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Union of two vector subspaces not a subspace?

$U,W\subseteq V$ are subspaces.

Prove that in order for $U \cup W$ to be a subspace as well, either $U\subseteq W$ or $W\subseteq U$

Can anyone please give me a lead on this ?

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Maybe think of the geometry of it. Pick $u\in U$ and $w\in W$ with $u\not\in W$ and $w\not\in U$; you can do this only if neither of $U$ and $W$ is a subspace of the other. Then consider $u+w$. If $U\cup W$ were a subspace, $u+w \in U\cup W$, but it isn't. –  MJD May 31 '12 at 17:44
See this answer of Gerry Myerson's. –  Dylan Moreland May 31 '12 at 17:46
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marked as duplicate by MJD, Arturo Magidin, Dylan Moreland, t.b., Asaf Karagila Jun 1 '12 at 5:45

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2 Answers

up vote 3 down vote accepted

Sure: supppose that $U$ is not contained in $W$ and $W$ is not contained in $U$. Let $x \in U \setminus W$, $y \in W \setminus U$, and consider $x+y$.

Note that this argument works for subgroups of an arbitrary group.

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The condition that $U\not\subseteq W$ and $W\not\subseteq U$ is equivalent to the statement that there are $v_1\in U\setminus W$ and $v_2\in W\setminus U$, from which the result easily follows.

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