# If a field $F$ is such that $\left|F\right|>n-1$ why is $V$ a vector space over $F$ not equal to the union of $n$ proper subspaces of $V$

If $U_1$, $U_2,\ldots,U_n$ are proper subspaces of a vector space $V$ over a field $F$, and $|F|\gt n-1$, why is $V$ not equal to the union of the subspaces $U_1$, $U_2,\ldots,U_n$?

• Covering Numbers in Linear Algebra by Pete L. Clark might help. Commented Aug 30, 2011 at 13:42
• @Pierre-Yves: yes, the answer to the question is contained in my note. Commented Sep 4, 2011 at 23:59
• I referred to this question in another thread, and wanted to make the title clearer. Sorry about the bump. Commented Nov 4, 2018 at 5:39
• @Pierre-YvesGaillard, the link is off
– user561334
Commented Mar 21, 2021 at 0:30
• @user1618 - You can use arxiv.org/abs/1208.0975 Commented Mar 21, 2021 at 11:29

If $|F|=q<\infty$, and $V$ is $m$-dimensional ($m<\infty$), then any proper subspace $U_i$ has at most $q^{m-1}-1$ non-zero elements. So to cover the $q^m-1$ non-zero vectors of $V\,$, the given $n\le q$ subspaces are not going to be enough, because $$n(q^{m-1}-1)\le q(q^{m-1}-1)<q^m-1.$$ So we need at least $|F|+1>n$ subspaces to get the job done.

If $m=\infty$, then we can extend all the subspaces to have codimension one (i.e. $\dim_F(V/U_i)=1$ for all $i$). In that case the intersection $U$ of all the $U_i$:s has finite codimension, and we can study $V/U$ instead of $V$ reducing the probelm to the previous case.

If $|F|=\infty, m<\infty$? Well, then we need some reinterpretation. The following argument shows that we need an infinite number of subspaces to cover $V$, and an uncountable number of subspaces to cover $\mathbf{R}^m$. Again, assume that all the subspaces have codimension one (w.l.o.g.), and that $m\geq 2$ (also w.l.o.g.). Identify $V$ with $F^m$, and consider the set $$S=\{(1,t,t^2,\ldots,t^{m-1})\in V\mid t\in F\}.$$ Any $U_i$ is now a hyperplane and consists of zeros $(x_1,x_2,\ldots,x_m)$ of a single non-trivial homogeneous linear equation $$a_{i1}x_1+a_{i2}x_2+\cdots+a_{in}x_m=0.$$ Therefore the number of elements of the intersection $S\cap U_i$ is equal to the number of solutions $t\in F$ of $a_{i1}+a_{i2}t+\cdots+a_{im}t^{m-1}=0$ and is thus $<m$, because a non-zero polynomial of degree $<m$ has less than $m$ solutions in a field. This shows that if $F$ is infinite, we need an infinite number of subspaces to cover all of $S$. Also, if $F$ is uncountable, then we need an uncountable number of subspaces to cover $S$. Obviously it is necessary to cover all of $S$ in order to cover all of $V$.

Hint  Let $$\rm\,U = U_1\! \cup \,\cdots\,\cup U_n,\,$$ wlog irredundant (i.e. no $$\rm\,U_i\,$$ lies in the union of the others). Choose $$\rm\,v\not\in U_1,$$ $$\rm\, u\in U_1,\, u\not\in U_{i>1}.\,$$ Put $$\rm\, L = v + u\:\! F.\,$$ Then $$\rm\,|L\cap U_1| = 0,\,$$ $$\rm |L\cap U_{i\,>1}| \le 1.\,$$ Therefore $$\rm\,|L\cap U| \le n-1 < |F| = |L|,\,$$ so the "generic" line $$\rm\,L\,$$ has a point not in $$\rm U.\$$

Proof $$\$$ First,  notice $$\rm\ |L\cap U_1| = 0\$$ since $$\rm\, u,v+cu \in U_1 \Rightarrow\, (v+cu)-cu\, =\, v \in U_1\,$$ contra choice of $$\rm\,v.\,$$ Second, $$\rm\,|L\cap U_{i\,>1}| \le 1\,$$ since if $$\rm\,v+cu,\, v+du\in U_i$$ then so too is their difference $$\rm\,(c-d)u.\,$$ Thus $$\rm\,c = d\$$ (else scaling by $$\rm\,(c-d)^{-1}$$ $$\Rightarrow$$ $$\rm\,u\in U_{i\,>1}\,$$ contra choice of $$\rm\,u).\,$$ Finally $$\rm\,v+cu\, =\, v+du\,$$ $$\Rightarrow$$ $$\rm\,(c-d)\,u = 0\,$$ $$\Rightarrow$$ $$\rm\,c=d,\,$$ so $$\rm\,c\,\mapsto\, v+c\,u\$$ is $$\,1$$-to-$$1,\,$$ thus $$\rm\,|F| = |L|.$$

The pivot of the standard proof (like the one Bill Dubuque gives) can be stated thus:

Proposition. Let $$V$$ be a vector space over a field $$F$$ and let $$S_1$$ and $$S_2$$ be subspaces of $$V$$. Suppose $$w\in S_1\setminus S_2$$, $$v\in V\setminus S_1$$ and put $$L=\{rw+v \mid r\in F\}$$, then $$L \cap S_1=\emptyset$$ and $$|L \cap S_2|\leq 1$$.