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 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).

Question
Vectors and spaces
asked 2020-11-30

Consider \(\displaystyle{V}={s}{p}{a}{n}{\left\lbrace{\cos{{\left({x}\right)}}},{\sin{{\left({x}\right)}}}\right\rbrace}\) 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 \(\displaystyle{\left\lbrace{\cos{{\left({x}\right)}}}+{\sin{{\left({x}\right)}}},{\cos{{\left({x}\right)}}}−{\sin{{\left({x}\right)}}}\right\rbrace}\) and determine if T is an isomorphism.

Answers (1)

2020-12-01
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 \(\displaystyle{B}={\left\lbrace{\cos{{x}}}+{\sin{{x}}},{\cos{{x}}}−{\sin{{x}}}\right\rbrace}.\)
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.
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