Let V and W be vector spaces, and let T

zakinutuzi 2022-01-07 Answered
Let V and W be vector spaces, and let T and U be nonzero linear transformations from V into W. If R(T) ∩ R(U) = {0}, prove that {T, U} is a linearly independent subset of L(V, W).

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zesponderyd
Answered 2022-01-08 Author has 5538 answers
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.
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