# Do every $3$ linearly independent vectors span all of $\mathbb{R}^3$?

I am given $3$ vectors that are linearly independent. I am trying to figure our if they span all of $\mathbb{R}^3$ to declare them as basis.

• I assume you mean, "do EVERY 3 linearly independent vectors span all of $\mathbb{R}^3$, because if you just needed to find one set of 3 vectors this would be trivial. Editing accordingly, please comment if you object. – 6005 Jun 30 '13 at 22:01
• @Goos I agree with the edit. thank you. – Surya Jun 30 '13 at 22:55

Yes, because $\mathbb R^3$ is $3$-dimensional (meaning precisely that any three linearly independent vectors span it). To see this, note that if we had $3$ linearly independent vectors which did not span $\mathbb R^3$, we could expand this to a collection of $4$ linearly independent vectors. Writing these in a matrix and performing row-reduction shows that this is impossible.

Your vectors span a subspace of $\mathbb{R}^{3}$ of dimension $3$.

Since the dimension of $\mathbb{R}^{3}$ is $3$ we conclude this subspace is all of $\mathbb{R}^{3}$.

ADDED: I have used the claim that if $V$ is a finite dimensional vector space, $S$ is a sub vector space with $\dim(S)=\dim(V)$, then $S=V$.

I did not include a proof for this, as I assumed it to be known (thus my answer is, as pointed out in the comments, is not self-contained).

• I actually don't buy this argument. Try it with $\mathbb{R}^3$ replaced by the vector space $\mathbb{R}[t]$ and its dimension $\aleph_0$: it doesn't work. We need some argument particular to the finite-dimensional case to tell us that we can't have a proper inclusion of vector spaces of the same dimension. (Possible arguments include row-reduction and the Steinitz Exchange Lemma.) – Pete L. Clark Jun 30 '13 at 22:06
• @PeteL.Clark - I believe that most of the basic linear algebra courses only treat the case of finite dimensional vector spaces (at least this is the case whre I study). It is often proved in those courses the property I used – Belgi Jun 30 '13 at 22:08
• I agree with both of the things you said. In fact the OP's question is certainly addressed in any basic linear algebra course, so in some sense a correct answer would be to refer him to the notes/text/videos of such a course. But it looks like you are claiming a self-contained proof of the result. I noted that your assertion doesn't hold in the infinite-dimensional case as proof that your argument is not in fact self-contained: something must be missing. If you don't in fact purport to give a self-contained proof, that's fine but it would be nice if you said so in your answer. – Pete L. Clark Jul 1 '13 at 0:54
• @PeteL.Clark -I have edited accordingly. I'm sorry I didn't do this earlier, my finals just started and I was incredibly busy. Thanks for pointing that out (I admit that at the time I didn't notice that myself , so your comment helped me understand what I was using and recall where it holds) – Belgi Jul 24 '13 at 17:21