Define the linear transformation \(T: C^3 \rightarrow C^2\)
\[ T( \begin{bmatrix} x_1\\ x_2\\ x_3\\ \end{bmatrix} ) = \begin{bmatrix} 2x_1 - x_2 + 5x_3\\ -4x_1 +2x_2 - 10x_3\\ \end{bmatrix} \]
Let \(U,V\) be arbitrary vectors in \(C^3\).
\[ u = \begin{bmatrix} 1\\ 0\\ 0\\ \end{bmatrix}\\ v = \begin{bmatrix} 0\\ 1\\ 0\\ \end{bmatrix} \]
We must show: 1. \(T(u+v) = T(u) + T(v)\) 2. \(T(\alpha u) = \alpha T(u)\)
First, we’ll show that \(T(u+v) = T(u) + T(v)\).
\[ T(u+v) = T( \begin{bmatrix} 1\\ 1\\ 0\\ \end{bmatrix} ) = \begin{bmatrix} 2(1) - (1) + 5(0)\\ -4(1) + 2(1) - 10(0)\\ \end{bmatrix} = \begin{bmatrix} 2 - 1\\ -4 + 2\\ \end{bmatrix} = \begin{bmatrix} 1\\ -2\\ \end{bmatrix} \\ T(u) = T( \begin{bmatrix} 1\\ 0\\ 0\\ \end{bmatrix} ) = \begin{bmatrix} 2(1) - (0) + 5(0)\\ -4(1) +2(0) -10(0)\\ \end{bmatrix} = \begin{bmatrix} 2\\ -4\\ \end{bmatrix}\\ T(v) = T( \begin{bmatrix} 0\\ 1\\ 0\\ \end{bmatrix} ) = \begin{bmatrix} 2(0) - (1) + 5(0)\\ -4(0) + 2(1) - 10(0\\ \end{bmatrix} = \begin{bmatrix} -1\\ 2\\ \end{bmatrix} \\ T(u) + T(v) = \begin{bmatrix} 2\\ -4\\ \end{bmatrix} + \begin{bmatrix} -1\\ 2\\ \end{bmatrix} = \begin{bmatrix} 1\\ -2\\ \end{bmatrix}\\ \]
Hence, \(T(u+v) = T(u) + T(v)\).
Now, we need to prove that \(T(\alpha u) = \alpha T(U)\).
\[ T(\alpha U) = \begin{bmatrix} 2(1 * \alpha) - (0 * \alpha) + 5(0 * \alpha)\\ -4 (1 * \alpha) + 2(0 * \alpha) - 10(0 * \alpha)\\ \end{bmatrix} = \begin{bmatrix} 2 \alpha \\ -4 \alpha \\ \end{bmatrix} = \alpha \begin{bmatrix} 2\\ -4\\ \end{bmatrix}\\ \alpha T(u) = \alpha \begin{bmatrix} 2(1) - 0 + 5(0)\\ -4(1) + 2(0) - 10(0)\\ \end{bmatrix} = \alpha \begin{bmatrix} 2\\ -4\\ \end{bmatrix} = \alpha \begin{bmatrix} 2\\ -4\\ \end{bmatrix}\\ \]
Hence, \(T(\alpha u) = \alpha T(u)\).
This is a linear transformation.