Let V and W be vector spaces, and let T

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).

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

zesponderyd
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.