0
$\begingroup$

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 $\vec w \notin \langle \vec v \rangle$.

$\endgroup$
  • 1
    $\begingroup$ $\vec{w} \not \in \left< \vec{v} \right>$ 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
1
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.