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Find the range and nullspace of the following linear transformation: $T:C^1(0,1)\rightarrow C(0,1)$ defined by $T(f)(x)=f'(x)e^x$

To get the rank and nullspace, I first have to identify the matrix of $T$. The matrix of a linear map $T:V\rightarrow W$ can be obtained from the basis of $V$ and $W$. I don't know if $C^1$ and $C$ is just a representation of a continuous domain/range or if its something more than that. If so, how can I proceed?

Edit: How can I find the rank?

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    These vector spaces are infinite dimensional, so you'll be hard pressed to find a matrix to represent $T$. – 2017-02-21
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    Those are infinite-dimensional vector spaces, so finding a basis and a matrix is the wrong way to go. – 2017-02-21
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    Also, you are asked for the *range*, not the *rank*. – 2017-02-21
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    @Leon Sot But to find rank, I need to convert a matrix to row echelon form and find the number of pivots right? – 2017-02-21
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    @GerryMyerson Is it possible to evaluate the rank? – 2017-02-21
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    The operator is surjective (so the range is the whole codomain) and the kernel is the constant functions. – 2017-02-21
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    Yes, but not at the level of an intro linear algebra course. – 2017-02-21
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    @GerryMyerson can I say the rank is infinity? Also, the range is C(0,1) and nullspace is R – 2017-02-21
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    Infinity isn't a number. You can say the rank is infinite – but why would you say anything about the rank, when the question didn't ask for it? – 2017-02-21

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Let's prove the assertions Leon Sot makes in the comments.

To find the nullspace of $T$ is to find the functions $f$ such that $T(f)$ is the zero function, so we need to solve $f'(x)e^x=0$ for $f$. Now $e^x$ is never zero, so we need to solve the differential equation $f'=0$. The solution is the set of all constant functions, so that's the nullspace.

Now to show $T$ is onto, we have to show that for every $g$ in the codomain there's an $f$ in the domain such that $T(f)=g$. That is, we have to solve $f'(x)e^x=g(x)$, or $f'(x)=e^{-x}g(x)$. So we can take $f$ to be any antiderivative of $e^{-x}g(x)$ (and note that any such $f$ is in the domain of $T$). Hence, $T$ is onto; its range is the whole of the codomain.