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.

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    $\begingroup$ i'd do it first for $\dim V=2$. $\endgroup$ – Angina Seng Nov 23 '18 at 7:27
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    $\begingroup$ 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, $V = V_\infty \cup \bigcup_{k \in K} V_k$. $\endgroup$ – 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 $x_2=cx_1$.


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