Incidence geometry and matrices Supposing I have a $0/1$ or $\pm1$ matrix $A$ of size $m\times n$, is there a minimum $d$ (that works for every $m\times n$ $A$) such that there exists $m$ lines $r_1,\dots,r_m$, $n$ lines $s_1,\dots,s_n$ in $\Bbb R^d$ such that $r_i$ intersects $s_j$ iff there are some permutations $\Pi_1,\Pi_2$ such that $\widetilde{A}[i,j]=1$ where $\widetilde{A}=\Pi_1A\Pi_2$?
If so what is $d$ as function of $m,n,r$?
It is clear this is possible for special $A$ with $d=2$. Is it possible for every $A$ at some $d$?
Related: When is a 0-1 matrix a one-intersection incidence matrix?
 A: As Will Sawin points out, this represents only a partial answer.
Theorem 5 of this paper

Laison, Joshua D., and Yulan Qing. "Subspace intersection graphs." Discrete Mathematics 310, no. 23 (2010): 3413-3416.
  Journal link

proves that this graph

                 


cannot be realized as the intersection graph of lines in $\mathbb{R}^d$ 
for any $d \ge 2$.
The corresponding incidence matrix is shown below,
using the ordering $(a,b,c,i,j,k)$.
$$
\left(
\begin{array}{cccccc}
 0 & 0 & 0 & 1 & 1 & 1 \\
 0 & 0 & 0 & 1 & 1 & 1 \\
 0 & 0 & 0 & 1 & 1 & 1 \\
 1 & 1 & 1 & 0 & 1 & 0 \\
 1 & 1 & 1 & 1 & 0 & 0 \\
 1 & 1 & 1 & 0 & 0 & 0 \\
\end{array}
\right)
$$
The logic of their proof is as follows (paraphrased from the paper).
Because $i$ and $j$ intersect (say at point $p$), they determine a plane $\pi$.
Since $a,b,c$ are mutually disjoint,
at most one could contain $p$.
Hence two of them are also contained in $\pi$,
as they intersect both $i$ and $j$; let these two be $a,b$.
Because $a,b$ are disjoint and lie in $\pi$, they must be parallel.
Now $k$ intersects both $a$ and $b$, so $k$ also lies in $\pi$.
But $k$ is disjoint from $i$ and $j$ (which cross in $\pi$),
which is impossible.
A: It seems that the matrix
$$
A = \left(\begin{array}{cccccc}
  1 & 1 & 1 & 1 & 1 & 1 \\
  1 & 1 & 1 & 1 & 1 & 1 \\
  1 & 1 & 1 & 1 & 1 & 1 \\
  1 & 1 & 1 & 0 & 1 & 1 \\
  1 & 1 & 1 & 1 & 0 & 1 \\
  1 & 1 & 1 & 1 & 1 & 0 \\
\end{array}\right)
$$
is not the incidence matrix for any distinct lines $r_1,\ldots,r_6$ and
$s_1,\ldots,s_6$ for any dimension $d$, over $\bf R$ or any other field,
and thus that there is no such $d(m,n)$ once $m,n \geq 6$.
The $5\times 5$ matrix $A_0$ formed by the first five rows and columns of $A$
is not enough: let $R,S$ be points on a plane $\Pi$;
let $r_1,r_2,r_3 \subset \Pi$ be lines containing $R$ but not $S$,
and $s_1,s_2,s_3 \subset \Pi$ lines containing $S$ but not $R$;
and then choose points $P,Q$ off $\Pi$ and collinear with neither $R$ not $S$,
and set
$$
r_4 = \overline{PS}, \quad
r_5 = \overline{QS}, \quad
s_4 = \overline{QR}, \quad
s_5 = \overline{PR}.
$$
But this seems to be the only way to realize $A_0$, and it does not extend to
$6+6$ lines with the intersections demanded by $A$ itself.
The key point is suggested by Boris Bukh's comment:
of any three $r_i$, two must intersect.  Indeed if three pairwise skew lines
$l,l',l''$ have more than one transversal then they determine a quadric surface
$Q$ with two rulings, one of which includes $l,l',l''$ and the other consists
of all the transversals.  Then $s_1,s_2,s_3$ are in the other ruling,
and are thus pairwise skew.  Then $r_1,r_2,r_3$ are in the first ruling,
and it follows that all five $r_i$ are in one ruling and all five $s_j$
in the other.  Hence $r_i$ meets $s_j$ for every $i,j$ $-$ contradiction.
Likewise, of any three $s_j$, two must intersect.
On the other hand, if any three of the $r_i$ meet pairwise,
but not at the same point, then they span a plane $\Pi$ and $s_1,s_2,s_3$
must be on $\Pi$; and then they must also coincide, else every $r_i \subset \Pi$
and we soon reach a contradiction ($s_4$ meets $r_1,r_2,r_3$ but not $r_4$,
so $r_1,r_2,r_3$ meet at some point $R$ and $s_4$ contains $R$ but is not
in $\Pi$; likewise the same is true of $s_5$; but then $r_4,r_5$ also
go through $R$, which contradicts both the assumption at the start of
this paragraph and the $A_0$ condition).  Likewise if any three $r_i$
meet at a point $R$ but are not coplanar then $s_1,s_2,s_3$ go through $R$
and are coplanar, else every $r_i$ contains $R$ and again a contradiction
soon ensues.
From here it should be just a matter of hunting down which configurations
of skew/coincident/coplanar lines are possible among the $r_i$ and $s_j$
until only the one configuration described earlier survives…
Another candidate counterexample, with $m=n=7$, is a "double seven"
with $r_i \cap s_j = \emptyset$ iff $i=j$.  Even $(m,n)=(6,7)$
would suffice if the only way to get $(m,n)=(6,6)$ is the famous
"double six" configuration on a smooth cubic surface, on which
no two $r_i$ meet but the $r_i$ are not all on the same quadric,
and likewise for the $s_j$.
