$\textbf{Exercise 7.}$ Suppose that $V$ is finite-dimensional and $S, T \in \mathcal{L}(V, W)$. Prove that $\operatorname{range} S = \operatorname{range} T$ if and only if there exists an invertible $E \in \mathcal{L}(V)$ such that $S = TE$. $\textbf{Solution 7.}$ First suppose there exists an invertible $E \in \mathcal{L}(V)$ such that $S = TE$. If $w \in \operatorname{range} S$, then there exists $v \in V$ such that $w = Sv$. Thus $\begin{equation*} w = Sv = (TE)(v) = T(Ev), \end{equation*}$ and hence $w \in \operatorname{range} T$. Thus $\operatorname{range} S \subseteq \operatorname{range} T$. We have $T = SE^{-1}$. Thus by the same reasoning as in the paragraph above, we conclude that $\operatorname{range} T \subseteq \operatorname{range} S$. Combining the conclusions of the two paragraphs above shows that $\begin{equation*} \operatorname{range} S = \operatorname{range} T, \end{equation*}$ as desired. To prove the other direction, now suppose that $\operatorname{range} S = \operatorname{range} T$. The fundamental theorem of linear maps (3.21) implies that $\begin{equation*} \dim \operatorname{null} S = \dim \operatorname{null} T. \end{equation*}$ Let $v_1, \ldots, v_m$ be a basis of $\operatorname{null} S$ and let $u_1, \ldots, u_m$ be a basis of $\operatorname{null} T$. Extend $v_1, \ldots, v_m$ to a basis $v_1, \ldots, v_m, \ldots, v_n$ of $V$. For $m < k \leq n$, we have $Sv_k \in \operatorname{range} S = \operatorname{range} T$, and thus there exists $u_k \in V$ such that $Sv_k = Tu_k$. Note that if $1 \leq k \leq m$, then we also have $Sv_k = Tu_k$ because both sides equal 0. Now use the linear map lemma (3.4) to define a linear map $E$ from $V$ to $V$ such that $Ev_k = u_k$ for each $k = 1, \ldots, n$. If $1 \leq k \leq n$, then $\begin{equation*} Sv_k = Tu_k = T(Ev_k) = (TE)(v_k). \end{equation*}$ Thus $S = TE$ because these two linear maps agree on a basis. To prove that $E$ is invertible, suppose $v \in V$ and $Ev = 0$. Then $Sv = TEv = 0$, and hence $v \in \operatorname{null} S$. Thus $v \in \operatorname{span}(v_1, \ldots, v_m)$, and we can write $\begin{equation*} v = a_1v_1 + \cdots + a_mv_m \end{equation*}$ for some $a_1, \ldots, a_m \in \mathbb{F}$. Applying $E$ to both sides of the equation above gives $\begin{equation*} 0 = a_1u_1 + \cdots + a_mu_m. \end{equation*}$ Because $u_1, \ldots, u_m$ is linearly independent, this implies $a_1 = \cdots = a_m = 0$. Hence $v = 0$. Thus $E$ is injective. Now 3.63 implies that $E$ is invertible, as desired. --- ![[sol-5.pdf#page=6]]