Whether T is a linear transformation, that is $T:{C}^{1}[-1,1]\to {R}^{1}$ defined by $T\left(f\right)={f}^{\prime}\left(0\right)$

postillan4
2021-01-28
Answered

Whether T is a linear transformation, that is $T:{C}^{1}[-1,1]\to {R}^{1}$ defined by $T\left(f\right)={f}^{\prime}\left(0\right)$

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falhiblesw

Answered 2021-01-29
Author has **97** answers

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:

$T(u\text{}+\text{}w)=T\left(u\right)\text{}+\text{}T\left(w\right)$

$T\left(ku\right)=kT\left(u\right)$

Calculation:

For T is a linear transformation, that is$T:{C}^{1}[-1,1]\to {R}^{1}$ defined by $T\left(f\right)={f}^{\prime}\left(0\right)$

Such that assume g is also present in the domain of$T:{C}^{1}[-1,1]\to {R}^{1}$ , that means $g\in {C}^{1}[-1,1]\to {R}^{1}$

Then,

$T\left(f\right)={f}^{\prime}\left(0\right)$

$T\left(g\right)={g}^{\prime}\left(0\right)$

Perform additional operation.

$T\left(f\right)\text{}+\text{}T\left(g\right)={f}^{\prime}\left(0\right)\text{}+\text{}{g}^{\prime}\left(0\right)$

And,

$T(f\text{}+\text{}g)={(f\text{}+\text{}g)}^{\prime}\left(0\right)$

$={f}^{\prime}\left(0\right)\text{}+\text{}{g}^{\prime}\left(0\right)$

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

Thus it shows that$T(f\text{}+\text{}g)=T\left(f\right)\text{}+\text{}T\left(g\right)$ , 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$T:{C}^{1}[-1,1]\to {R}^{1}$ defined by $T\left(f\right)={f}^{\prime}\left(0\right)$ is linear transformation.

Answer:

As for$T:{C}^{1}[-1,1]\to {R}^{1}$ , the property of addition and scalar are satisfied. Thus the expression of $T\left(f\right)={f}^{\prime}\left(0\right)$ is linear transformation.

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:

Calculation:

For T is a linear transformation, that is

Such that assume g is also present in the domain of

Then,

Perform additional operation.

And,

Thus it shows that

Now, take any scalar be k

And

Thus is also satisfies the scalar property.

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

Hence, the expression of

Answer:

As for

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