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?
Best Answer
You've got the right general idea but your proof is overly complicated. Just observe that $Tv_1,\ldots,Tv_m$ are linearly independent since if $$\alpha_1Tv_1+\cdots+\alpha_m Tv_m=0$$ then $$T(\alpha_1v_1+\cdots+\alpha_mv_m)=0$$ by linearity of $T$, which implies that $\alpha_1v_1+\cdots+\alpha_mv_m$ is in the kernel of $T$. But this means $$\alpha_1v_1+\cdots+\alpha_mv_m=\alpha_{m+1}v_{m+1}+\cdots+\alpha_nv_n$$ for some scalars $\alpha_{m+1},\cdots,\alpha_n$. Subtracting and using linear independence of $v_1,\ldots,v_n$ it follows that $\alpha_1=\cdots\alpha_m=0$.