I'm trying to prove Axler 3.36, which is as follows (paraphrased):
Suppose $S$ and $T$ are linear maps from $V$ to $W$. Then the matrix of $(S + T)$ is equal to the matrix of $S$ plus the matrix of $T$. (Assume that the same bases have been chosen for all three maps.)
He defines the sum of two matrices in terms of the mechanical operation, $$ (A + C)_{j,k} = A_{j,k} + C_{j,k}. $$ It seems that the matrix sum isn't given, by definition, the interpretation of a linear transformation relative to a basis, but this formula gives us that interpretation as the transformation $(S + T)$ relative to the bases. Is that the correct way to interpret this? (So prior to this theorem, matrix addition was a purely mechanical process.)
Here is my attempt at proving it.
Let $v_1, \ldots, v_n$ be a basis for $V$ and $w_1, \ldots, w_m$ a basis for $W$. We will prove that the $k$th column of the matrices $\mathcal{M}(S + T)$ is equal to the $k$th column of the matrix $\mathcal{M}(S) + \mathcal{M}(T)$. Relative to the chosen basis, we have $$ Sv_k = \sum\limits_{i=1}^m S_{i,k} u_i \qquad Tv_k = \sum\limits_{i=1}^m T_{i,k} u_i. $$ That is, the $k$th column of $S$ is given by $S_{1,k}, S_{2,k}, \ldots, S_{m,k}$ and the $k$th column of $T$ is given by $T_{1,k}, T_{2,k}, \ldots, T_{m,k}$. By the definition of matrix addition, the $k$th column of $\mathcal{M}(S) + \mathcal{M}(T)$ consists of $S_{1,k} + T_{1,k}, S_{2,k} + T_{2,k}, \ldots, S_{m,k} + T_{m,k}$.
Now, we consider $\mathcal{M}(S + T)$. We have \begin{align*} (S+T)v_k &= Sv_k + Tv_k \\ &= \sum\limits_{i=1}^m S_{i,k} u_i + \sum\limits_{i=1}^m T_{i,k} u_i \\ &= \sum\limits_{i=1}^m (S_{i,k} + T_{i,k}) u_i, \end{align*} so the $k$th column of $\mathcal{M}(S + T)$ consists of $S_{1,k} + T_{1,k}, \ldots, S_{m,k} + T_{m,k}$. As this holds for all $1 \leq k \leq n$, we conclude that $\mathcal{M}(S + T) = \mathcal{M}(S) + \mathcal{M}(T)$.
How does this attempt look?