--- Commentary: $\textbf{Exercise 2.}$ Suppose $S, T \in \mathcal{L}(V)$ are such that $\operatorname{range} S \subseteq \operatorname{null} T$. Prove that $(ST)^2 = 0$. $\textbf{Solution 2.}$ Suppose $v \in V$. Then $(ST)^2v = ST(S(Tv)).$ Now $S(Tv) \in \operatorname{range} S \subseteq \operatorname{null} T$. Thus $T(S(Tv)) = 0$. Hence the equation above shows that $(ST)^2v = 0$. Thus $(ST)^2 = 0$. $\textit{Commentary:}$ This exercise explores a condition under which the square of the product of two linear operators is zero. The condition is that the range of $S$ is contained in the null space of $T$. Geometrically, this means that $S$ maps every vector into the null space of $T$, so that $T$ then maps these vectors to zero. When we apply $ST$ twice, the first application of $T$ maps everything to zero, so the second application of $ST$ is just zero. This property is useful in various contexts, such as in the study of nilpotent operators (operators $N$ such that $N^k = 0$ for some positive integer $k$), and in the study of upper triangular matrices (which have this property when the diagonal entries are zero). $\textit{Examples:}$ 1. Let $V = \mathbb{R}^3$, and define $S, T \in \mathcal{L}(\mathbb{R}^3)$ by $S(x, y, z) = (x, y, 0)$ and $T(x, y, z) = (0, 0, z)$. Then $\operatorname{range} S = {(x, y, 0) : x, y \in \mathbb{R}}$ and $\operatorname{null} T = {(x, y, 0) : x, y \in \mathbb{R}}$, so $\operatorname{range} S \subseteq \operatorname{null} T$. Indeed, $(ST)^2 = 0$. 2. Let $V = \mathcal{P}_3(\mathbb{C})$, and define $S, T \in \mathcal{L}(\mathcal{P}_3(\mathbb{C}))$ by $S(p(z)) = p(0)$ and $T(p(z)) = zp(z)$. Then $\operatorname{range} S = {constant polynomials}$ and $\operatorname{null} T = {constant polynomials}$, so $\operatorname{range} S \subseteq \operatorname{null} T$. Indeed, $(ST)^2 = 0$. 3. Let $V = \mathbb{F}_2^3$, and define $S, T \in \mathcal{L}(\mathbb{F}_2^3)$ by $S(x, y, z) = (x, 0, 0)$ and $T(x, y, z) = (0, y, z)$. Then $\operatorname{range} S = {(x, 0, 0) : x \in \mathbb{F}_2}$ and $\operatorname{null} T = {(x, 0, 0) : x \in \mathbb{F}_2}$, so $\operatorname{range} S \subseteq \operatorname{null} T$. Indeed, $(ST)^2 = 0$. These examples illustrate the principle in various vector spaces over different fields. In each case, $S$ maps vectors into a subspace that is annihilated by $T$, leading to $(ST)^2$ being zero. The geometric interpretation is the same in each case: $S$ maps vectors into a subspace where $T$ is zero, so applying $ST$ twice yields zero.