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Let $L,M$, two subspaces of $V$ where $L,M\ne V$. Prove $L\cup M \ne V$

My Try:

Obliviously, $L\cup M \subseteq V$. It's left to show $L\cup M \subsetneq V$.
$L,M$ have bases as subspaces. Let:
$$l = \sum\limits_{i=1}^{j} = \alpha_il_i$$ $$m = \sum\limits_{i=1}^{k} = \beta_im_i$$

Now, I want to claim that $v=l+m$, is of course $\in V$ but $\notin L\cup M$

Is that how am I suppose to do that? How to proceed then?

Thanks.

Elimination
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1 Answers1

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If $L\subset M$ or $M\subset L$ then the result is obvious. Now assume that $L\not\subset M$ and $M\not\subset L$ and let $l\in L\setminus M$ and $m\in M\setminus L$ so $u=l+m\not\in L\cup M$. In fact, if not then $u\in L$ or $u\in M$ and in both cases we find easily a contradiction: $$m=u-l\in L\qquad\text{or}\qquad l=u-m\in M$$