--- Commentary: $\textbf{Exercise 12.}$ Suppose $U$ is a subspace of $V$ with $U \neq V$. Suppose $S \in \mathcal{L}(U, W)$ and $S \neq 0$ (which means that $Su \neq 0$ for some $u \in U$). Define $T : V \to W$ by $Tv = \begin{cases} Sv & \text{if } v \in U, \ 0 & \text{if } v \in V \text{ and } v \notin U. \end{cases}$ Prove that $T$ is not a linear map on $V$. $\textbf{Solution 12.}$ Let $u \in U$ be such that $Su \neq 0$. Let $w \in V$ be such that $w \notin U$. Then $u + w \notin U$ [because $u + w \in U$ would imply that $(u + w) + (-u)$, which equals $w$, is in $U$]. Thus $T(u + w) = 0.$ However, $Tu + Tw = Tu + 0 = Tu = Su \neq 0.$ Thus $T(u + w) \neq Tu + Tw$. Hence $T$ is not linear. $\textit{Commentary:}$ This exercise provides an example of a function $T : V \to W$ that is defined piecewise from a linear map $S : U \to W$, where $U$ is a proper subspace of $V$. The function $T$ agrees with $S$ on $U$, but maps all vectors outside of $U$ to 0. The exercise asks to prove that $T$ is not linear. The proof uses a specific property of subspaces: if $U$ is a proper subspace of $V$, and $u \in U$ and $w \notin U$, then $u + w \notin U$. This is because if $u + w$ were in $U$, then $w = (u + w) - u$ would also be in $U$, contradicting the assumption that $w \notin U$. Using this property, we choose a vector $u \in U$ such that $Su \neq 0$ (which exists because $S \neq 0$), and a vector $w \notin U$. Then $u + w \notin U$, so $T(u + w) = 0$. But $Tu + Tw = Su + 0 = Su \neq 0$. This shows that $T$ does not satisfy additivity, and hence is not linear. This exercise highlights the fact that piecewise-defined functions, even when they are based on linear maps, are not necessarily linear. It also provides practice in working with the properties of subspaces and constructing counterexamples. $\textit{Examples:}$ 1. Let $V = \mathbb{R}^2$, $U = {(x, 0) : x \in \mathbb{R}}$, $W = \mathbb{R}$, and $S : U \to W$ be defined by $S(x, 0) = x$. Define $T : V \to W$ by $T(x, y) = x$ if $y = 0$ and $T(x, y) = 0$ if $y \neq 0$. Then $T$ is not linear. 2. Let $V = \mathcal{P}(\mathbb{R})$, $U = {p \in \mathcal{P}(\mathbb{R}) : p(0) = 0}$, $W = \mathbb{R}$, and $S : U \to W$ be defined by $Sp = p'(0)$. Define $T : V \to W$ by $Tp = p'(0)$ if $p(0) = 0$ and $Tp = 0$ if $p(0) \neq 0$. Then $T$ is not linear. 3. Let $V = \mathbb{C}^2$, $U = {(z, 0) : z \in \mathbb{C}}$, $W = \mathbb{C}$, and $S : U \to W$ be defined by $S(z, 0) = iz$. Define $T : V \to W$ by $T(z, w) = iz$ if $w = 0$ and $T(z, w) = 0$ if $w \neq 0$. Then $T$ is not linear. These examples illustrate the construction of non-linear functions from linear maps on subspaces in various contexts, including $\mathbb{R}^n$, polynomial spaces, and $\mathbb{C}^n$. In each case, the function agrees with a linear map on a subspace, but maps all vectors outside the subspace to zero. The non-linearity of these functions can be demonstrated by finding vectors $u$ in the subspace and $w$ outside the subspace such that $T(u + w) \neq Tu + Tw$.