Question about constructing proper matrices (Prove that $[T+U]^\gamma_\beta=[T]^\gamma_\beta+[U]^\gamma_\beta$) 
Let $T,U:V \rightarrow W$ be linear and assume $\beta$, $\gamma$ are finite ordered bases for V and W respectively. Prove that $[T+U]^\gamma_\beta=[T]^\gamma_\beta+[U]^\gamma_\beta$

I have question with construction. So can I say that let $X \in T:V \rightarrow W$ and $Y \in U: V \rightarrow W$, let $\beta$ be basis {$v_1,v_2...v_n$} for V and $\gamma$ be basis {$w_1,w_2...,w_m$}, then I say by linearity, we prove that $[T+U]^\gamma_\beta=[T]^\gamma_\beta+[U]^\gamma_\beta$. My question is mainly constructing two matrices and basis. Any help is appreciated
 A: Assume that $\beta = \{v_1,\dots,v_n\}$, $\gamma = \{w_1,\dots,w_m\}$ and the field what we are working in it is $F$. 
How do you get the $j$-th column of the matrix $A = [\textsf{T}]_\beta^\gamma$? Well, we write the vector $\textsf{T}(v_j)$ as a linear combination of the vectors in $\gamma$ and then put the coefficients in a column vector, right?
That is, if $\textsf{T}(v_j) = a_{1j}w_1 + a_{2j}w_2 + \cdots + a_{mj}w_m$, then the $j$-th column of $A$ is simply
$$\begin{pmatrix} a_{1j} \\ a_{2j} \\ \vdots \\ a_{mj} \end{pmatrix}.$$
Similarly, if $\textsf{U}(v_j) = \displaystyle \sum_{i=1}^m b_{ij}w_j$, then the $j$-th column of $B = [\textsf{U}]_\beta^\gamma$ is 
$$\begin{pmatrix} b_{1j} \\ b_{2j} \\ \vdots \\ b_{mj} \end{pmatrix},$$
but also (see this), since we can write
$$(\textsf{T} + \textsf{U})(v_j) = \textsf{T}(v_j) + \textsf{U}(v_j) = \sum_{i=1}^m (a_{ij} + b_{ij})w_j$$
then the $j$-th column of $[\textsf{T} + \textsf{U}]_\beta^\gamma$ is the sum of the previous ones, right?
Therefore, $[\textsf{T} + \textsf{U}]_\beta^\gamma = [\textsf{T}]_\beta^\gamma + [\textsf{U}]_\beta^\gamma$.
