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$L_1$ and $L_2$ are vector subspaces of the vector space $V$ with finite dimension.

Prove: If $\dim(L_1+L_2) = 1 + \dim(L_1\cap L_2)$ than the sum $L_1+L_2$ equals to one of the subspaces and the intersection $L_1\cap L_2$ equals the other one.

I can see why it's true, and I've tried to use the dimension theorem but couldn't evaluate it.

Any ideas?

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Hint: ($L_1\cap L_2)\subset L_i \subset (L_1 + L_2)$. What if you apply $\dim(\cdot)$?

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  • $\begingroup$ So I should do something like the squeeze theorem to the dimensions and to conclude that the dimension of one of the subspaces equals the dimension intersection? How can I explain it ? And can I conclude from it that the subspace equals the intersection? $\endgroup$
    – Itay4
    Jan 16, 2017 at 9:28
  • $\begingroup$ If a containment is strict, then the dimensions can't be equal ( can you prove this?), and if the dimensions are equal, the containment is not strict (spaces are equal; can you prove this too?). Note this is only true for finite-dimensional spaces $\endgroup$
    – MPW
    Jan 16, 2017 at 9:38
  • $\begingroup$ I can prove these propositions, but how do I conclude from the fact the dimensions are equal and the containment is not strict that the spaces are equal? $\endgroup$
    – Itay4
    Jan 16, 2017 at 9:45

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