Q: Suppose $\{\vec v_1, \vec v_2, ..., \vec v_k\}$ is linearly independent and $\vec w \notin \left<\vec v_1, \vec v_2, ..., \vec v_k\right>$. Proving that $\{\vec v_1, \vec v_2, ..., \vec v_k, \vec w\}$ is linearly independent.

I have chosen to prove by contradiction (maybe, I'm stuck right now so maybe this wasn't the right choice).

So suppose $\{\vec v_1, \vec v_2, ..., \vec v_k, \vec w\}$ is linearly dependent. This means there are scalars such that: $\{\,c_1\vec v_1+c_2\vec v_2+...+c_k\vec v_k + c_l\vec w_l=\vec 0\mid c \in R\,\}$

But this would mean $\vec w$ is in our span. I think I am missing some steps and may not even be coming to the correct conclusion.

EDIT: so I can state that $c_l \neq 0$ bc otherwise $ \vec w$ would be the zero vector and any space with the zero vector in the span cannot be linearly independent?

Beyond this, I've been trying a calculation to prove a vector is in a span.

Q: In $R^5$ let $S=\left<(1,0,1,0,1), (2,1,0,1,1), (4,1,2,1,3), (0,0,1,1,1)\right>$. I need to determine if $(1,2,3,4,5)$ and $(2,1,1,1,1)$ are in $S$.

I made an augmented matrix with $(1,2,3,4,5)$ but ended up with three rows giving me difference answers as to what $x_4$ would be? Am I messing up my math?

  • 1
    $\begingroup$ You are right. The proof by contradiction is just definition (and noticing that some $c_k$ is not zero) $\endgroup$ – Will M. Mar 10 '17 at 20:50
  • $\begingroup$ If $c_l = 0$, consider the first $c_j$ in $(c_1, \dots, c_k)$ such that $c_j = 0$, but then this would mean $(v_1,\dots, v_k)$ is linearly dependent. $\endgroup$ – IAmNoOne Mar 10 '17 at 22:04
  • $\begingroup$ I've edited the question typographically (though you probably also want $\Bbb R$, typed \Bbb{R}, instead of $R$). But $\{\,c_1\vec v_1+c_2\vec v_2+...+c_k\vec v_k + c_l\vec w_l=\vec 0\mid c \in R\,\}$ is still mathematically wrong, having an equation at the left in $\{\mid\}$ is meaningless (you need elements there, not conditions). What you probably meant is just the equation, without the $\{\mid c\in R\}$ (and note that no plain $c$ occurs on the left in the first place). $\endgroup$ – Marc van Leeuwen Mar 17 '17 at 7:37
  • $\begingroup$ @MarcvanLeeuwen I apologize for my formatting, this is how we set things up in our class. I understand your comment though. However, if I let the script on my $c=0$ i.e. let $c_l\vec w$ where $l=0$ then $\vec w$ would be the zero vector, thus making it linearly dependent. Otherwise, any $c_l$ where $l \ne 0$ will be linearly independent, yes? $\endgroup$ – K Math Mar 20 '17 at 18:54

What you know is that there are scalars $c_1,\dots,c_k,c_ld$ such that $c_1\vec{v}_1+\cdots+c_k\vec{v}_k+c_l\vec{w}=\vec{0}$ and not all of the scalars are zero. You have to consider the cases $c_l=0$ and $c_l\neq 0$ separately. In particular, if $c_l=0$, you cannot rearrange the equation to show that $\vec{w}$ is in the span of $\vec{v}_1,\dots,\vec{v}_k$. But in that case, you can say something about the linear (in)dependence of $\vec{v}_1,\dots,\vec{v}_k$.


Since you write $R$ for your ring of scalar, I am assuming that you are working in more general context than a field, for example including $\mathbb{Z}$. In that case the claim is wrong, for example, $1\notin\langle 2\rangle$ but $\{2,1\}$ is not linearly independent.

However, if you are indeed assuming a field. Then what you need to do next is as follow:

First $c_{l}\not=0$ since otherwise you will have a nontrivial linear combination not involving $w$, contradicting a hypothesis that the rest are linearly independent.

Then since you are in a field, $c_{l}$ is invertible, so by dividing by $c_{l}$ and move $w$ to the other side, you can write $w$ as a linear combination of the rest, also violating the span condition.


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