--- $\textbf{Exercise 25 Commentary:}$ This exercise characterizes when the null space of a linear map $T: V \to W$ is contained in the null space of another linear map $S: V \to W$. The condition is that there exists a linear operator $E: W \to W$ such that $T = ES$. Geometrically, this means that $T$ is obtained by "rescaling" $S$ in different directions in the codomain $W$. If $T = ES$, then $\text{null}(T) = \{v \in V \mid Tv = E(Sv) = 0\} \subseteq \{v \in V \mid Sv = 0\} = \text{null}(S)$, since $Sv = 0$ implies $E(Sv) = 0$. Conversely, if $\text{null}(S) \subseteq \text{null}(T)$, then we can define $E$ on $\text{range}(S)$ by $E(Sv) = Tv$, and this definition is independent of the choice of $v$ because if $Sv = Sv'$, then $v - v' \in \text{null}(S) \subseteq \text{null}(T)$ implies $Tv = Tv'$. $\textbf{Exercise 25 Examples:}$ 1. Let $V = \mathbb{R}^3, W = \mathbb{R}^2$, and define $S, T: V \to W$ by $S(x,y,z) = (x+y, y+z), T(x,y,z) = (2x+3y, 4y+5z)$. Then $T = ES$ where $E: W \to W$ is given by $E(a,b) = (2a+3b, 4b+5(b-a))$. Note that $\text{null}(S) = \{(x,-x,0) \mid x \in \mathbb{R}\} \subseteq \text{null}(T) = \{(0,0,0)\}$. 2. Let $V = \mathcal{P}_2(\mathbb{R}), W = \mathbb{R}^3$, and define $S: V \to W$ by $S(a+bx+cx^2) = (a,b,c), T: V \to W$ by $T(a+bx+cx^2) = (2a+b, 3b+c, 4c)$. Then $T = ES$ where $E: W \to W$ is given by $E(x,y,z) = (2x+y, 3y+z, 4z)$. Note that $\text{null}(S) = \{0\} \subseteq \text{null}(T) = \{0\}$. 1. Let $V = \mathbb{R}^2, W = \mathbb{F}_3^2$, and define $S(x,y) = (x,y), T(x,y) = (2x,y)$. Then there does not exist $E: W \to W$ with $T = ES$, since $\text{null}(S) = \{(0,0)\} \not\subseteq \text{null}(T) = \{(0,y) \mid y \in \mathbb{F}_3\}$. 2.