Define the linear transformation
\(T:\mathbb{C}^3\rightarrow\mathbb{C}^2,\quad T\left(\begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix}\right)=\begin{bmatrix}2x_1-x_2+5x_3\\-4x_1+2x_2-10x_3\end{bmatrix}\)
Verify that \(T\) is a linear transformation.
The two conditions of a linear transformation are \(T(x+y)=T(x)+T(y)\) (in more colloquial terms, “the output of the sum is the sum of the outputs”) and \(T(\alpha x)=\alpha T(x)\) (in more colloquial terms, “the output of a scalar times an input is equal to the scalar times the output for that input”). We can verify both.
\(T(x+y)\\=T\left(\begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix}+\begin{bmatrix}y_1\\y_2\\y_3\end{bmatrix}\right)\\=T\left(\begin{bmatrix}x_1+y_1\\x_2+y_2\\x_3+y_3\end{bmatrix}\right)\\=\begin{bmatrix}2(x_1+y_1)-(x_2+y_2)+5(x_3+y_3)\\-4(x_1+y_1)+2(x_2+y_2)-10(x_3+y_3)\end{bmatrix}\\=\begin{bmatrix}2x_1+2y_1-x_2-y_2+5x_3+5y_3\\-4x_1-4y_1+2x_2+2y_2-10x_3-10y_3\end{bmatrix}\\=\begin{bmatrix}2x_1-x_2+5x_3+2y_1-y_2+5y_3\\-4x_1+2x_2-10x_3-4y_1+2y_2-10y_3)\end{bmatrix}\\=\begin{bmatrix}2x_1-x_2+5x_3\\-4x_1+2x_2-10x_3\end{bmatrix}+\begin{bmatrix}2y_1-y_2+5y_3\\-4y_1+2y_2-10y_3\end{bmatrix}\\=T\left(\begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix}\right)+T\left(\begin{bmatrix}y_1\\y_2\\y_3\end{bmatrix}\right)\\=T(x)+T(y)\)
\(T(\alpha x)\\=T\left(\alpha\begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix}\right)\\=T\left(\begin{bmatrix}\alpha x_1\\\alpha x_2\\\alpha x_3\end{bmatrix}\right)\\=\begin{bmatrix}2\alpha x_1-\alpha x_2+5\alpha x_3\\-4\alpha x_1+2\alpha x_2-10\alpha x_3\end{bmatrix}\\=\begin{bmatrix}\alpha (2x_1-x_2+5x_3)\\\alpha (-4x_1+2x_2-10x_3)\end{bmatrix}\\=\alpha \begin{bmatrix}2x_1-x_2+5x_3\\-4x_1+2x_2-10x_3\end{bmatrix}\\=\alpha T(x)\)
Thus, both conditions of linear transformations are satisfied.