(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)Prove that the converse of(i) is also true.That is to say, if there exists a constant c such that v1 = c v2 or v2 = c v1.1, then{v1,v2}is linearly dependent.

(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)Prove that the converse of(i) is also true.That is to say, if there exists a constant c such that v1 = c v2 or v2 = c v1.1, then{v1,v2}is linearly dependent.

Question
Alternate coordinate systems
asked 2020-11-06
(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)Prove that the converse of(i) is also true.That is to say, if there exists a constant c such that v1 = c v2 or v2 = c v1.1, then{v1,v2}is linearly dependent.

Answers (1)

2020-11-07
Recall:Linear Dependence:A finite set of vectors \(\displaystyle{\left\lbrace{a}_{{1}},{a}_{{2}},\ldots,{a}_{{n}}\right.}\) pf a vector space V over a field F is said to be linearly dependent in V if there exist scalars \(\displaystyle{c}{1},{c}{2},\ldots{c}_{{n}}\), not all zero, in F such that
\(\displaystyle{c}_{{1}}{a}_{{1}}+{c}_{{2}}{a}_{{2}}+\ldots+{c}_{{n}}{a}_{{n}}={0}\)
θ is the zero element of the vector space V.
(i)Given\(\displaystyle{\left\lbrace{v}_{{1}},{v}_{{2}}\right\rbrace}\)is linearly independent i.e \(\displaystyle{v}_{{1}},{v}_{{2}}\) are linearly dependent. Let 0 be the zero element. So there exist scalars \(\displaystyle{c}_{{1}},{c}_{{2}}\) such that
\(\displaystyle{c}_{{1}}{v}_{{1}}+{c}_{{2}}{v}_{{2}}={0}\),
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 \(\displaystyle{c}_{{1}},{c}_{{2}}\) have to be non-zero.Therefore,
\(\displaystyle{c}_{{1}}{v}_{{1}}+{c}_{{2}}{v}_{{2}}={0}\)
\(\displaystyle\Rightarrow{c}_{{1}}{v}_{{1}}=-{c}_{{2}}{v}_{{2}}\)
\(\displaystyle\Rightarrow{v}_{{1}}={\left(-\frac{{c}_{{2}}}{{c}_{{1}}}\right)}{v}_{{2}}\)
\(\displaystyle\Rightarrow{v}_{{1}}={c}{v}_{{2}}\)
Similarly,
\(\displaystyle{c}_{{1}}{v}_{{1}}+{c}_{{2}}{v}_{{2}}={0}\)
\(\displaystyle\Rightarrow{c}_{{2}}{v}_{{2}}=-{c}_{{1}}{v}_{{1}}\)
\(\displaystyle\Rightarrow{v}_{{2}}={\left(-\frac{{c}_{{1}}}{{c}_{{2}}}\right)}{v}_{{1}}\)
\(\displaystyle\Rightarrow{v}_{{2}}={c}{v}_{{1}}\).
So it is proved that if \(\displaystyle{\left\lbrace{v}_{{1}},{v}_{{2}}\right\rbrace}\) is linearly dependent, then there exist a constant c such that either \(\displaystyle{v}_{{1}}={c}{v}_{{2}}{\quad\text{or}\quad}{v}_{{2}}={c}{v}_{{1}}\).
(ii)Let there exist a constant c not equal to 0 such that either \(\displaystyle{v}_{{1}}={c}{v}_{{2}}{\left({\quad\text{or}\quad}{v}_{{2}}={c}{v}_{{1}}\right)}\).
Then it can be written as,
\(\displaystyle{v}_{{1}}={c}{v}_{{2}}\)
\(\displaystyle\Rightarrow{v}_{{1}}-{c}{v}_{{2}}={0}\)
Hence by the definition of linear dependence we can say that \(\displaystyle{v}_{{1}},{v}_{{2}}\) are linearly dependent i.e.
\(\displaystyle{\left\lbrace{v}_{{1}},{v}_{{2}}\right\rbrace}\) is linearly dependent
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