Let u and v be distinct vectors of a vector

Kathleen Rausch 2022-01-07 Answered
Let u and v be distinct vectors of a vector space V. Show that if {u, v} is a basis for V and a and b are nonzero scalars, then both {u+v, au} and {au, bv} are also bases for V.

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Expert Answer

Orlando Paz
Answered 2022-01-08 Author has 767 answers
Given, u and v be distinct vectors of a vector space V and {u, v} is a basis for V and a and b are nonzero scalars.
We have to show: both {u + v, au} and {au, bv} are also bases for V.
\(\displaystyle{u}\ne{v},{a},{b}\ne{0}\)
\(\displaystyle{\left\lbrace{u},{v}\right\rbrace}\) is a basis for V \(\displaystyle\Rightarrow{\left\lbrace{u}+{v},{a}{u}\right\rbrace}\) and \(\displaystyle{\left\lbrace{a}{u},{b}{v}\right\rbrace}\) are bases for V
From basis \(\displaystyle{\left\lbrace{u},{v}\right\rbrace}\) we get dimension of V
\(\displaystyle{\left\lbrace{u},{v}\right\rbrace}\) is a basis for V \(\displaystyle\Rightarrow\) V is a 2-dimensional vector space and \(\displaystyle{\left(\cdot\right)}\alpha{u}+\beta{v}={0}\Rightarrow\alpha=\beta={0}\)
Show \(\displaystyle{\left\lbrace{u}+{v},{a}{u}\right\rbrace}\) is a basis for V:
\(\displaystyle{\left\lbrace{u}+{v},{a}{u}\right\rbrace}\):
\(\displaystyle{A}{\left({u}+{v}\right)}+{B}{\left({a}{u}\right)}={0}\Leftrightarrow{A}{u}+{A}{v}+{B}{a}{u}={0}\Leftrightarrow\)
\(\displaystyle{\left({A}+{B}{a}\right)}{u}+{\left({B}\right)}{v}={0}\)
By the step (in second step)
\(\displaystyle{A}+{B}{a}={0},{a}\ne{0}\)
\(\displaystyle{B}={0}\)
\(\displaystyle\Rightarrow{A}={B}={0}\Rightarrow{\left\lbrace{u}+{v},{a}{u}\right\rbrace}\) is linearly independent set of two vectors
\(\displaystyle\Rightarrow{\left\lbrace{u}+{v},{a}{u}\right\rbrace}\) is a basis for V
(Proved)
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xandir307dc
Answered 2022-01-09 Author has 598 answers

here is the continuation of the solution:
Show \(\displaystyle{\left\lbrace{a}{u},{b}{v}\right\rbrace}\) is a basis for V
\(\displaystyle{\left\lbrace{a}{u},{b}{v}\right\rbrace}:\)
\(A(au)+B(bv)=0 \Leftrightarrow (Aa)u+(bb)v=0\)
By the step (in second step)
\(\displaystyle{A}{a}={0},{a}\ne{0}\)
\(\displaystyle{B}{b}={0},{b}\ne{0}\)
\(\displaystyle\Rightarrow{A}={B}={0}\Rightarrow{\left\lbrace{a}{u},{b}{v}\right\rbrace}:\) is linearly independent set of two vectors
\(\displaystyle\Rightarrow{\left\lbrace{a}{u},{b}{v}\right\rbrace}:\) is a basis for V
(Proved)

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