Fano variety of lines on the Segre and the Grassmannian 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.
 A: 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.
A: This is a comment, but I cannot comment as my reputation is too low :/
I'm very sorry: why is this the top class?
I.e. why we do not get additionally $pr^*_{Grass}[c_2(S^\vee)^2c_1(S^\vee)^2]$ (giving $30=5*6$) and $c_2(S^\vee)^3$ (giving $10$)? Did I miscompute that the top class is $(c_2(S^\vee)+c_1(T^\vee)c_1(S^\vee)+c_1(T^\vee)^2)^5$ or I have an error elswhere?
