Is the Following argument correct?
Suppose $V$ is finite-dimensional and $T_1,T_2\in\mathcal{L}(V,W)$. Prove that $\operatorname{range}T_1\subset\operatorname{range}T_2$ if and only if there exists $S\in\mathcal{L}(V,V)$ such that $T_1 = T_2S$.
Proof. Since $V$ is finite-dimensional we may therefore invoke a basis $v_1,v_2,\dots,v_n$ for $V$.
$(\Rightarrow)$. Assume that $\operatorname{range}T_1\subset\operatorname{range}T_2$, then for each $T_1v_j$ there exists a $T_2u_j$ such that $T_1v_j=T_2u_j$. Now using theorem $\textbf{3.5}$ we may define a linear map $T:V\to V$ as follows. $$Sv_j = u_j$$ Given the above definition, it is evident that the images of the basis vectors under $T_2S$ and $T_1$ are the same implying that $T_1=T_2S$
$(\Leftarrow)$. Conversely assume $T_1=T_2S$ and let $Tv_1\in\operatorname{range}T_1$ and $Sv_1=u$, then $T_1v=T_2S = T_2u$ implying $T_1v\in\operatorname{range}T_2$.
$\blacksquare$
Note:
- Theroem $\textbf{3.5}$ says that a linear transformation is uniquely determined by its action on a basis.