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So I have a set of 2 vectors, $v1,v2$ which are linearly independent and the dimension of the subspace they belong to is 3 ($dim(W) =3)$. How can I prove there is some vector $W$ that is not a linear combination of $v1,v2$?

edit: my attempt so far, I assumed that some vector $W$ is in $span(v1,v2)$

I show some vector in the subspace, $W$ can be written as a linear combination of $v1,v2$

then i showed $v1 + v2 -W = 0$ and hence $W,v1,v2$ and linearly dependent which contradicts the original assumption that $W$ is in $span(v1,v2)$

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Suppose for all $v \in W$, the vectors $ v_1,v_2,v$ are linearly dependent. Then we can write $v$ as a linear combination of $v_1,v_2$ (why?). But that means $v_1,v_2$ span $W$. Hence $v_1,v_2$ is a basis and hence dimension of $W$ is 2 (contradiction).

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  • $\begingroup$ Thanks for the reply! I probably should have put in my attempt at the proof in the original question but I edited it in. Does it look okay? $\endgroup$ – lohboys Dec 15 '18 at 2:04

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