Is the Following Proof Correct ?
Theorem. Given that $V$ and $W$ are finite dimensional vectors spaces and $T\in\mathcal{L}(V,W)$ i.e. a linear transformation from $V$ to $W$ then there exists bases for $V$ and $W$ such that all entries of $\mathcal{M}(T)$ i.e. the matrix of $T$ are $0$ except that the entries in row $j$ and column $j$ equal 1 for $1\leq j\leq\dim\operatorname{range}T$.
Proof. Since $W$ is finite dimensional and $\operatorname{range}T\leqslant W$ it follows that $\operatorname{range}T$ must have a basis, we may therefore invoke the existence of vectors $r_1,r_2,...,r_l\in\operatorname{range}T$ where $l=\dim\operatorname{range}T$ such that they act as a basis for $\operatorname{range}T$.
Consider now the list of vectors $v_1,v_2,...,v_l$ such that $\forall j\in\{1,2,...,l\}(Tv_j=r_j)$, furthermore it is not difficult to see that the linear independence of $r_1,r_2,...,r_l$ implies that linear independence of $v_1,v_2,...v_l$.
Extending the list $r_1,r_2,...,r_l$ to form a basis for $W$ we have $$r_1,r_2,...,r_l,w_1,w_2,...,w_m\in W$$
furthermore by letting $u_1,u_2,...,u_n$ be a basis for $\operatorname{null}T$ and invoking the rank-nullity theorem we may form a basis for $V$ i.e.
$$v_1,v_2,...,v_l,u_1,u_2,....u_n\in V$$
Given the above choice of basis we see that $\mathcal{M}(r_k)$ i.e the matrix of $r_k$ is a $(l+m)\times 1$ column vector with the $k$th entry equal to $1$ and all others equal to $0$, consequently $\mathcal{M}(T)$ is as specified in the statement of the theorem.
$\blacksquare$