Suppose $V$ and $W$ are finite-dimensional and $T \in \mathcal L(V,W)$. Prove that there exists a basis of $V$ and a basis of $W$ such that with respect to these bases, all entries of $\mathcal M(T)$ are $0$ except that the entries in row $j$, column $j$ equal $1$ for $1 \le j \le \dim \text{range }T$.

(This generalizes 3c/2)

Let $v_1,\dots,v_n$ be a basis of $V$ such that $Tv_1,\dots,Tv_r$ is a basis for $\text{range }T$ and $v_{r+1},\dots,v_n$ is a basis for $\text{null }T$. The existance of such a basis is shown in the proof of 3.22 though it isn't in the theorem statement. (todo: ugly! construct it using book results)
Let $w_j = Tv_j$ for $1 \le j \le r$ then extend to a basis $w_1,\dots,w_m$ of $W$.

Clearly $\mathcal M(T)$ will have ones in the diagonal up to $r = \dim \text{range }T$ since writing $Tv_k$ in terms of the matrix, where $1 \le k \le r$ gives

Now since $w_k = Tv_k$ and the $w$'s are independent we must have

To finish we need to show that the $(r+1),\dots,m$ columns are all zero. This is because $Tv_k = 0$ for $k > r$ and

Implies (by independence of $w_1,\dots,w_m$) that $A_{j,k} = 0$ for all $j$.

This was intuitively obvious to me after doing 3c/2, but took a while to make rigorous. I need to improve my "rigorous articulation"