1. **Problem:** Let $T: \mathbb{R}^3 \to \mathbb{R}^2$ be defined by $T(x,y,z) = (2x - y, 3z)$. Check if $T$ is linear, find $N(T)$ and $R(T)$, and verify the Dimension theorem. 2. **Check linearity:** A transformation $T$ is linear if for all vectors $\mathbf{u}, \mathbf{v}$ and scalar $c$, we have: $$T(\mathbf{u} + \mathbf{v}) = T(\mathbf{u}) + T(\mathbf{v})$$ $$T(c\mathbf{u}) = cT(\mathbf{u})$$ 3. **Verification:** Let $\mathbf{u} = (x_1, y_1, z_1)$ and $\mathbf{v} = (x_2, y_2, z_2)$. $$T(\mathbf{u} + \mathbf{v}) = T(x_1 + x_2, y_1 + y_2, z_1 + z_2) = (2(x_1 + x_2) - (y_1 + y_2), 3(z_1 + z_2))$$ $$= (2x_1 - y_1 + 2x_2 - y_2, 3z_1 + 3z_2) = T(\mathbf{u}) + T(\mathbf{v})$$ Similarly, $$T(c\mathbf{u}) = T(cx_1, cy_1, cz_1) = (2cx_1 - cy_1, 3cz_1) = c(2x_1 - y_1, 3z_1) = cT(\mathbf{u})$$ Thus, $T$ is linear. 4. **Find $N(T)$ (null space):** $N(T) = \{(x,y,z) \in \mathbb{R}^3 : T(x,y,z) = (0,0)\}$ So, $$2x - y = 0 \implies y = 2x$$ $$3z = 0 \implies z = 0$$ Hence, $$N(T) = \{(x, 2x, 0) : x \in \mathbb{R}\} = \text{span}\{(1,2,0)\}$$ 5. **Find $R(T)$ (range):** $R(T) = \{(a,b) \in \mathbb{R}^2 : \exists (x,y,z) \text{ with } T(x,y,z) = (a,b)\}$ Since $a = 2x - y$ and $b = 3z$, $a$ and $b$ can be any real numbers because $x,y,z$ are free variables. Thus, $$R(T) = \mathbb{R}^2$$ 6. **Verify Dimension theorem:** Dimension theorem states: $$\dim(N(T)) + \dim(R(T)) = \dim(\mathbb{R}^3) = 3$$ We have: $$\dim(N(T)) = 1$$ $$\dim(R(T)) = 2$$ Sum is $1 + 2 = 3$, which matches $\dim(\mathbb{R}^3)$. Hence, the Dimension theorem is verified. --- **Final answers:** - $T$ is linear. - $N(T) = \text{span}\{(1,2,0)\}$. - $R(T) = \mathbb{R}^2$. - Dimension theorem holds: $\dim(N(T)) + \dim(R(T)) = 3$.