# proving $\text{span}\{v1, v2\} = \text{span}\{v1, v3\} = \text{span}\{v2, v3\}$

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.

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.