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.

  • $\begingroup$ 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. $\endgroup$ – 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 $\mathcal{L} \{ \mathbf{v_1,v_2,v_3} \} = \mathcal{L} \{ \mathbf{v_1,v_2} \};$

  • $\begingroup$ I thought about this, but I got confused because we can isolate any vector and all of them could be lin. comb. of each other, so each is an element of the span of the other three. $\endgroup$ – DC_AC Nov 6 '18 at 17:20
  • $\begingroup$ 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 $v_1$ . $\endgroup$ – 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


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.


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