 # 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) Bevan Mcdonald 2021-01-10 Answered
Whether T is a linear transformation, that is $T:C\left[0,1\right]\to C\left[0,1\right]$ defined by $T\left(f\right)=g$, where $g\left(x\right)={e}^{x}f\left(x\right)$
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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 $T:C\left[0,1\right]\to C\left[0,1\right]$ defined by $T\left(f\right)=g\right]where\left[g\left(x\right)={e}^{x}f\left(x\right)$
Such that assume g is also present in the domain of $T:\left[0,1\right]\to C\left[0,1\right]$, that means $T\left(g\right)=f$ where $f\left(x\right)={e}^{x}g\left(x\right)$
Then,
$T\left(f\right)={e}^{x}g$
$T\left(g\right)={e}^{x}f$

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

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