# Minimum number of vectors that span a linear subspace

There was this question that I got wrong when doing some practicing problems for my freshman Linear Algebra course:

Let $$\mathbf{v_1}, \mathbf{v_2}, \mathbf{v_3}, \mathbf{v_4}$$ be non-zero vectors of a given vector space and $$\mathcal{L} \{ \mathbf{v_1,v_2,v_3,v_4} \}$$ the linear subspace $$V$$ generated by those vectors. Given:

• $$\mathbf{v_2} \notin \mathcal{L} \{ \mathbf{v_1} \} ;$$
• $$\mathbf{v_3} \in \mathcal{L} \{ \mathbf{v_1,v_2} \} ;$$
• $$2 \mathbf{v_4} + 2 \mathbf{v_3} + 7 \mathbf{v_2} + 4 \mathbf{v_1} = 0$$

What is the minimum number of linearly independent vectors that still span $$V$$?

I can conclude that because $$\mathbf{v_2}$$ is not included on the span of $$\mathbf{v_1}$$, they are not a linear combination of one another and thus linearly independent. So, we need both of them to generate $$V$$.

$$\mathbf{v_3}$$, however, is not needed because it can be represented as a linear combination of $$\mathbf{v_1}$$ and $$\mathbf{v_2}$$.

I said that the minimum number of required vectors was 3 and I was wrong. I know that it has something to do with the third equation but I don't understand how to get there.

Thanks for your help.

• By #2, $\boldsymbol v_3$ is a linear combination of $\boldsymbol v_1, \boldsymbol v_2$, then plug this into #3 you get that $\boldsymbol v_4 \in \mathcal L\{\boldsymbol v_1, \boldsymbol v_2\}$ as well. So actually, $\boldsymbol v_3, \boldsymbol v_4$ are not needed. – xbh Nov 6 '18 at 16:14

the last equation tell you that also $$\mathbf{v_4}$$ is not necessary since $$\mathbf{v_4} = -\frac{2 \mathbf{v_3} + 7 \mathbf{v_2} + 4 \mathbf{v_1}}{2}$$
hence it lies in $$\mathcal{L} \{ \mathbf{v_1,v_2,v_3} \} = \mathcal{L} \{ \mathbf{v_1,v_2} \};$$
• yes, sure you can use different vectors which define the same space, the invariant is the number of vector you need (the dimension of a basis). In this case we choose $v_1$ and $v_2$ but the same is true with $v_1$ and $v_4$ for example if $v_4$ is not in the span of $v_1$ . – ALG Nov 6 '18 at 17:23
In the last equation you in fact have only three "interesting" vectors: $$\;v_1,v_2,v_4\;$$ , since $$\;v_3\;$$ is a linear combination of the first two, so we in fact could write
$$Av_1+Bv_2+4v_4=0\;$$
Assuming you're working on a vector space over a field of characteristic $$\;\neq2\;$$ ( most probably, over the reals $$\;\Bbb R\;$$) , the last equality means $$\;v_1,v_2,v_4\;$$ are linearly dependent since $$\;4\neq0\;$$ , and since we already know $$\;v_1,v_2\;$$ are linearly independent, this means $$\;v_4\;$$ is lin. dep. on $$\;v_1,v_2\;$$ so we only need these first two vectors to span that space.