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Let $M=(E,\mathcal{I})$ be a matroid with rank function $r$. A hyperplane is a maximal set with rank $r(E)-1$.

What can we say about the rank of the intersection of $k$ hyperplanes? Is it true it has rank at least $r(E)-k$ or $0$? This is true for an arrangement of hyperplanes in $\mathbb{R}^n$.

This seems true for linear matroids.

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  • $\begingroup$ What about, say, the linear matroid on $E = \{(\pm 1,\pm 1,\pm 1)\} \cup \{(0,0,0)\} \subset \mathbb{R}^3$, and then the intersection of the hyperplanes $H_1:x=y$, $H_2:x+y=0$. Then $H_1 \cap H_2 = \{(0,0,0)\}$ instead of a line (rank $1$). $\endgroup$ – Zach Teitler Oct 24 '17 at 7:10
  • $\begingroup$ @ZachTeitler Thanks. Indeed there are matroids with disjoint hyperplanes so dimension 0 is possible. I've updated the question. $\endgroup$ – Chao Xu Oct 24 '17 at 23:16
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    $\begingroup$ No, it still fails for linear matroids with essentially the same example. Pick any integers $0 \leq r \leq s \leq n-k \leq n$ and $d \geq k$, any $s$-dimensional subspace of $\Bbbk^n$, $d$ hyperplanes that contain the $s$-plane, and whose intersection is the $s$-plane; in each hyperplane, a basis that's disjoint from the $s$-plane; in the $s$-plane, an $r$-dimensional subspace, and a basis of it. The resulting matroid on $d(n-1)+r$ vectors (go ahead and add more if you want) has $k$ hyperplanes whose intersection has rank $r$. Of course other examples have intersections of rank $>n-k$. $\endgroup$ – Zach Teitler Oct 25 '17 at 0:08
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In the rank $4$ projective geometry $\text{PG}(3,2)$, every hyperplane is isomorphic to the Fano matroid. Now delete two points on the same line. The two hyperplanes that share that line intersect at a single point, so the rank of their intersection is $1$.

And of course, any minor of a projective geometry is a linear matroid.

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