# Prove $V$ over finite field of $q$ elements can be written as union of $q + 1$ proper subspaces

Let $$V$$ be a vector space (can be finite or infinite) over finite field $$K$$, such that $$\dim V > 1$$ and $$|K| = q < \infty$$. Prove there exist proper subspaces $$V_0, \dots, V_q$$ such that $$V = V_0 \cup \dots \cup V_q$$. I have no idea where to start from.

• i'd do it first for $\dim V=2$. – Angina Seng Nov 23 '18 at 7:27
• Pick two linearly independent maps $a, b : V \to K$. For each $k \in K$, let $V_k = \left\{v \in V \mid a\left(v\right) = k b\left(v\right)\right\}$. Also, let $V_\infty = \left\{v \in V \mid b\left(v\right) = 0 \right\}$. Then, $V = V_\infty \cup \bigcup_{k \in K} V_k$. – darij grinberg Nov 23 '18 at 18:23

We claim that if $$V$$ is a vector space over a finite field $$K$$ of order $$q$$ such that $$\dim V>1$$, and $$m$$ is a non-negative integer, then $$V$$ can be written as a union of $$m$$ proper subspaces of $$V$$ if and only if $$m\geq q+1$$. (From the proof below, it also follows that if $$K$$ is not finite, then there is no way to cover a vector space $$V$$ over $$K$$ with $$\dim V>1$$ by finitely many proper subspaces.)

First suppose that $$m\geq q+1$$. It suffices to assume that $$m=q+1$$. Pick a basis $$\mathcal{B}$$ of $$V$$. Let $$a,b\in\mathcal{B}$$ be two distinct elements (noting that $$|\mathcal{B}|>1$$ since $$\dim V>1$$). For each $$k\in K$$, we define $$V_k$$ to be the span of $$\{a+kb\}\cup\big(\mathcal{B}\setminus\{a,b\}\big)$$, and $$U$$ is the span of $$\mathcal{B}\setminus\{a\}$$. Show that $$V=U\cup \bigcup_{k\in K}V_k$$.

Conversely, suppose that $$V$$ can be written as a union of $$m$$ proper subspaces $$W_1,W_2,\ldots,W_m$$ with $$m$$ being smallest possible (from the previous paragraph we know $$m$$ exists, so taking the smallest one is possible). It is easy to see that $$m>1$$. By minimality of $$m$$, for any $$i$$, we have $$W_i\not\subseteq \bigcup_{j\neq i}W_j$$.

Take $$u\in W_1\setminus\bigcup_{j\neq1}W_j$$ and $$v\in W_2\setminus\bigcup_{j\neq 2}W_j$$. Since $$u+sv\in V$$ for all $$s\in K$$ such that $$s\neq 0$$, we must have $$u+sv\in W_j$$ for some $$j$$. We claim that the assignment $$s\in K\setminus\{0\}$$ to the smallest $$j$$ such that $$u+sv\in W_j$$ is an injective function from $$K\setminus\{0\}$$ to $$\{3,4,\ldots,m\}$$. From here, it follows that $$q-1=\big|K\setminus\{0\}\big|\leq \big|\{3,4,\ldots,m\}\big|=m-2,$$ establishing our claim.

Now, to prove the assertion in the previous paragraph, we first note that $$u+sv\notin W_1$$ and $$u+sv\notin W_2$$ for $$s\ne 0$$. If $$u+sv\in W_1$$, then $$v=s^{-1}\big((u+sv)-u\big)\in W_1$$ since $$u\in W_1$$, which is a contradiction. If $$u+sv\in W_2$$, then $$u=(u+sv)-sv\in W_2$$ since $$v\in W_2$$, which is also a contradiction. So, $$u+sv\in W_j$$ for some $$j\in\{3,4,\ldots,m\}$$.

Now, suppose that there are two non-zero $$s,t\in K$$ such that $$u+sv$$ and $$u+tv$$ are in the same $$W_i$$, where $$i\in\{3,4,\ldots,m\}$$. Then, $$v=(s-t)^{-1}\big((u+sv)-(u+tv)\big)\in W_i.$$ But $$v\in W_2\setminus \bigcup_{j\neq 2}W_j$$, so we have another contradiction. The assertion is now proven.

Hint: If $$(x_1,x_2,\ldots)\in V$$ then either $$x_1=0$$ or there exists $$c\in K$$ with $$x_2=cx_1$$.