That is let \(\displaystyle{\left\lbrace{T},{U}\right\rbrace}\) is not linearly independent subset of L(V,W) then there exist some \(\displaystyle\alpha,\beta\) not both such that \(\displaystyle\alpha{T}\). \(\displaystyle\beta{U}={0}\), where 0 is the transformation that takes every element of V to zero.

Let Assume that a #0 then there exist xeV such that x e kemel of T.

Let \(\displaystyle{y}={T}{\left({x}\right)}\ne{0}\).

It is known that,

\(\displaystyle{0}={\left(\alpha{T}+\beta{U}\right)}{\left({x}\right)}\)

\(\displaystyle=\alpha{T}{\left({x}\right)}+\beta{U}{\left({x}\right)}\)

This can also be written as follows.

\(\displaystyle{y}={T}{\left({x}\right)}\)

\(\displaystyle={\frac{{-\beta}}{{\alpha}}}{U}{\left({x}\right)}\)

\(\displaystyle={U}{\left({\frac{{-\beta}}{{\alpha}}}{\left({x}\right)}\right)}\)

Which implies y is the image of U.

So, \(\displaystyle{y}\in{I}{m}{T}\cap{I}{m}{U}\ne{\left\lbrace{0}\right\rbrace}\) which proves the contrapositive. Thus, \(\displaystyle{\left\lbrace{T},{U}\right\rbrace}\) is linearly independent subset of L (V, W). Hence proved.

Let Assume that a #0 then there exist xeV such that x e kemel of T.

Let \(\displaystyle{y}={T}{\left({x}\right)}\ne{0}\).

It is known that,

\(\displaystyle{0}={\left(\alpha{T}+\beta{U}\right)}{\left({x}\right)}\)

\(\displaystyle=\alpha{T}{\left({x}\right)}+\beta{U}{\left({x}\right)}\)

This can also be written as follows.

\(\displaystyle{y}={T}{\left({x}\right)}\)

\(\displaystyle={\frac{{-\beta}}{{\alpha}}}{U}{\left({x}\right)}\)

\(\displaystyle={U}{\left({\frac{{-\beta}}{{\alpha}}}{\left({x}\right)}\right)}\)

Which implies y is the image of U.

So, \(\displaystyle{y}\in{I}{m}{T}\cap{I}{m}{U}\ne{\left\lbrace{0}\right\rbrace}\) which proves the contrapositive. Thus, \(\displaystyle{\left\lbrace{T},{U}\right\rbrace}\) is linearly independent subset of L (V, W). Hence proved.