Define the linear transformation
\[ T: \mathbb{C}^{3} \rightarrow \mathbb{C}^{2}, T(\begin{bmatrix} x \\ y \\ z \end{bmatrix} ) = \begin{bmatrix} 2x_{1} - x_{2} + 5x_{3} \\ -4x_{1} + 2x_{2} -10x_{3} \end{bmatrix} \]
Verify that T is a linear transformation.
First the addition property. \(T(u_1+u_2) = T(u_1) +T(u_2)\)
Let \(x = \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix}\) and \(y = \begin{bmatrix} y_1 \\ y_2 \\ y_3 \end{bmatrix}\)
\[ x + y = \begin{bmatrix} x_1+y_1 \\ x_2 +y_2 \\ x_3 +y_3 \end{bmatrix} \]
\[ T(x+y) = \begin{bmatrix} 2(x_{1} + y_1) - (x_{2} + y_2) + 5(x_{3} +y_3) \\ -4(x_{1}+ y_1) + 2(x_{2} +x_2) -10(x_{3} + y_3) \end{bmatrix} \]
Rearranging terms:
\[ T(x+y) = \begin{bmatrix} 2x_{1} + 2y_1 - x_{2} - y_2 + 5x_{3} +5y_3 \\ -4x_{1} - 4y_1 + 2x_{2} + 2y_2 -10x_{3} - 10y_3 \end{bmatrix} \] \[ T(x+y) = \begin{bmatrix} (2x_{1}- x_{2} + 5x_{3}) + (2y_1 - y_2 +5y_3) \\ (-4x_{1} + 2x_{2} -10x_{3}) + (- 4y_1 + 2y_2 - 10y_3) \end{bmatrix} \]
\[ T(x+y) = \begin{bmatrix} (2x_{1}- x_{2} + 5x_{3}) \\ (-4x_{1} + 2x_{2} -10x_{3}) \end{bmatrix} + \begin{bmatrix} (2y_1 - y_2 +5y_3) \\ (- 4y_1 + 2y_2 - 10y_3) \end{bmatrix} \]
Therefore,
\[T(x+y) = T(x) +T(y)\]
\(T(x)\) follows the addition property of Linear Transformations
Next the multiplication property
\(T(\alpha x) = \alpha T(x)\)
\[\alpha x = \begin{bmatrix} \alpha x_1 \\ \alpha x_2 \\ \alpha x_3 \end{bmatrix}\]
$$ T(x) = \[\begin{bmatrix} 2(\alpha x_{1}) - (\alpha x_{2}) + 5(\alpha x_{3}) \\ -4(\alpha x_{1}) + 2(\alpha x_{2}) - 10(\alpha x_{3}) \end{bmatrix}\]=
\[\begin{bmatrix} \alpha (2x_{1} - x_{2} + 5 x_{3}) \\ \alpha (-4 x_{1} + 2x_{2} - 10 x_{3}) \end{bmatrix}\]$$
\[ T(\alpha x) = \alpha \begin{bmatrix} 2x_{1} - x_{2} + 5 x_{3} \\ -4 x_{1} + 2x_{2} - 10 x_{3} \end{bmatrix} \] \[T(\alpha x) = \alpha T(x)\]
\(T(x)\) also follows the multiplication rule so it is a Linear Transformation.