# (i)Prove that if{v1,v2}is linearly dependent, then are multiple of each other, that is, there exists a constant c such that v1 = c v2 or v2=cv1. (ii)P

Trent Carpenter 2020-11-06 Answered

(i)Prove that if${v}_{1},{v}_{2}$is linearly dependent, then are multiple of each other, that is, there exists a constant c such that .
(ii)Prove that the converse of(i) is also true.That is to say, if there exists a constant c such that , 1. then${v}_{1},{v}_{2}$is linearly dependent.

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## Expert Answer

hosentak
Answered 2020-11-07 Author has 100 answers
Recall:Linear Dependence:A finite set of vectors $\left\{{a}_{1},{a}_{2},\dots ,{a}_{n}$ pf a vector space V over a field F is said to be linearly dependent in V if there exist scalars $c1,c2,\dots {c}_{n}$, not all zero, in F such that
${c}_{1}{a}_{1}+{c}_{2}{a}_{2}+\dots +{c}_{n}{a}_{n}=0$
θ is the zero element of the vector space V.
(i)Given$\left\{{v}_{1},{v}_{2}\right\}$is linearly independent i.e ${v}_{1},{v}_{2}$ are linearly dependent. Let 0 be the zero element. So there exist scalars ${c}_{1},{c}_{2}$ such that
${c}_{1}{v}_{1}+{c}_{2}{v}_{2}=0$,
if any one of the scalar is 0,then the other scalar also becomess 0(since the vectors are non zero).
So for linear dependency both the scalars ${c}_{1},{c}_{2}$ have to be non-zero.Therefore,
${c}_{1}{v}_{1}+{c}_{2}{v}_{2}=0$
$⇒{c}_{1}{v}_{1}=-{c}_{2}{v}_{2}$
$⇒{v}_{1}=\left(-\frac{{c}_{2}}{{c}_{1}}\right){v}_{2}$
$⇒{v}_{1}=c{v}_{2}$
Similarly,
${c}_{1}{v}_{1}+{c}_{2}{v}_{2}=0$
$⇒{c}_{2}{v}_{2}=-{c}_{1}{v}_{1}$
$⇒{v}_{2}=\left(-\frac{{c}_{1}}{{c}_{2}}\right){v}_{1}$
$⇒{v}_{2}=c{v}_{1}$.
So it is proved that if $\left\{{v}_{1},{v}_{2}\right\}$ is linearly dependent, then there exist a constant c such that either ${v}_{1}=c{v}_{2}\phantom{\rule{1em}{0ex}}\text{or}\phantom{\rule{1em}{0ex}}{v}_{2}=c{v}_{1}$.
(ii)Let there exist a constant c not equal to 0 such that either ${v}_{1}=c{v}_{2}\left(\phantom{\rule{1em}{0ex}}\text{or}\phantom{\rule{1em}{0ex}}{v}_{2}=c{v}_{1}\right)$.
Then it can be written as,
${v}_{1}=c{v}_{2}$
$⇒{v}_{1}-c{v}_{2}=0$
Hence by the definition of linear dependence we can say that ${v}_{1},{v}_{2}$ are linearly dependent i.e.
$\left\{{v}_{1},{v}_{2}\right\}$ is linearly dependent
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