# Covering vector space over finite field by subspaces

In the course of investigating this problem, I needed to use the following fact: If $T,W$ are proper subspaces of a vector space $V$, then $T\cup W$ is not the entire space $V$. Provided $V$ is a vector space over the $\mathbb{R}$ or $\mathbb{C}$, this is trivial, and in fact the analogous statement holds for any finite number of subspaces. You can see this to be true using e.g. the fact that the measure of a proper subspace intersected with a ball of radius 1 is zero, so the union of any finite (or countable) number of subspaces can't fill the ball.

However, to prove this fact using only ideas from linear algebra is surprisingly subtle. The reason is that the generalized claim is false for vector spaces over general fields, e.g. finite fields. For instance, the vector space $\mathbb{F}^2_2$ over the finite field $\mathbb{F}_2$ can be decomposed as a union of three proper subspaces.

This leads me to the following question: For a vector space $V$ of dimension $n$ over a finite field $\mathbb{F}_{p^q}$, what is the minimum number $m$ of subspaces $W_1,\dots,W_m$, each of dimension $k$ (with $k<n$), such that $V = \cup_{j=1}^m W_j$?

Edit: For what it's worth, here's a proof that two proper subspaces $T,W$ of $V$ can never have union equal to $V$:

$T$ proper means there is some $v\in V$ with $v\notin T$. Then $T$ and $T+v$ are disjoint. If $T\cup W = V$, then all $w\in v+T$ lie in $W$, and in particular $v\in W$. But this means that for any $t\in T$, both $v\in W$ and $t+v\in W$, hence $t\in W$, hence $T\subset W$, hence $T\cup W = W = V$, contradicting that $W$ is a proper subspace.

This shows that the $m$ sought above must be greater than 2.

• I'm inclined to call this a duplicate of this older question. But, because I answered that one I should not initiate the closure process. Jan 14, 2018 at 10:38

If a vector space over a field $F$ is the union of $n$ proper subspaces, then $|F|\le n-1$.
Thus for a finite field $\mathbf F_q$, where $q$ is some prime power $p^r$, we must have
$$n\ge p^r+1.$$
• Very cool. To put this in the notation of the OP, it proves that $m\geq p^q+1$. I wouldn't call this an answer, but it is a nice bound.