--- $\textbf{Exercise 26 Commentary:}$ This exercise gives a condition for when the range of a linear map $S: V \to W$ is contained in the range of another linear map $T: V \to W$. The condition is that there exists a linear operator $E: V \to V$ such that $S = TE$. Geometrically, this means that $S$ is obtained by "rescaling" $T$ in different directions in the domain $V$. If $S = TE$, then for any $v \in V$, we have $Sv = T(Ev) \in \text{range}(T)$, so $\text{range}(S) \subseteq \text{range}(T)$. Conversely, if $\text{range}(S) \subseteq \text{range}(T)$, then for any $v \in V$, we can find $u \in V$ such that $Sv = Tu$. Defining $E: V \to V$ by $Ev = u$ gives $S = TE$. $\textbf{Exercise 26 Examples:}$ 1. Let $V = \mathbb{R}^3, W = \mathbb{R}^2$, and define $S(x,y,z) = (x+2y, 2y+3z), T(x,y,z) = (x,y)$. Then $S = TE$ where $E(x,y,z) = (x+2y, 2y+3z, 0)$. 2. Let $V = \mathcal{P}_3(\mathbb{R}), W = \mathcal{P}_2(\mathbb{R})$, and let $S, T: V \to W$ be differentiation maps. Then $S = TE$ where $E: V \to V$ is the identity map. 1. Let $V = \mathbb{F}_2^3, W = \mathbb{F}_2^2$, and define $S(x,y,z) = (x+y,y+z), T(x,y,z) = (x,y)$. Then there does not exist $E: V \to V$ with $S = TE$, since $(1,1,0) \in \text{range}(S)$ but $(1,1,0) \notin \text{range}(T)$.