Writing the most general form of a linear mapping Let S and T be linear transformations from $V$ to $W$. Let $\alpha =$ {$x_1, ..., x_n$} and $\beta = ${$w_1,...,w_m$} be bases for $V$ and $W$. 
I want to show that for $c$ real, $\;\;\;[cS]^\beta_\alpha\;$ = $\;\;\;c[S]^\beta_\alpha\;$.
I think my issue is that I don't know how to write the linear mapping in the most general form, so I don't know how to proceed. For instance, we can't just write it as a summation from $i = 1$ to $m$, can we? Because linear mappings also include things like taking derivatives.
 A: You could simply note that the linear map $cS:V \to W$ is defined by $(c \cdot S)(v) = c \cdot S(v)$ and that the collection of ($m \times n$) matrices forms a vector space and hence is scalar multiplicative.
However if this is unsatisfactory we consider the matter further. What we want to show is that for all $v \in V$ we have the follwing identity 
$$ [cS]^\beta_\alpha v = c[S]^\beta_\alpha v $$
To achieve this use the definition of matrix multiplication and consider the $i$-th component of the equation above:
\begin{align} 
\left( [cS]^\beta_\alpha v \right)_i 
&= \left(c \sum_{k=1}^n s_{ik}v_k\right)_i \\
&= \left( \sum_{k=1}^n cs_{ik}v_k\right)_i \\
&= \left( [cS]^\beta_\alpha v \right)_i
\end{align}
where $s_{ij}$ is the entry in the $i$-th row and $j$-th column of $[S]$ and $v = v_1 e_1 +  \ldots + v_n e_n$.
A: Well, the most general way to write a linear map between $V$ and $W$ is precisely to simply write it as $S:V\to W$. And this is enough to prove what you want!
Indeed, as you said yourself, in order to obtain $[cS]^\beta_\alpha$ you need to take the vectors in $\alpha$, apply $cS$ to each, and write the result as a linear combination of the vectors in $\beta$. But applying $cS$ is just applying $S$ and then multiplying by $c$. So if you have that
\begin{equation}Sx_j=\sum_{i=1}^m s_{ij}w_i,\tag{1}\label{1}\end{equation}
for all $1\leq j\leq m$ (where the $s_{ij}\in \Bbb R$), then
\begin{equation}(cS)x_j=c(Sx_j)=c\sum_{i=1}^m s_{ij}w_i=\sum_{i=1}^m (cs_{ij})w_i.\tag{2}\label{2}\end{equation}
Now equation \eqref{1} tells you that $[S]^\beta_\alpha=(s_{ij})_{1\leq i\leq m\\ 1\leq j\leq n}$, and equation \eqref{2} tells you that
$$[cS]^\beta_\alpha=(cs_{ij})_{1\leq i\leq m\\ 1\leq j\leq n}=c(s_{ij})_{1\leq i\leq m\\ 1\leq j\leq n}=c[S]^\beta_\alpha.$$
