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Let $L,M$ two subspaces of the vector space, $V$ such that both $L,M \ne V$.
Prove: $L\cup M \ne V$.

I think this is a case of a proof by contradiction.
Lets assume $L \cup M = V$.

$$\dim(V) = \dim(L) + \dim(M) - \dim(L\cap M)$$

How to proceed?

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marked as duplicate by egreg, Lost1, T. Bongers, Your Ad Here, Zev Chonoles Feb 26 '14 at 8:45

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.

See this – angryavian Feb 15 '14 at 15:58
@angryavian: The idea is similar, but it's not the exact same question. – Najib Idrissi Feb 15 '14 at 15:59
So, is it suffice to show a contradiction example? @angryavian – AndrePoole Feb 15 '14 at 15:59
The simplest method to show what you're looking for is to consider $L\oplus M \subset V$, and show that $L\cup M$ just won't cut it for the direct sum, let alone the whole vector space. – FireGarden Feb 15 '14 at 16:00
The dimensional argument is not going to work: You can take two one-dimensional subspaces of $\Bbb R^2$ and union them together; the equation will work, but the union is not a vector space (as shown in the link) so won't equal $V$. – tabstop Feb 15 '14 at 16:01
up vote 5 down vote accepted

Let's break down the proof in two cases:

  • If $L \subset M$ (resp. $M \subset L$) then $L \cup M = M \neq V$ (resp. $L \cup M = L \neq V$);
  • Otherwise choose $x \in L \setminus M$, $y \in M \setminus L$. Then $x + y$ is neither in $L$ (for then $y$ would be in $L$) nor in $M$ (same reason). Therefore $x + y \not\in L \cup M$.
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So simple! Thanks. The main key was using the properties of a vector space. Cool. – AndrePoole Feb 15 '14 at 16:04

Suppose $W \not\subset L $ and $L \not\subset W $, otherwise is trivial.

Let $v \in L-W $ and $u \in W - L $, then if $ L \cup W = V $ we have $v + u \in L$ or $v + u \in L$, and in both cases we obtain a contradiction.

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