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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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2Maybe 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
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1See this answer of Gerry Myerson's. – Dylan Moreland May 31 '12 at 17:46
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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.
Pete L. Clark
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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.
tomasz
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