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.

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.