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 \(\displaystyle{T}:{C}^{{1}}{\left[-{1},{1}\right]}\rightarrow{R}^{{1}}\) defined by \(\displaystyle{T}{\left({f}\right)}={f}'{\left({0}\right)}\)

Such that assume g is also present in the domain of \(\displaystyle{T}:{C}^{{1}}{\left[-{1},{1}\right]}\rightarrow{R}^{{1}}\), that means \(\displaystyle{g}\in{C}^{{1}}{\left[-{1},{1}\right]}\rightarrow{R}^{{1}}\)

Then,

\(\displaystyle{T}{\left({f}\right)}={f}'{\left({0}\right)}\)

\(\displaystyle{T}{\left({g}\right)}={g}'{\left({0}\right)}\)

Perform additional operation.

\(\displaystyle{T}{\left({f}\right)}\ +\ {T}{\left({g}\right)}={f}'{\left({0}\right)}\ +\ {g}'{\left({0}\right)}\)

And,

\(\displaystyle{T}{\left({f}\ +\ {g}\right)}={\left({f}\ +\ {g}\right)}'{\left({0}\right)}\)

\(\displaystyle={f}'{\left({0}\right)}\ +\ {g}'{\left({0}\right)}\)

\(\displaystyle={T}{\left({f}\right)}\ +\ {T}{\left({g}\right)}\)

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

\(\displaystyle{T}{\left({k}{f}\right)}={\left({k}{f}\right)}'{\left({0}\right)}\)

\(\displaystyle={k}{f}'{\left({0}\right)}\)

\(\displaystyle={k}{\left({f}'{\left({0}\right)}\right)}\)

\(\displaystyle={k}{T}{\left({f}\right)}\)

Thus is also satisfies the scalar property.

So, it satisfies all the condition of the linear transformation.

Hence, the expression of \(\displaystyle{T}:{C}^{{1}}{\left[-{1},{1}\right]}\rightarrow{R}^{{1}}\) defined by \(\displaystyle{T}{\left({f}\right)}={f}'{\left({0}\right)}\) is linear transformation.

Answer:

As for \(\displaystyle{T}:{C}^{{1}}{\left[-{1},{1}\right]}\rightarrow{R}^{{1}}\), the property of addition and scalar are satisfied. Thus the expression of \(\displaystyle{T}{\left({f}\right)}={f}'{\left({0}\right)}\) 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 \(\displaystyle{T}:{C}^{{1}}{\left[-{1},{1}\right]}\rightarrow{R}^{{1}}\) defined by \(\displaystyle{T}{\left({f}\right)}={f}'{\left({0}\right)}\)

Such that assume g is also present in the domain of \(\displaystyle{T}:{C}^{{1}}{\left[-{1},{1}\right]}\rightarrow{R}^{{1}}\), that means \(\displaystyle{g}\in{C}^{{1}}{\left[-{1},{1}\right]}\rightarrow{R}^{{1}}\)

Then,

\(\displaystyle{T}{\left({f}\right)}={f}'{\left({0}\right)}\)

\(\displaystyle{T}{\left({g}\right)}={g}'{\left({0}\right)}\)

Perform additional operation.

\(\displaystyle{T}{\left({f}\right)}\ +\ {T}{\left({g}\right)}={f}'{\left({0}\right)}\ +\ {g}'{\left({0}\right)}\)

And,

\(\displaystyle{T}{\left({f}\ +\ {g}\right)}={\left({f}\ +\ {g}\right)}'{\left({0}\right)}\)

\(\displaystyle={f}'{\left({0}\right)}\ +\ {g}'{\left({0}\right)}\)

\(\displaystyle={T}{\left({f}\right)}\ +\ {T}{\left({g}\right)}\)

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

\(\displaystyle{T}{\left({k}{f}\right)}={\left({k}{f}\right)}'{\left({0}\right)}\)

\(\displaystyle={k}{f}'{\left({0}\right)}\)

\(\displaystyle={k}{\left({f}'{\left({0}\right)}\right)}\)

\(\displaystyle={k}{T}{\left({f}\right)}\)

Thus is also satisfies the scalar property.

So, it satisfies all the condition of the linear transformation.

Hence, the expression of \(\displaystyle{T}:{C}^{{1}}{\left[-{1},{1}\right]}\rightarrow{R}^{{1}}\) defined by \(\displaystyle{T}{\left({f}\right)}={f}'{\left({0}\right)}\) is linear transformation.

Answer:

As for \(\displaystyle{T}:{C}^{{1}}{\left[-{1},{1}\right]}\rightarrow{R}^{{1}}\), the property of addition and scalar are satisfied. Thus the expression of \(\displaystyle{T}{\left({f}\right)}={f}'{\left({0}\right)}\) is linear transformation.