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 $span(v1,v2)$


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).

| cite | improve this answer | |
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.