# Finite unions of subspaces

2021-12-06

This post collects some proofs of the following statement.

## Theorem

Let $$V$$ be a vector-space over a field $$F$$ with infinitely many elements, and let $$V_1, \ldots, V_n$$ be finitely many proper subspaces. Then $$\bigcup_{i=1}^n V_i \ne V$$.

## Induction

This is the first proof I have ever seen of this statement.

It proceeds by induction:

If $$n=1$$, then the statement holds obviously.

Suppose it holds for $$n-1$$.

Now let the notations be as in the statement. If $$V_1 \subseteq \bigcup_{i=2}^{n-1} V_i$$, then we are reduced to the $$n-1$$ case: $$V_2, \ldots, V_n$$. So we can assume that $$\exists x\in V_1 \setminus \bigcup_{i=2}^n V_i$$.

Similarly, we can assume that $$\exists V_i$$ such that $$V_i \not\subseteq V_1$$. By renumbering if necessary, we can assume that $$i=2$$. Let $$y\in V_2 \setminus V_1$$.

Then the set $$\{ x + \lambda y \mid \lambda \in F \}$$ is infinite by the hypothesis. So if the union of $$V_i$$ is the whole $$V$$, then by the pigeon-hole principle, there exist $$\lambda \ne \mu \in F$$ such that $$x + \lambda y \in V_j$$ and $$x + \mu y \in V_j$$ for some $$j=1,\ldots,n$$.

This means $$(\lambda - \mu ) \cdot y \in V_j$$, and hence $$y\in V_j$$. It follows that $$x\in V_j$$ as well. But this is impossible: $$x\in V_j$$ implies that $$j=1$$ by our assumption of $$x$$, while $$y\notin V_1$$.

## Vandermonde matrix

This is quite a surprising proof: I have never thought that the Vandermonde matrices can have a play here!

First, if the field $$F$$ has characteristic $$0$$, then define $$a_i:=i,\,\forall i\in \mathbb N_{\geq1}$$. If the characteristic is a prime $$p$$, then define $$a_i := \sum_{j=1}^i f_j$$, where for every $$j\in\mathbb N_{\geq1}$$, we choose some $$f_j$$ in $$\mathbb F_{p^{j+1}}^*$$ but not in $$\mathbb F_{p^j}^*$$, which belongs to $$F$$ since it is infinite.

Let $$v_1, \ldots, v_n$$ be a basis for $$V$$.

For every $$i = 1,\ldots$$, define $\alpha_i := v_1 + a_i v_2 \cdots + {a_i}^{n-1} v_n.$

Then for any $$n$$ numbers $$i_1, \ldots, i_n$$, we know that the matrix

\begin{bmatrix} 1 & a_{i_1} & \cdots & {a_{i_1}}^{n-1} \\ \vdots & \vdots & \cdots & \vdots \\ 1 & a_{i_n} & \cdots & {a_{i_n}}^{n-1} \end{bmatrix}

has determinant $$\prod_{1\leq a \lt b \leq n} (a_{i_a} - a_{i_b}) \ne 0$$, and hence the vectors $$\{\alpha_{i_1}, \ldots, \alpha_{i_n} \}$$ form a basis for $$V$$.

This shows that every sub-space of $$V$$ contains at most $$n-1$$ of theses $$\alpha_i$$. Therefore the union of finitely many sub-spaces contains finitely many these $$\alpha_i$$. Consequently there are infinitely many $$\alpha_i$$ not contained in the union.

## Polynomials

I had the idea of this proof earlier, but I was thinking about using schemes, which did not end up practical. But that idea can be modified to give the following proof.

Since $$V$$ has finite dimension, it has a finite basis, say $$\beta_1, \ldots, \beta_n$$, where $$n$$ is the dimension of $$V$$. Then a subspace has a basis as well, say $$\gamma_1, \ldots, \gamma_m$$, where $$m\lt n$$ is the dimension of the subspace. Then we can find linearly independent vectors $$\gamma_{m+1}, \ldots, \gamma_n$$ such that $$\{\gamma_i, i=1,\ldots,n\}$$ is a basis for $$V$$.

We can express $$\beta_i$$ as a linear combination $$\sum_{i=1}^n A_{ij} \gamma_j$$ of $$\gamma_j$$. So the subspace consists exactly of the vectors which are linear combinations of $$\gamma_i,\, i=1,\ldots,m$$, which are exactly the vectors $$\sum_{i=1}^n a_i \beta_i$$ such that $\sum_{i=1}^n a_i \cdot A_{ij} = 0,\, \forall j = m+1,\ldots,n.$ So a subspace consists of points whose coordinates satisfy a finite set of linear equations.

Moreover, if $$m\lt n-1$$, then the given subspace is contained in the subspace generated by $$\gamma_1, \ldots, \gamma_{n-1}$$, so we may assume that $$m=n-1$$ and each subspace under consideration is given by one linear equation.

Then the intersection of finitely many subspaces (each of codimension $$1$$) is given as the points whose coordinates satisfy the product of the linear equations. Thus if an intesection of finitely many subspaces is the whose vector-space $$V$$, then every point of $$V$$ satisfies the product of the linear equations. But this is impossible, as the prodoct of linear equations is a polynomial equation, and a polynomial has at most $$d$$ roots over a field, where $$d$$ is the degree of the polynomial over the field, while the field $$F$$ has infinitely many elements.

## Others

If I found other proofs, I will supply them here.