Q: Working in $V$ = any vector space. Prove that if $\vec v \not = \vec 0$ and $\vec w \notin \langle \vec v \rangle$ then $\{\vec v, \vec w \}$ is linearly independent.

So since $\vec w \notin \langle \vec v \rangle$ this means that $\vec w$ is not our set of vectors. I'm not sure if I need to figure out how to get $\vec w$ in our vector space $V$ or not? Any help?

EDIT: So a vector equation would be $x_1\vec v_1 + x_2\vec v_2 + ... + x_k\vec v_k = \vec 0_V$

Could I say $\vec w$ is in the linear combination of V such that $\vec w = c_1\vec v_1 + c_2\vec v_2 + ... + c_k\vec v_k$ ?

I'm still not sure I could put $\vec v$ and $\vec w$ in the same vector equation since $\vec w \notin \langle \vec v \rangle$.

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    $\begingroup$ $\vec{w} \not \in \left< \vec{v} \right>$ means $\vec{w}$ is not in the span of $\vec{v}$ (it is not a scalar multiple of $\vec{v}$). Try a proof by contradiction. If they were linearly dependent, then there would be constants $a$ and $b$ so that $a \vec{v} + b \vec{w} = \vec{0}$. Can you see what to do from here? $\endgroup$ – Nick Mar 8 '17 at 21:22
  • $\begingroup$ If they were linearly dependent and such scalars existed that would mean there would be some scalar $\vec v$=$\vec w$ (similar to what mvmath has done) thus $\vec w$ would be in our span. So since we've reached this contradiction we can state that they must be linearly independent, yes? $\endgroup$ – K Math Mar 10 '17 at 20:18
  • $\begingroup$ Sorry to bring up an old question but if we changed from $\vec w \notin <\vec v>$ to $\vec v \in <\vec w>, \vec w \in <\vec v>$ We would be linearly independent right? $\endgroup$ – K Math Apr 10 '17 at 21:39
  • $\begingroup$ No. If something is in the span of something else, then they are linearly DEPENDENT (not linearly independent). $\endgroup$ – Nick Apr 11 '17 at 13:51

Let's assume, that they are linear dependent. It means that there are two constants $a_1, a_2$ : $a_1v+a_2w=0 \Leftrightarrow w = -\frac{a_1}{a_2}v$, which means, that $w \in \left< v \right>$, because $v \neq 0$. Contradiction.

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  • $\begingroup$ Note that your equivalence is only true if $a_2\neq0$. In general, that doesn't have to be true. But it is so in this question because we know that $\overrightarrow{v}\neq\overrightarrow{0}$. $\endgroup$ – zipirovich Mar 8 '17 at 22:14
  • $\begingroup$ Yes, you'r right. I'll fix it. $\endgroup$ – MOVZBL Mar 8 '17 at 22:20

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