# Suppose $V$ is finite-dimensional and $T_1,T_2∈L(V,W)$. Prove that $rangeT_1 ⊂ rangeT_2$ if and only if there exists $S∈L(V,V)$ such that $T_1=T_2S$

Suppose $V$ is finite-dimensional and $T_1,T_2∈L(V,W)$. Prove that $\text{range}(T_1) ⊂ \text{range}(T_2)$ if and only if there exists $S∈L(V,V)$ such that $T_1=T_2S$

• What's $T_2S$? The domain doesn't match. (Is $S\in L(V, V)$?)
• Nota Bene: What John said: $S \in L(V, W)$ cannot work, since then $S:V \to W$ and $T_2: V \to W$; so for $v \in V$, $Sv \in W$ but $T_2$ acts on $V$ so things jam up, like badly meshed gears. What about $S \in L(V, V)$? Cheers! Commented May 11, 2015 at 3:00
• Hint: Pick any arbitrary base of $V$ and express $T_1$, $T_2$ in terms of the values at this base. Then think about what $\mathrm{range}T_1\subset\mathrm{range}T_2$ means. Commented May 11, 2015 at 3:02
Let $\{e_1,e_2,\ldots, e_n\}$ be a basis for $V$, then your hypothesis for each $1\leq i\leq n, \exists f_i \in V$ such that $$T_1(e_i) = T_2(f_i)$$ Define $S(e_i) := f_i$ and extend it linearly to all of $V$. Now $$T_1(e_i) = T_2S(e_i) \quad\forall 1\leq i\leq n$$ and so $T_1 = T_2S$ must hold on $V$.