What is the maximal number of sub spaces of a fixed dimension such that there is another sub space which intersects them are all null

Question

Let $\mathbb F_q$ be the finite field with $q$ elements. Suppose $V$ is a linear space of dimension $n$ over $\mathbb F_q$, and $r<n$. What is the maximal $k$ such that for arbitrary $k$ subspaces $W_1,W_2,\dots,W_k$ of $V$ of dimension $r$, there always exists a subspace $U$ of $V$ of dimension $n-r$ which satisfies $U\cap W_i=\{0\}$ for each $W_i$? Thank you.

PS. The choice of $W_i$ is arbitrary, and I didn't say it clearly in the former version, sorry.

PS. I guess the number might be something like $q^{n-r}$, but I don't know how to prove it.

PS. The answer is $k=q$.

Answer 1

I think we have $k = q$ (for all $r$).

First, suppose $r = n - 1$. The vector space $V$ can be covered by $q + 1$ codimension-one hyperplanes $W_1,W_2, \ldots,W_{q+1}$ (but no fewer). (If we think of the projective space $[V]$, the most efficient covering is by taking the hyperplane at infinity, plus a set of $q$ parallel hyperplanes in affine space.) Then every $1$-dimensional subspace intersects (in fact, is contained in) some $W_i$. So $k = q$ when $r = n-1$.

To prove $k \le q$ in general, fix a subspace $Z$ of $V$ that has dimension $r + 1$, and use the previous case to choose $q + 1$ subspaces $W_1,\ldots,W_{q+1}$ of dimension $r$ in $Z$, such that every $1$-dimensional subspace of $Z$ is contained in some $W_i$. Since every subspace $U$ of dimension $n - r$ in $V$ contains a subspace of $Z$ that has dimension (at least) $1$, we conclude that $U$ intersects some $W_i$. So $k < q + 1$, as desired.

Now we prove the matching lower bound: $k \ge q$. Let $W_1,W_2, \ldots,W_q$ be any $r$-dimensional subspaces of $V$. Since (as mentioned above), it takes $q + 1$ hyperplanes to cover all of $V$, we know there is some $1$-dimensional subspace $L$ that is not contained in $\bigcup_i W_i$. Let $\overline{V} = V/L$. By induction on $n - r$, there is an $(n-r-1)$-dimensional subspace $\overline{U}$ of $\overline{V}$ that has trivial intersection with every $\overline{W_i}$. Lift $\overline{U}$ to an $(n-r)$-dimensional subspace of $V$. Then $U$ has trivial intersection with every $W_i$.

Answer 2

EDIT: This answers an earlier version of the question:

Your question seems to be equivalent to fixing $V$ and an $(n-r)$-dimensional subspace $U$ and then asking how many $r$-dimensional subspaces $W$ have $\{0\}$ intersection with that $U$. The number is (unless I've made a silly mistake) $$ \frac {(q^n-q^{n-r})(q^n-q^{n-r+1})\cdots(q^n-q^{n-1})} {(q^r-1)(q^r-q)\cdots(q^r-q^{r-1})}. $$ Here the numerator counts the number of bases $\{b_1,\dots,b_r\}$for such subspaces $W$. The first vector $b_1$ can be any vector $\notin U$; then $b_2$ can be any vector not in the space spanned by $U$ and $b_1$; etc. Similarly, the denominator counts the number of bases for any single $W$, so the quotient counts the $W$'s.