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.
Best Answer
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 $\mathcal{L} \{ \mathbf{v_1,v_2,v_3} \} = \mathcal{L} \{ \mathbf{v_1,v_2} \};$