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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?

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    $\begingroup$ n or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers. $\endgroup$ – José Carlos Santos Jan 21 at 0:01
  • $\begingroup$ This is not true for vector spaces over finite fields, but presumably these are $\Bbb R$ vector spaces $\endgroup$ – Omnomnomnom Jan 21 at 0:03
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    $\begingroup$ It would be helpful if you at least summarized your proof for $n=2$, so that we could perhaps lead you along the solution that you've already found $\endgroup$ – Omnomnomnom Jan 21 at 0:04
  • $\begingroup$ sorry I'm a new user. will edit $\endgroup$ – battacor hero Jan 21 at 0:04
  • $\begingroup$ Maybe induct on the dimension. Then the case would be V $\cup$ W, where W has dimension 1. $\endgroup$ – Joel Pereira Jan 21 at 1:48
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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.

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