0
$\begingroup$

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.

$\endgroup$
  • 5
    $\begingroup$ i'd do it first for $\dim V=2$. $\endgroup$ – Angina Seng Nov 23 '18 at 7:27
  • 3
    $\begingroup$ 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
3
$\begingroup$

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.

$\endgroup$
4
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.