Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Possible Duplicate:
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 ?

share|improve this question
2  
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
1  
See this answer of Gerry Myerson's. –  Dylan Moreland May 31 '12 at 17:46
add comment

marked as duplicate by MJD, Arturo Magidin, Dylan Moreland, t.b., Asaf Karagila Jun 1 '12 at 5:45

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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.

share|improve this answer
add comment

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.

share|improve this answer
add comment

Not the answer you're looking for? Browse other questions tagged or ask your own question.