Let $L_1,L_2$, two subspaces of a finite dimensional vector space.
Prove that if $\dim(L_1 + L_2) = 1 + \dim (L_1 \cap L_2)$ then $L_1\subseteq L_2$ or $L_2 \subseteq L_1$.

Well, I've read a proof which was done by assuming the contradiction. But I was thinking, why not just claim:

$$\dim(L_1 + L_2) = 1 + \dim(L_1 \cap L_2)$$

$$\dim L_1 + \dim L_2 = 1$$

Because dimensions are integers, one must be $0$. WLOG, $\dim L_1 = 0$. Therefore, $L_1 \subseteq L_2$. The other case is similar of course.

Am I right here? Because it seems pretty much easy (instead of proving by contradiction)

  • $\begingroup$ Your argument is not correct. For instance it doesn't work with the $xy$-plane inside $\Bbb R^3$. $\endgroup$
    – Arthur
    Jun 29, 2014 at 15:59
  • $\begingroup$ What makes it incorrect then? $\endgroup$
    – AnnieOK
    Jun 29, 2014 at 16:05
  • $\begingroup$ $L_1$ is the xy plane and $L_2$ is a line within it. These subspaces of $R^3$ satisfy the proposition, but neither has dimension 0 as in your argument. $\endgroup$ Jun 29, 2014 at 16:29

2 Answers 2


I think you made a sign error when simplifying and accidentally cancelled your $\dim(L_1 \cap L_2)$. I assume you were using $\dim(L_1 + L_2) = \dim(L_1) + \dim(L_2) - \dim(L_1 \cap L_2)$. Then you have

$$\dim(L_1 + L_2) = 1 + \dim(L_1 \cap L_2)$$ $$\dim(L_1) + \dim(L_2) - \dim(L_1 \cap L_2) = 1 + \dim(L_1 \cap L_2)$$ $$\dim(L_1) + \dim(L_2) = 1 + 2\dim(L_1 \cap L_2)$$

which does not imply that one the spaces is trivial.


Without loss of generality, suppose $L_1$ is not a subset of $L_2$. Then $L_1 \cap L_2$ is a proper subspace of $L_1$. And hence $\text{dim}(L_1\cap L_2)+1\leq \text{dim}(L_1)$. Since $\text{dim}(L_1+L_2)=\text{dim}(L_1\cap L_2)+1$ and $L_1$ is a subspace of $L_1+L_2$, $L_1=L_1+L_2$. This implies that $L_2\subset L_1$


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