--- Commentary: $\textbf{Exercise 16.}$ Suppose $V$ is finite-dimensional with $\dim V > 1$. Prove that there exist $S, T \in \mathcal{L}(V)$ such that $ST \neq TS$. $\textbf{Solution 16.}$ Let $v_1, \ldots, v_n$ be a basis of $V$. Use the linear map lemma (3.4) to define $S, T \in \mathcal{L}(V)$ such that $Sv_k = \begin{cases} v_2 & \text{if } k = 1, \ 0 & \text{if } k \neq 1, \end{cases}$ and $Tv_k = \begin{cases} v_1 & \text{if } k = 2, \ 0 & \text{if } k \neq 2. \end{cases}$ Then $(ST)(v_1) = S(Tv_1) = S0 = 0$ but $(TS)(v_1) = T(Sv_1) = Tv_2 = v_1.$ Thus $ST \neq TS$. $\textit{Commentary:}$ This exercise demonstrates that in a finite-dimensional vector space $V$ with dimension greater than 1, there always exist linear operators $S$ and $T$ on $V$ that do not commute, i.e., $ST \neq TS$. The proof uses the linear map lemma to construct such operators. Given a basis $v_1, \ldots, v_n$ of $V$, we define $S$ to map $v_1$ to $v_2$ and all other basis vectors to 0, and we define $T$ to map $v_2$ to $v_1$ and all other basis vectors to 0. Then, for the basis vector $v_1$, we have: $(ST)(v_1) = S(Tv_1) = S(0) = 0,$ but $(TS)(v_1) = T(Sv_1) = T(v_2) = v_1.$ Thus, $ST$ and $TS$ differ in their action on $v_1$, so they are not equal. This exercise highlights the non-commutativity of linear operators, which is a fundamental feature of linear algebra. In general, the composition of two linear maps depends on the order in which they are applied. This non-commutativity underlies many important constructions in linear algebra, such as the commutator of two operators and the Lie bracket of two matrices. The exercise also provides practice in working with the linear map lemma and constructing linear operators with specific properties. $\textit{Examples:}$ 1. In $\mathbb{R}^2$, let $S(x, y) = (y, 0)$ and $T(x, y) = (0, x)$. Then $(ST)(1, 0) = (0, 0)$ but $(TS)(1, 0) = (1, 0)$, so $ST \neq TS$. 2. In $\mathcal{P}_2(\mathbb{R})$, let $S(a + bx + cx^2) = bx$ and $T(a + bx + cx^2) = cx$. Then $(ST)(1 + x + x^2) = 0$ but $(TS)(1 + x + x^2) = x$, so $ST \neq TS$. 3. In $\mathbb{C}^3$, let $S(z_1, z_2, z_3) = (z_2, 0, 0)$ and $T(z_1, z_2, z_3) = (0, z_1, 0)$. Then $(ST)(1, 0, 0) = (0, 0, 0)$ but $(TS)(1, 0, 0) = (1, 0, 0)$, so $ST \neq TS$. These examples demonstrate non-commuting linear operators in various finite-dimensional vector spaces, including $\mathbb{R}^n$, polynomial spaces, and $\mathbb{C}^n$. In each case, $S$ and $T$ are chosen to "shift" basis vectors in a cyclic manner, which results in their compositions being different depending on the order of application. This illustrates the general principle that, unless there are specific conditions forcing them to commute (such as one of the operators being a scalar multiple of the identity), linear operators usually do not commute.