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)

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)