Exercise: Exercise: Suppose $V$ and $W$ are finite-dimensional and $T \in L(V, W)$. Prove that there exist a basis of $V$ and a basis of $W$ such that with respect to these bases, all entries of $M(T)$ are $0$ except that the entries in row $j$ , column $j$ , equal $1$ for $1 \le j \le \dim \text{range }T$.
Proof: Let dim range $T=m$. By the fundamental theorem of linear maps we have that dim null $T=n-m$ where dim $V=n$. Thus there exist $m$ basis vectors of $V$ that do not get mapped to $0$. Let $v_1,\dots,v_m, \dots,v_n$ be that basis in which those $m$ vector exist and where the remaining $n-m$ vectors get mapped to zero. Using the theorem that if $v_1,\dots,v_n$ spans $V$ then $Tv_1,\dots,Tv_n$ spans range $T$, we see that $Tv_1,\dots,Tv_n$ is a spanning list in range $T$. Using the theorem that a spanning list can be reduced to a basis, we can reduce the list $Tv_1,\dots,Tv_n$ to a basis of range $T$ by removing those vectors that are in the span of the previous ones. In doing so, all the $n-m$ vectors that map to $0$ get removed.
The remaining vectors $Tv_{1},\dots,Tv_m$ span range $T$. To show that this is a basis of range $T$, it suffices to show that $Tv_j$ is not in the span of the previous vectors.
Suppose there exist scalars $a_{1},\dots,a_{j-1}$ such that $a_{1}Tv_{1}+\dots+a_{j-1}Tv_{j-1}=Tv_j$. Subtracting $Tv_j$ from both sides and using the linearity of $T$ we get $T(a_{1}v_{1}+\dots+a_{j-1}v_{j-1}-v_j)=0$. Because none of the vectors in the list $Tv_{1},\dots,Tv_m$ get mapped to zero, the independence of $v_{1},\dots,v_m$ and the above representation imply that the scalars $a_{1}=\dots=a_m=0$. Thus, the list $Tv_{1},\dots,Tv_m$ is a basis of range $T$.
Extending this basis to a basis of $W$ we see that $Tv_1,\dots,Tv_m$ is part of the basis of $W$. For the basis $v_1,\dots,v_m,\dots,v_n$ of $V$ and the basis $Tv_1,\dots,Tv_m,\dots,w_p$ of $W$ we have that $M(T)$ has a $1$ on row $j$ and column $j$ for $1\le j\le \dim \text{range }T$ as $Tv_j=1Tv_j$. All the other entries equal zero as the rest of the vectors in the basis get mapped to zero.
Is this proof correct?