# Consider V= span{cos(x),sin(x)} a subspace of the vector space of continuous functions and a linear transformation T:Vrightarrow V where T(f) =f(0)timescos(x)−f(π2)timessin(x).

Consider $$\displaystyle{V}={\cos{{\left({x}\right)}}},{\sin{{\left({x}\right)}}}$$ a subspace of the vector space of continuous functions and a linear transformation $$\displaystyle{T}:{V}\rightarrow{V}$$ where $$\displaystyle{T}{\left({f}\right)}={f{{\left({0}\right)}}}\times{\cos{{\left({x}\right)}}}−{f{{\left(π{2}\right)}}}\times{\sin{{\left({x}\right)}}}.$$

Find the matrix of T with respect to the basis $$\displaystyle{\cos{{\left({x}\right)}}}+{\sin{{\left({x}\right)}}},{\cos{{\left({x}\right)}}}−{\sin{{\left({x}\right)}}}$$ and determine if T is an isomorphism.

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Aamina Herring

The linear transformation is given by
$$\displaystyle{T}{\left({f}\right)}={f{{\left({0}\right)}}}{\cos{{x}}}−{f{{\left(\pi\right)}}}{\sin{{x}}}.$$
Now the basis is given by $$\displaystyle{B}=\cos x+\sin x,\cos x−\sin{{x}}.$$
Now when $$\displaystyle{f{{\left({x}\right)}}}={\cos{{x}}}+{\sin{{x}}}{f{{\left({x}\right)}}}={\cos{{x}}}+{\sin{{x}}}$$ then
$$\displaystyle{T}{\left({\cos{{x}}}+{\sin{{x}}}\right)}={\cos{{x}}}+{\cos{{x}}}={2}{\cos{{x}}}={\left({\cos{{x}}}+{\sin{{x}}}\right)}+{\left({\cos{{x}}}−{\sin{{x}}}\right)}.$$
Again when $$\displaystyle{f{{\left({x}\right)}}}={\cos{{x}}}−{\sin{{x}}}{f{{\left({x}\right)}}}={\cos{{x}}}−{\sin{{x}}}$$ then
$$\displaystyle{T}{\left({\cos{{x}}}−{\sin{{x}}}\right)}={\cos{{x}}}+{\cos{{x}}}={2}{\cos{{x}}}={\left({\cos{{x}}}+{\sin{{x}}}\right)}+{\left({\cos{{x}}}−{\sin{{x}}}\right)}.$$
Thus the matrix representation of T is given by
$$\displaystyle{\left[{T}\right]}{b}={\left[{1},{1},{1},{1}\right]}$$
Since $$\displaystyle{\det{{\left({\left[{T}\right]}{B}\right)}}}{)}={0}$$, thus T is not an isomorphism.