Does every $\mathbb{P}^{19}\subset \mathbb{P}(\mathbb{C}^5\otimes\mathbb{C}^5)$
intersect the Segre variety of rank one matrices in at least a $\mathbb{P}^1$?
A naive dimension count suggests this is possible. The intersection is a
$3$-fold and I would be happy for any qualitative information about it.



Fano variety of lines on the Segre and the Grassmannian
Although dhy's comment does solve this problem, I thought I would point out another approach. Let $U$, $V$ and $W$ be $5$-dimensional $k$-vector spaces. Let $$B: U\times V \to W,$$ be a bilinear pairing. Landsberg's question, essentially, asks whether there exists a $2$-dimensional subspace, $S\subset U$, and a $1$-dimensional subspace, $T\subset V$, such that $B$ is zero on $S\times T$. On the parameter space $\text{Grass}(2,U)\times \mathbb{P}V$ for pairs $([S],[T])$, this is a straightforward "degeneracy locus" problem. Denoting by $$ \sigma: U^\vee\otimes\_k \mathcal{O}\_{\text{Grass}} \to \mathcal{S}^\vee$$ the universal rank $2$ locally free quotient on $\text{Grass}(2,U)$, and denoting by $$ \tau: V^\vee \otimes\_k \mathcal{O}\_{\mathbb{P}V} \to \mathcal{T}^\vee$$ the universal rank $1$ locally free quotient on $\mathbb{P}V$, then $B$ induces a global section, say $\widetilde{B}$, of the rank $10$ locally free sheaf on $\text{Grass}(2,U)\times \mathbb{P}V$, $$\textit{Hom}\_{\mathcal{O}}(\text{pr}\_{\text{Grass}}^\*\mathcal{S} \otimes\_{\mathcal{O}} \text{pr}\_{\mathbb{P}V} \mathcal{T}, W\otimes\_k \mathcal{O}).$$ This rank $10$ locally free sheaf on the $10$-dimensional projective scheme $\text{Grass}(2,U)\times \mathbb{P}V$ has top Chern class $$5\text{pr}\_{\text{Grass}}^\*[c\_1(\mathcal{S}^\vee)^4\cap c\_2(\mathcal{S}^\vee)] \cap \text{pr}\_{\mathbb{P}V}^\*[c\_1(\mathcal{T}^\vee)^4].$$ The total degree of this Chern class is $10$. Since it is nonzero, the zero scheme of $\widetilde{B}$ is nonempty, i.e., there exists a line $\mathbb{P}S$ in $\mathbb{P}U$ and a singleton $\mathbb{P}T$ in $\mathbb{P}V$ such that the image under the Segre morphism of $\mathbb{P}S\times \mathbb{P}T$ is a line in $\mathbb{P}(U\otimes\_k V)$ that is contained in the zero scheme of the linear form $B$.


Of course, by a symmetric argument, there also exists a $1$-dimensional subset $S$ of $U$ and a $2$-dimensional subspace $T$ of $V$ such that $\mathbb{P}S\times \mathbb{P}T$ is in the zero locus of $B$. Thus, we know that there are at least $2$ lines. If $B$ is general, there should be precisely $20$ lines contained in the intersection of the Segre locus and the zero locus of $B$.


**Edit**. As Wajcha correctly points out, I had the wrong computation for $c\_2(\text{pr}\_{\text{Grass}}^\*\mathcal{S}^\vee\otimes\_{\mathcal{O}}\text{pr}\_{\mathbb{P}V}^\*\mathcal{T}^\vee)$. The correct formula is $$ \alpha = \text{pr}\_{\text{Grass}}^\*c\_2(\mathcal{S}^\vee) + \text{pr}\_{\text{Grass}}^\*c\_1(\mathcal{S}^\vee)\cap\text{pr}\_{\mathbb{P}V}^\*c\_1(\mathcal{T}^\vee) + \text{pr}\_{\mathbb{P}V}^\*c\_1(\mathcal{T}^\vee)^2. $$ Therefore, the top Chern class of the rank 10 locally free sheaf equals
$$
\alpha^5 = 5\text{pr}\_{\text{Grass}}^\*[c\_1(\mathcal{S}^\vee)^4\cap c\_2(\mathcal{S}^\vee) +6c\_1(\mathcal{S}^\vee)^2\cap c\_2(\mathcal{S}^\vee)^2+ 2c\_2(\mathcal{S}^\vee)^3]\cap \text{pr}\_{\mathbb{P}V}^\*c\_1(\mathcal{T})^2. 
$$
Via Schubert calculus, the term in square brackets is $10$ times the class of a point (rather than just $2$, as I computed before). Therefore $\alpha^5$ has total degree $50$. So, if $B$ is generic, there are $50$ lines in the intersection lying in the first "ruling" of the Segre variety by linear $\mathbb{P}^4$s. By a symmetric argument, there are also $50$ lines lying in the second ruling. Therefore, altogether, there should be $100$ lines, just as in the article cited by dhy.


