# A union of two subspaces not equal to the vector space. [duplicate]

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 $L \cup M = V$.

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

How to proceed?

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

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

• 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 $x + y \not\in L \cup M$.
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.