Show that any pair of vectors jointly lies in the same number of subspaces

Question

Assume a finite field $F_q=GF(q)$ of order $q$. Let $V$ be the $(n+1)$-dimensional vector space over $F_q$. Let $P$ be the set of 1-dimensional subspaces of $V$ i.e. $P=V\setminus\{\mathbf{0}_{(n+1)\times1}\}$. Let $B$ be the set of all $(i+1)$-dimensional subspaces of $V$ for some $i\in\{2,3,\dots,n-1\}$.

I want to show that any pair of vectors in $P$ jointly lies in the same number of elements of $B$. In other words, if we pick any two vectors $\{\mathbf{p}_1,\mathbf{p}_2\}\in P$ then the following quantity should be constant $$|\{S\in B:\mathbf{p}_1\in S \text{ and } \mathbf{p}_2\in S\}|$$

I want to show this for all possible values of $i$.

Any ideas or a sketch of the proof?

Answer 1

Let $\mathbf{e_j},j=1,2,\ldots,n+1,$ be the vectors in the natural basis. Pick arbitrary linearly independent vectors $\mathbf{p_1}$ and $\mathbf{p_2}$, and show that $$|\{S\in B:\mathbf{p}_1\in S \text{ and } \mathbf{p}_2\in S\}|= |\{S\in B:\mathbf{e}_1\in S \text{ and } \mathbf{e}_2\in S\}|.$$

This can be done as follows:

• The group $GL_{n+1}(\Bbb{F}_q)$ acts doubly transitively on $P$. In other words, given $\mathbf{p}_1$ and $\mathbf{p}_2$ there exists a bijective linear transformation $T:V\to V$ such that $T(\mathbf{e_1})=\mathbf{p}_1$ and $T(\mathbf{e_2})=\mathbf{p}_2$, This is because we can extend $\mathbf{p_1},\mathbf{p_2}$ to a basis $\mathbf{p_1},\mathbf{p_2},\ldots,\mathbf{p_{n+1}}$ of $V$, when the matrix with $\mathbf{p_j}, j=1,2,\ldots,n+1,$ as columns is invertible. Fix one such linear transformation $T$.
• $S$ contains $\mathbf{e_1},\mathbf{e_2}$ if and only if $T(S)$ contains $\mathbf{p_1},\mathbf{p_2}$.
• And because $T$ is bijective, $S$ contains $\mathbf{p_1},\mathbf{p_2}$ if and only if $T^{-1}(S)$ contains $\mathbf{e_1},\mathbf{e_2}$.
• Leaving it to you to produce a bijection between the prescribed two subsets of $B$.