C25, Page 349
For a transformation to be linear, it must satisfy both additivity and homogeneity.
\(T: \mathbb{C}^3 \rightarrow \mathbb{C}^2\)
\[ T\left(\begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix}\right) = \begin{bmatrix} 2x_1 - x_2 + 5x_3 \\ -4x_1 + 2x_2 - 10x_3 \end{bmatrix} \]
The matrix representation of \(T\) is:
\[ A = \begin{bmatrix} 2 & -1 & 5 \\ -4 & 2 & -10 \end{bmatrix} \]
Let \(\mathbf{u} = \begin{bmatrix} u_1 \\ u_2 \\ u_3 \end{bmatrix}\) and \(\mathbf{v} = \begin{bmatrix} v_1 \\ v_2 \\ v_3 \end{bmatrix}\) be vectors in \(\mathbb{C}^3\).
Then:
\[\begin{align*} T(\mathbf{u} + \mathbf{v}) &= T\left(\begin{bmatrix} u_1 + v_1 \\ u_2 + v_2 \\ u_3 + v_3 \end{bmatrix}\right) \\ \\ &= \begin{bmatrix} 2(u_1+v_1) - (u_2+v_2) + 5(u_3+v_3) \\ -4(u_1+v_1) + 2(u_2+v_2) - 10(u_3+v_3) \end{bmatrix} \\ \\ &= \begin{bmatrix} 2u_1 - u_2 + 5u_3 \\ -4u_1 + 2u_2 - 10u_3 \end{bmatrix} + \begin{bmatrix} 2v_1 - v_2 + 5v_3 \\ -4v_1 + 2v_2 - 10v_3 \end{bmatrix} \\ \\ &= T(\mathbf{u}) + T(\mathbf{v}) \end{align*}\]
Let \(c \in \mathbb{C}\) be a scalar. Then:
\[\begin{align*} T(c\mathbf{u}) &= T\left(\begin{bmatrix} cu_1 \\ cu_2 \\ cu_3 \end{bmatrix}\right) \\ \\ &= \begin{bmatrix} 2(cu_1) - (cu_2) + 5(cu_3) \\ -4(cu_1) + 2(cu_2) - 10(cu_3) \end{bmatrix} \\ \\ &= c\begin{bmatrix} 2u_1 - u_2 + 5u_3 \\ -4u_1 + 2u_2 - 10u_3 \end{bmatrix} \\ \\ &= cT(\mathbf{u}) \end{align*}\]