Question

Whether T is a linear transformation, that is T : C[0,1] rightarrow C[0,1] defined by T(f)=g, where g(x)=e^{x}f(x)

Transformation properties
ANSWERED
asked 2021-01-10
Whether T is a linear transformation, that is \(\displaystyle{T}:{C}{\left[{0},{1}\right]}\rightarrow{C}{\left[{0},{1}\right]}\) defined by \(\displaystyle{T}{\left({f}\right)}={g}\), where \(\displaystyle{g{{\left({x}\right)}}}={e}^{{{x}}}{f{{\left({x}\right)}}}\)

Answers (1)

2021-01-11
Approach:
For U and V vector spaces and T is a function from U to V, then T would be considered as linear transformation if for all u, w lies in U and all scalars k as if satisfies the following properties:
\(\displaystyle{T}{\left({u}\ +\ {w}\right)}={T}{\left({u}\right)}\ +\ {T}{\left({w}\right)}\)
\(\displaystyle{T}{\left({k}{u}\right)}={k}{T}{\left({u}\right)}\)
Calculation:
For T is a linear transformation, that is \(\displaystyle{T}:{C}{\left[{0},{1}\right]}\rightarrow{C}{\left[{0},{1}\right]}\) defined by \(\displaystyle{T}{\left({f}\right)}={g}{]}{w}{h}{e}{r}{e}{\left[{g{{\left({x}\right)}}}={e}^{{{x}}}{f{{\left({x}\right)}}}\right.}\)
Such that assume g is also present in the domain of \(\displaystyle{T}:{\left[{0},{1}\right]}\rightarrow{C}{\left[{0},{1}\right]}\), that means \(\displaystyle{T}{\left({g}\right)}={f}\) where \(\displaystyle{f{{\left({x}\right)}}}={e}^{{{x}}}{g{{\left({x}\right)}}}\)
Then,
\(\displaystyle{T}{\left({f}\right)}={e}^{{{x}}}{g}\)
\(\displaystyle{T}{\left({g}\right)}={e}^{{{x}}}{f}\)
Perform additional operation.
\(\displaystyle{T}{\left({f}\right)}\ +\ {T}{\left({g}\right)}={e}^{{{x}}}{g}\ +\ {e}^{{{x}}}{f}\)
\(\displaystyle={e}^{{{x}}}{\left({f}\ +\ {g}\right)}\)
\(\displaystyle={T}{\left({f}\ +\ {g}\right)}\)
Thus it shows that \(\displaystyle{T}{\left({f}\ +\ {g}\right)}={T}{\left({f}\right)}\ +\ {T}{\left({g}\right)}\), that means it satisfies the addition property
Now, take any scalar be k
And,
\(\displaystyle{T}{\left({k}{f}\right)}={e}^{{{x}}}{\left({k}{f}\right)}\)
\(\displaystyle={k}{e}^{{{x}}}{f}\)
\(\displaystyle={k}{\left({e}^{{{x}}}{f}\right)}\)
\(\displaystyle={k}{T}{\left({f}\right)}\)
Thus is also satisfies the scalar property.
So, it satisfies all the condition of the linear transformation.
Hence, the expression of \(\displaystyle{T}:{C}{\left[{0},{1}\right]}\rightarrow{C}{\left[{0},{1}\right]}\) defined by \(\displaystyle{T}{\left({f}\right)}={g}\) where \(\displaystyle{g{{\left({x}\right)}}}={e}^{{{x}}}{f{{\left({x}\right)}}}\) is linear transformation.
Answer:
As for \(\displaystyle{T}:{C}{\left[{0},{1}\right]}\rightarrow{C}{\left[{0},{1}\right]}\), the property of addition and scalar are satisfied. Thus the expression of \(\displaystyle{T}{\left({f}\right)}={g}\) where \(\displaystyle{g{{\left({x}\right)}}}={e}^{{{x}}}{f{{\left({x}\right)}}}\) is linear transformation.
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