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I'm trying to prove the following: Given $\{v_1, v_2\}, \{v_1, v_3\}$, and $\{v_2, v_3\}$ are all linearly independent but $\{v_1, v_2, v_3\}$ is linearly dependent that

$$\text{span}\{v_1, v_2\} = \text{span}\{v_1, v_3\} = \text{span}\{v_2, v_3\}.$$

Any help would be appreciated.

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If $\{u,v,w\}$ is linearly dependent, then $\text{dim}(\text{span}\{u,v,w\})<3$. But since $\{u,v\}$ is linearly independent, $\text{dim}(\text{span}\{u,v,w\})\ge 2$. So in particular the dimension is $2$. But then any two lineraly independent vectors span the whole space, so in your case with $u=v_1, v=v_2, w=v_3$ we have that $\text{span}\{v_i, v_j\}=\text{span}\{u,v,w\}$ for all $i\ne j$, hence they are in particular all equal to one another.

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  • $\begingroup$ Perfect. Many thanks. $\endgroup$ – H.Rappeport Dec 15 '16 at 13:52

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