Whether T is a linear transformation, that is T : C^1[-1,1] rightarrow R^1 defined by T(f)=f'(0)

Whether T is a linear transformation, that is $T:{C}^{1}\left[-1,1\right]\to {R}^{1}$ defined by $T\left(f\right)={f}^{\prime }\left(0\right)$
You can still ask an expert for help

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

falhiblesw
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:

$T\left(ku\right)=kT\left(u\right)$
Calculation:
For T is a linear transformation, that is $T:{C}^{1}\left[-1,1\right]\to {R}^{1}$ defined by $T\left(f\right)={f}^{\prime }\left(0\right)$
Such that assume g is also present in the domain of $T:{C}^{1}\left[-1,1\right]\to {R}^{1}$, that means $g\in {C}^{1}\left[-1,1\right]\to {R}^{1}$
Then,
$T\left(f\right)={f}^{\prime }\left(0\right)$
$T\left(g\right)={g}^{\prime }\left(0\right)$

And,

Thus it shows that , that means it satisfies the addition property.
Now, take any scalar be k
And
$T\left(kf\right)={\left(kf\right)}^{\prime }\left(0\right)$
$=k{f}^{\prime }\left(0\right)$
$=k\left({f}^{\prime }\left(0\right)\right)$
$=kT\left(f\right)$
Thus is also satisfies the scalar property.
So, it satisfies all the condition of the linear transformation.
Hence, the expression of $T:{C}^{1}\left[-1,1\right]\to {R}^{1}$ defined by $T\left(f\right)={f}^{\prime }\left(0\right)$ is linear transformation.
As for $T:{C}^{1}\left[-1,1\right]\to {R}^{1}$, the property of addition and scalar are satisfied. Thus the expression of $T\left(f\right)={f}^{\prime }\left(0\right)$ is linear transformation.