C25 pg. 443
Define the linear transformation
\[T: C^3 ->C^2, T(\begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix}) = \begin{bmatrix} 2x_1-x_2+5x_3 \\ -4x_1+2x_2 -10x_3 \end{bmatrix}\]
Verify that \(T\) is a linear transformation.
Solution
\(1.proof: T(x+y) = T(x) + T(y)\)
\[T(x+y) = \begin{bmatrix} 2(x_1+y_1)-(x_2+y_2)+5(x_3+y_3) \\ -4(x_1+y_1)+2(x_2+y_2)-10(x_3+y_3)\end{bmatrix}\]
\[=\begin{bmatrix} 2x_1+2y_1-x_2-y_2+5x_3+5y_3\\-4x_1-4y_1+2x_2+2y_2-10x_3-10y_3\end{bmatrix}\]
\[=\begin{bmatrix} (2x_1-x_2+5x_3)+(2y_1-y_2+5y_3)\\(-4x_1+2x_2-10x_3)+(-4y_1+2y_2-10y_3) \end{bmatrix}\]
\[=\begin{bmatrix} 2x_1-x_2+5x_3 \\ -4x_1+2x_2 -10x_3 \end{bmatrix} + \begin{bmatrix}2y_1-y_2+5y_3\\-4y_1+2y_2-10y_3\end{bmatrix}\]
\[=T(x) + T(y)\]
\(2.proof: T(\alpha x) = \alpha T(x)\)
\[T(\alpha x) = \begin{bmatrix}2\alpha x_1 -\alpha x_2+5\alpha x_3 \\ -4\alpha x_1+2\alpha x_2-10\alpha x_3\end{bmatrix}\]
\[=\alpha \begin{bmatrix} 2x_1-x_2+5x_3 \\ -4x_1+2x_2 -10x_3 \end{bmatrix}\]
\[=\alpha T(x) \]
Conclusion
\(T\) is a linear transformation