeneous polynomials of degree $1$, such that for any $[\lambda, \mu] \in \mathbb{P}$, the linear forms $\lambda L_1 + \mu M_1,\dots,\lambda L_{n-1}+\mu M_{n-1}$ are linearly independent. For each $[\lambda, \mu]$, the equations $\lambda L_1 + \mu M_1=\dots=\lambda L_{n-1}+\mu M_{n-1}=0$ define a line $\ell_{\lambda,\mu}$ of $\mathbb{P}^n$. The union of all the lines as $[\lambda,\mu]$ varies is precisely the rank-1 locus variety of $\Omega$.

Question: Does there exist an open set $\mathcal{U}$ of hyperplanes of $\mathbb{P}^n$, such that every hyperplane $H \in \mathcal{U}$ does not contain any $\ell_{\lambda,\mu}$? (Here we parametrize the hyperplanes of $\mathbb{P}^n$ by elements of $\mathbb{P}^n$: each such element defines the normal to a hyperplane.) In other words, can we say that a general hyperplane $H$ of $\mathbb{P}^n$ does not contain any of the lines $\ell_{\lambda,\mu}$?

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    $\begingroup$ Inside the dual projective space $(\mathbb{P}^n)^\vee$ parameterizing hyperplanes $H$ in $\mathbb{P}^n$, for every line $\ell \subset \mathbb{P}^n$, the subvariety $S_\ell \subset (\mathbb{P}^n)^\vee$ parameterizing hyperplanes $H$ that contain $\ell$ is a codimension $2$ linear subspace, $S_\ell \cong \mathbb{P}^{n-2}$. Thus, for a one-parameter family $[\ell_{\lambda,\mu}]_{(\lambda,\mu)\in \mathbb{P}^1}$, the union of the subvarieties $S_{\ell_{\lambda,\mu}}$ is a codimension $1$ subvariety of $(\mathbb{P}^n)^\vee$. The dense open complement is $\mathcal{U}$. $\endgroup$ – Jason Starr Jun 24 '15 at 16:37
  • $\begingroup$ @JasonStarr: Very instructive thank you. How do you see though that the codimension of the union is $1$? $\endgroup$ – Manos Jun 24 '15 at 18:36
  • $\begingroup$ Thiso sounds like the kind of question people interested in hyperplane arrangements would know about. You might try looking into the literature in that area. $\endgroup$ – Avi Steiner Jun 25 '15 at 3:31
  • $\begingroup$ @AviSteiner: Actually Jason's comment does it. The only thing i can't see is why the codimension is $1$. $\endgroup$ – Manos Jun 25 '15 at 5:18

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