Is the following Proof Correct ? If so please do point out any extraneous details . Remarks showing how to reduce the length of the argument would also be appreciated.
Theorem. Given that $V$ and $W$ are finite dimensional vector spaces and $T\in\mathcal{L}(V,W)$ (where $\mathcal{L}(V,W)$ denotes the set of all Linear transformations from $V$ to $W$). There exists bases for $V$ and $W$ such that with respect to theses bases all entries of $\mathcal{M}(T)$ are $0$ except the entries in row $j$, column $j$, equal $1$ for $1\leq j\leq\dim\operatorname{range}T$.
Proof. We know that $\operatorname{range}T$ is a subspace of $W$ an thus there is a basis for $\operatorname{range}T$. Let $w_1,w_2,...,w_m$ be such a basis where $\dim\operatorname{range}T = m$. Since $w_1,w_2,...,w_m\in\operatorname{range}T$ it follows that there exist a list vectors $v_1,v_2,...,v_m$ such that $Tv_j = w_j$ for $1\leq j\leq m$.
We now show that this list is linearly independent. Let $c_1,c_2,...,c_m$ be arbitrary scalars in $\mathcal{F}$ and assume that $c_1v_1+c_2v_2,...,c_mv_m = 0$ applying $T$ to this sum we have
$$T\left(\sum_{j=1}^{m}c_jv_j\right) = \sum_{j=1}^{m}Tc_jv_j = 0$$ but $Tv_1,Tv_2,...Tv_m$ is linearly independent implying that $c_1=c_2=...c_m = 0$.
Now Since $v_1,v_2,...v_m$ is linearly independent we may extend it to obtain a basis $v_1,v_2,...,v_m,v_{m+1},v_{m+2},...,v_n$ for $V$ and summarize our findings thus far by presenting a definition of $T:V\to W$ as follows $$T\left(\sum_{j=1}^{n}a_jv_j\right) = \sum_{j=1}^{m}a_jw_j$$ which implies that $v_{m+1},v_{m+2},...v_n$ is a basis for $\operatorname{null}T$. Now let $w_1,w_2,...,w_m,w_{m+1},w_{m+2},...,w_{q}$ be a basis for $W$ obtained by extending the list $w_1,w_2,...,w_m$ and for the basis of $V$ we retain our choice of the list $v_1,v_2,...,v_m,v_{m+1},v_{m+2},...,v_n$.
Given the above choices for bases of $V$ and $W$ it follows the column vector representing $Tv_j$ for $1\leq j\leq m$ will have all entries $0$ except for $1$ in the $j^{th}$ row and all column vectors following the $j^{th}$ column vector will have entries $0$ since $v_{m+1},v_{m+2},...v_n$ is a basis for $\operatorname{null}T$.
$\blacksquare$
Looks good, if you can use everything you stated as obvious in the proof. E.g. that you can extend every linear independent set to a basis.
You should maybe pull the $c_j$ out of T in the first equation line.