Let $L_1,L_2$ be sub-spaces of finitely generated vector space.
Prove that if $\dim(L_1+L_2)=1+\dim(L_1 \cap L_2)$, then $L_1 \subseteq L_2$ or $L_2 \subseteq L_1$.
Unfortunately, I don't have a clue where to start...
Please help, thank you!
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2 Answers
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We know that $$\dim(L_1+L_2)=\dim L_1+\dim L_2-\dim(L_1\cap L_2)$$ so with the hypothesis we have $$\dim L_1+\dim L_2=2\dim(L_1\cap L_2)+1\tag{*}$$ but by contradiction if $L_1\not\subset L_2$ and $L_2\not\subset L_1$ then $$\dim L_1\ge \dim(L_1\cap L_2)+1$$ and $$\dim L_2\ge \dim(L_1\cap L_2)+1$$ hence $$\dim L_1+\dim L_2\ge2\dim(L_1\cap L_2)+2$$ which contradicts $(*)$.
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$\begingroup$ You're welcome. $\endgroup$– user63181Feb 14, 2014 at 13:37
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You know $L_1 ⊂ L_2$ is equivalent to $L_1 ∩ L_2 = L_1$.
So assume $L_1 \not\subset L_2$. Therefore $L_1 ∩ L_2 \subsetneq L_1$. You arrive at: $$\dim (L_1 ∩ L_2) < \dim L_1 ≤ \dim (L_1 + L_2) = \dim (L_1 ∩ L_2) + 1.$$ Since all dimensions are integers and they already increase once with the “$<$”, you can conclude $\dim L_1 = \dim (L_1 + L_2)$. Else they’d increase twice. Especially, you get $L_1 = L_1 + L_2$, since both spaces have the same dimension and one is included in the other. Now $L_2 ⊂ L_1 + L_2 = L_1$.
All in all: Either $L_1 ⊂ L_2$ or $L_2 ⊂ L_1$.
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$\begingroup$ Very clear answer, thanks. The conclusion needs to be fixed - $L_1 \subseteq L_2$ or $L_2 \subseteq L_1$ $\endgroup$– Galc127Feb 14, 2014 at 13:38