If union of n subspaces of V is a subspace of V, then one of the n subspaces must contain the other n-1 subspaces

Question

Help prove this? I can prove for n=2, but I'm stuck on proving it for general n. Thanks!

My proof for n=2

Forward direction: Consider A and B and A $\cup$ B is a subspace of V. We prove by contradiction by assuming $\exists$ x $\in$ A s.t x $\notin$ B and $\exists$ y $\notin$ s.t y $\in$ B. Consider x+y. Since x $\in$ A and y $\in$ B, if A$\cup$B is a subspace, we expect x+y $\in$ A$\cup$B. However, since y$\notin$A, then x+y $\notin$A because of closure under addition and multiplication of scalar. If, -x+x+y$\in$A then y$\in$A, but that is a contradiction. Thus x+y $\notin$ A and by similar logic, same for B. Since x+y $\notin$ A and x+y $\notin$ B, x+y $\notin$ A $\cup$ B which is a contradiction. Thus one of the subspaces must contain the other

Reverse direction: WLOG, A $\subset$ B so A $\cup$ B is equivalent to B. Since A and B are subspaces, they both have the 0 vector so A$\cup$B also does. For a1, a2 $\in$ A$\cup$B, a1, a2 $\in$ B, so a1+a2 $\in$A$\cup$B-->closed under addition. Consider b$\in$ A$\cup$B and scalar $\lambda$. b$\in$ B, and $\lambda$b $\in$ B so $\lambda$b $\in$ A$\cup$B.--> so closed under scalar multiplication. Thus A$\cup$B is a subspace.

Edit: Does it work that since I proved it for n=2, we can use induction to just look at V$\cup$W for dim(W)=1 as one of the comments pointed out? Doesn't that just reduce it to the 2 subspace case again? And since we know that the union is a subspace, is that too easy?

Answer 1

Suppose $k$ is an infinite field. In fact, we must suppose that $k$ is an infinite field, since every finite field is the union of its one dimensional subspaces, of which there are finitely many.

The idea of the proof is to consider an affine line passing through multiple subspaces at once, and to use the fundamental fact that any affine space containing two distinct points on a line contains that line.

Lemma: Let $V$ be a vector space over a field with infinitely many elements. Let $U_1, ..., U_n$ be proper subspaces of $V$ whose union is $V$. Then one of $U_i$ is equal to $V$.

Proof: Suppose one of these- $U_1$ is not contained in the union of the other spaces, and take $u \in U_1$ not contained in any of the others. I claim that $U_1$ must in fact contain the union of the other spaces. Suppose for a contradiction that it does not. Then take a point $v \in V$ not contained in $U$ and let $l$ be the affine line containing $u$ and $v$. There are infinitely many points on this line and only finitely many of $U_1, ..., U_n$. Some $U_i$ for $i > 1$ must contain more than one point on this line ($U_1$ cannot, as it would then contain $v$). But then some $U_i$ contains two distinct points on this affine line and therefore contains the line. Hence this choice of $U_i$ contains $u$, a contradiction.

Of course, this can be changed around a little into the question you're asking.

Answer 2

Let $U_1, U_2, \dots, U_n$ be subspaces of $V$ such that their union is also a subset of $V$.

We can suppose that $U_1 \nsubseteq U_2 \nsubseteq \dots \nsubseteq U_{n-1} \nsubseteq U_n \nsubseteq U_1$, because if $U_i \subseteq U_j$ for $i \ne j$ then we use induction on $n$ to conclude that there is $k \in \{1, \dots, n\} \setminus \{i\}$ such that $U_k$ contains all the other subsets.

Let $u_k \in U_k \setminus U_{k+1}$ for $k \in \{1, \dots, n\}$, where the indices of subspaces are modulo $n$. Suppose we have $N = n^2 - n + 1$ vectors in $\mathbb{F}^n$ such that any $n$ of then are linearly independent. Using them as coefficients, we construct the following

$$ x_1 = a_{11} u_1 + a_{12} u_2 + \dots + a_{1n} u_n\\ x_2 = a_{21} u_1 + a_{22} u_2 + \dots + a_{2n} u_n\\ \dots\\ x_N = a_{N1} u_1 + a_{N2} u_2 + \dots + a_{Nn} u_n\\ $$

As all $u_i$ are in $U_1 \cup U_2 \cup \dots \cup U_n$, so are all $x_i$. By the pigeonhole principle, there is $U_p$ that contains $n$ different $x_i$. WLOG, $x_1, \dots, x_n \in U_p$.

$$\begin{bmatrix} a_{11} & a_{12} & \dots & a_{1n}\\ a_{21} & a_{22} & \dots & a_{2n}\\ \dots & \dots & \dots & \dots\\ a_{n1} & a_{n2} & \dots & a_{nn}\\ \end{bmatrix} \begin{bmatrix}u_1\\u_2\\\dots\\u_n\end{bmatrix} = \begin{bmatrix}x_1\\x_2\\\dots\\x_n\end{bmatrix} $$

As we chose the vectors such that $x_1, \dots x_n$ are linearly independent, the matrix above is invertible. Therefore, each $u_i$ is written as a linear combination of the $x_i$, and so each $u_i \in U_p$. This is a contradiction, since $u_{p-1} \notin U_p$.

Therefore, we could not suppose that $U_1 \nsubseteq U_2 \nsubseteq \dots \nsubseteq U_{n-1} \nsubseteq U_n \nsubseteq U_1$, and so we conclude by induction that there is a subspace that contains all others.