Is the matrix of the sum of two linear maps equal to the sum of the matrices of the two maps? Is the matrix of the sum of two linear maps equal to the sum of the matrices of the two maps?
We have bases $\{v_1,\dots,v_n\}$ of $V$ and $\{w_1,\dots,w_m\}$ of $W$.
$M(T+S) = M(T) + M(S)$?
Whenever $T,S\in \mathcal{L}(V,W)$

What does this look like(in matrix form)? I can't imagine it for the general case, hence I am not sure how to verify it(For axler $2_{\mathbf{nd}}$ Edition, page $50$).
Is it something like:
$A\times M(T+S)=A\times M(T) + A\times M(S)$

I will fix the question after class, don't have time now(just the quality and my little effort shown)
 A: The answer is yes.
Here is some lazy but hopefully clear notation.  If $x$ is a vector, then $[x]_B$ is the column vector relative to the basis $B$ (aka the coordinate vector of $x$ relative to the basis).  Let $B_V$ and $B_W$ denote your choices of basis for the respective spaces.
From the definition of the matrix corresponding to a transformation, we have
$$
[Tx]_{B_W} = M(T) [x]_{B_V}
$$
Moreover, for any vectors $x,y$ in the same space, we have
$$
[x + y]_B = [x]_B + [y]_B
$$
It now follows that for any $x \in V$, we have
$$
M(S + T)[x]_{B_V} = \\
[(S + T)x]_{B_W} = \\
[Sx + Tx]_{B_W} = \\
[Sx]_{B_W} + [Tx]_{B_W} =\\
M(S)[x]_{B_V} + M(T)[x]_{B_V} = \\
(M(S) + M(T))[x]_{B_V}
$$
Since $M(S + T)[x]_{B_V} = (M(S) + M(T))[x]_{B_V}$ for any choice of $x$, we can conclude that $M(S+T)$ and $M(S) + M(T)$ must be the same matrices.
I think this is probably the clearest way to show this directly, but perhaps it is usually taken as a consequence of a broader statement.  At any rate, I hope this helps.
