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Problem

Show that if vectors $(\overline{v},\overline{w}) \in V$ are linearly independent and neither of them is zero vector then they are not parallel

Attempt to solve

vectors $\overline{v},\overline{w}$ are linear independent if

$$ \exists(c_1,c_2)\in \mathbb{R} : c_1\overline{v} + c_2\overline{w} = \overline{0} \implies c_1=0,c_2=0 $$

Now it follows from this that they are not parallel when this condition is satisfied.


However, I'm having trouble connecting the fact that these vectors cannot be parallel when they are linearly independent. This is intuitive to me at some level by the definition.

One way would be to find a connection with cross product and the fact that when

$$ \overline{v} \times \overline{w} = 0 \implies \text{ parallel} $$

then since I wanted to show that they are not parallel use negation

$$ \overline{v} \times \overline{w} \neq 0 \implies \text{ not parallel } $$

But it's problematic since it limits me to $\mathbb{R}^3$ vector space?

Better option is possibly to try to find

$$ \forall(a,b)\in \mathbb{R} : a \overline{v} - b \overline{w} \neq \overline{0} $$

which implies they cannot be parallel since by scaling them with arbitrary $(a,b)$ they cannot be the same.

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    $\begingroup$ What is the exact definition of "parallel" in this context? Is it $\exists a, b \in \mathbb{R}: a \bar{v} = b \bar{w}$? $\endgroup$ – Connor Harris Aug 12 '19 at 19:55
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    $\begingroup$ well, the problem doesn't give a definition for "parallel" so I made one by myself. I guess any definition that is correct is fine. @ConnorHarris $\endgroup$ – Tuki Aug 12 '19 at 19:59
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    $\begingroup$ In that case, suppose that $\bar{v}$ and $\bar{w}$ are parallel, i.e. that there exist $a, b \in \mathbb{R} \setminus \{0\}$ such that $a \bar{v} = b \bar{w}$. This means $a \bar{V} - b \bar{w} = 0$, which contradicts linear independence. $\endgroup$ – Connor Harris Aug 12 '19 at 20:03
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    $\begingroup$ So use the fact that $$ q \rightarrow p \iff \neg p \rightarrow \neg q $$ so indirect proof it is? $\endgroup$ – Tuki Aug 12 '19 at 20:06
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    $\begingroup$ That's correct. $\endgroup$ – Connor Harris Aug 12 '19 at 20:09
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I assume $V$ is a vector space over the field of scalars $\Bbb F$.

If

$0 \ne \bar v, \bar w \in V \tag 1$

are parallel, then

$\exists 0 \ne \alpha \in \Bbb F, \; \bar v = \alpha \bar w; \tag 2$

therefore,

$1_{\Bbb F} \bar v - \alpha \bar w = \bar v - \alpha \bar w = 0, \tag 3$

which is a relationship of the form

$c_1 \bar v + c_2 \bar w = 0, \tag 4$

with

$c_1 = 1_{\Bbb F} \ne 0 \ne -\alpha = c_2; \tag 5$

then by definition, $\bar v$ and $\bar w$ are linearly dependent. By contraposition, this implies that linearly independent $\bar v$, $\bar w$ cannot be parallel. $OE\Delta.$

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    $\begingroup$ Hey what do you mean by scalar field exactly? I'm having a bit trouble understanding your answer. Also I find the "which is a relationship of the form" a bit vague . $\endgroup$ – Tuki Aug 13 '19 at 0:31
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    $\begingroup$ A vector space is specified over some field $\Bbb F$, which acts on vectors according the usual rules. This is the "field of scalars" of $V$; it is not a "scalar field" in the sense of, say, electrodynamics. "A relationship of the form $c_1 \bar v + c_2 \bar w = 0$" is simply a linear dependence 'twixt $\bar v$ and $\bar w$, an abstracion of which (3) is a concrete example. $\endgroup$ – Robert Lewis Aug 13 '19 at 1:06
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By the contrapositive, if they are parallel, then there must exist a scalar $\alpha$ such that $\bar{v} - \alpha \bar{w} = 0$.

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