Exercise 1 Instructor’s Solutions Manual, Section 3A Suppose 𝑏,𝑐 ∈ 𝐑. Define 𝑇∶𝐑 →𝐑 by 𝑇(𝑥, 𝑦, 𝑧) = (2𝑥 − 4𝑦 + 3𝑧 + 𝑏, 6𝑥 + 𝑐𝑥𝑦𝑧). Show that 𝑇 is linear if and only if 𝑏 = 𝑐 = 0. SOLUTION: First suppose 𝑏 = 𝑐 = 0. Then 𝑇(𝑥, 𝑦, 𝑧) = (2𝑥 − 4𝑦 + 3𝑧, 6𝑥), which easily implies that 𝑇 is linear. Conversely, now suppose that 𝑇 is linear. Note that 𝑇(0,0,0) = (𝑏,0). Note that 𝑇(1,1,1 )= (1,6+𝑐) and 𝑇(2,2,2)=(2,12+8𝑐). >[!error] arithmetic error > T(1,1,1) = (1 + b, 6+c) > T(2,2,2) = (2 + b, 12 + 8c) Now 3.10 implies that 𝑏 = 0. The equation 𝑇(2, 2, 2) = 2𝑇(1, 1, 1) implies that 12+8𝑐 = 12+2𝑐, which implies that 𝑐 = 0. --- Suppose $( b, c \in \mathbb{R} )$. Define $\( T: \mathbb{R} \to \mathbb{R} \)$ by \[ T(x, y, z) = (2x - 4y + 3z + b, 6x + cx yz). \] Show that \( T \) is linear if and only if \( b = c = 0 \). **Solution:** First suppose \( b = c = 0 \). Then \[ T(x, y, z) = (2x - 4y + 3z, 6x), \] which easily implies that \( T \) is linear. Conversely, now suppose that \( T \) is linear. Note that \[ T(0,0,0) = (b,0). \] Note that \( T(1,1,1) = (1,6+c) \) and \( T(2,2,2) = (2,12+8c) \). >[!error] arithmetic error > T(1,1,1) = (1 + b, 6+c) > T(2,2,2) = (2 + b, 12 + 8c) Now \( 3.10 \) implies that \( b = 0 \). The equation \( T(2, 2, 2) = 2T(1, 1, 1) \) implies that \( 12+8c = 12+2c \), which implies that \( c = 0 \). --- Commentary: $\textbf{Section 3A}$ $\textbf{Exercise 1.}$ Suppose $b, c \in \mathbb{R}$. Define $T : \mathbb{R}^3 \to \mathbb{R}^2$ by $T(x, y, z) = (2x - 4y + 3z + b, 6x + cxyz).$ Show that $T$ is linear if and only if $b = c = 0$. $\textbf{Solution 1.}$ First suppose $b = c = 0$. Then $T(x, y, z) = (2x - 4y + 3z, 6x),$ which easily implies that $T$ is linear. Conversely, now suppose that $T$ is linear. Note that $T(0, 0, 0) = (b, 0).$ Now 3.10 implies that $b = 0$. Note that $T(1, 1, 1) = (1, 6 + c) \text{ and } T(2, 2, 2) = (2, 12 + 8c).$ The equation $T(2, 2, 2) = 2T(1, 1, 1)$ implies that $12 + 8c = 12 + 2c$, which implies that $c = 0$. $\textit{Commentary:}$ This exercise explores the conditions under which a given function $T : \mathbb{R}^3 \to \mathbb{R}^2$ is a linear map. The function $T$ is defined in terms of two parameters $b$ and $c$, and the exercise asks to prove that $T$ is linear if and only if both $b$ and $c$ are zero. The proof of the forward direction is straightforward: if $b = c = 0$, then $T(x, y, z) = (2x - 4y + 3z, 6x)$, which is clearly linear in $x$, $y$, and $z$. The proof of the reverse direction uses the properties of linear maps. First, it uses the fact that linear maps send 0 to 0 (property 3.10) to conclude that $b$ must be 0. Then, it considers the specific vectors $(1, 1, 1)$ and $(2, 2, 2)$ and uses the fact that linear maps preserve scalar multiplication (i.e., $T(2v) = 2T(v)$) to conclude that $c$ must be 0. This exercise provides practice in using the defining properties of linear maps to prove results about them. It also illustrates that a function that looks "almost linear" (like $T$ with nonzero $b$ or $c$) can fail to be linear. $\textit{Examples:}$ 1. Define $T : \mathbb{R}^2 \to \mathbb{R}^2$ by $T(x, y) = (2x - y + 3, xy)$. Then $T$ is not linear, because $T(0, 0) = (3, 0) \neq (0, 0)$. 2. Define $T : \mathbb{C}^2 \to \mathbb{C}$ by $T(z, w) = iz + w + 1$. Then $T$ is not linear, because $T(0, 0) = 1 \neq 0$. 3. Define $T : \mathcal{P}_2(\mathbb{R}) \to \mathbb{R}^3$ by $T(a + bx + cx^2) = (a, b + c, ac)$. Then $T$ is not linear, because $T(2p) \neq 2T(p)$ in general (consider $p(x) = 1 + x + x^2$). These examples demonstrate various ways in which a function between vector spaces can fail to be linear. The failures can be due to not sending 0 to 0, not preserving scalar multiplication, or not preserving addition. Understanding these failures helps to build intuition about the stringent requirements of linearity.