I have a set of independent vectors {${u, v}$} and I need to determine if the set of vectors {$u - v, v, 4v$} is also linearly independent. I originally saw the question here which seems to be exactly what I need to do, however I'm slightly confused about the combination of $v$ and $4v$. It seems that those two are already linearly dependent so it seems that I could simply leave out the $4v$ and show that $u-v$ and $v$ are linearly independent:

$$a*(u - v) + b(v) = 0\\ (a) * u + (-a + b) * v = 0 $$

which leads to

$$ \begin{bmatrix} 1 & 0 & 0 \\ -1 & 1 & 0\\ \end{bmatrix} $$

from which I can say that the two vectors are linearly independent, because there is no non-trivial solution. Is it possible to leave out the $4v$?

  • $\begingroup$ You are right: $\{u-v,v\}$ is a linearly independent set but $\{u-v,v,4v\}$ is not. $\endgroup$ – A. Goodier Jan 1 '18 at 15:31
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    $\begingroup$ The subsystem $\{v,4v\}$ is linearly dependent, hence, the whole system is dependent as well. $\endgroup$ – A.Γ. Jan 1 '18 at 15:32
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    $\begingroup$ No three vectors in a two dimensional system can be linearly independent. $\endgroup$ – lulu Jan 1 '18 at 15:38

You can just choose one of $v$ or $4v$ for the purposes of this question. As you remark, $v$ and $4v$ are linearly dependent because $$ 4v+(-4)v=0$$ while the coefficients of $4v$ and $v$ are nontrivial. For your original question of "is $\{u-v,v,4v\}$ linearly independent?" this is sufficient. Take $$ 0(u-v)+4v+(-4)v=0.$$ If you wish to proceed anyway to learn about $\{u-v,v\}$, then from here you can just row reduce the unaugmented system $$ \begin{bmatrix} 1&0\\ -1&1 \end{bmatrix}$$ to find the dimension of $\text{span}(u-v,v)$. It's basically true by inspection that the dimension of this span is $2$, so that they are indeed linearly independent.


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