Page 349 Question C25
Define the linear transformation \(T: \mathbb{C}^3 \rightarrow \mathbb{C}^2\) \[ T\left(\begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix}\right) = \begin{bmatrix} 2x_1 - x_2 + 5x_3 \\ -4x_1 + 2x_2 - 10x_3 \end{bmatrix} \]
Let \(U, V\) be arbitrary vectors in \(\mathbb{C}^3\)
\[ u = \begin{bmatrix} u_1 \\ u_2 \\ u_3 \end{bmatrix} , v = \begin{bmatrix} v_1 \\ v_2 \\ v_3 \end{bmatrix} \]
To show that this is a Linear Transformation we must show 2 things:
1.) \(T(u+v) = T(u) + T(v)\)
2.) \(T(\alpha u) = \alpha T(u)\)
To show that \(T(u+v) = T(u) + T(v)\):
\[ T(u+v) = T( \begin{bmatrix} u_1 \\ u_2 \\ u_3 \end{bmatrix} + \begin{bmatrix} v_1 \\ v_2 \\ v_3 \end{bmatrix} ) \]
Substituting in we get:
\[ T( \begin{bmatrix} u_1 + v_1 \\ u_2 + v_2 \\ u_3 + v_3 \end{bmatrix} ) = \begin{bmatrix} 2(u_1+v_1) - (u_2 + v_2) + 5(u_3+v_3) \\ -4(u_1+v_1) + 2(u_2+v_2) - 10(u_3+v_3) \end{bmatrix} = \begin{bmatrix} (2u_1-u_2+5u_3) + (2v_1-v_2+5v_3) \\ (-4u_1 +2u_2 - 10u_3) + (-4v_1 +2v_2-10v_3) \end{bmatrix} = \begin{bmatrix} (2u_1-u_2+5u_3) \\ (-4u_1 +2u_2 - 10u_3) \end{bmatrix} + \begin{bmatrix} (2v_1-v_2+5v_3) \\ (-4v_1 +2v_2-10v_3) \end{bmatrix} \]
Therefore: \(T(u+v) = T(u) + T(v)\):
Now, we need to show \(T(\alpha u) = \alpha T(u)\):
Let \(u\) be an arbitrary vector \[ u= \begin{bmatrix} u_1 \\ u_2 \\ u_3 \end{bmatrix} \]
\[ T(\alpha u) = T( \alpha \begin{bmatrix} u_1 \\ u_2 \\ u_3 \end{bmatrix} ) = T( \begin{bmatrix} \alpha u_1 \\ \alpha u_2 \\ \alpha u_3 \end{bmatrix} ) \]
Substituting in we get:
\[ T(\alpha u )= \begin{bmatrix} 2(\alpha u_1) - \alpha u_2 + 5(\alpha u_3)\\ -4(\alpha u_1) + 2(\alpha u_2) - 10(\alpha u_3) \end{bmatrix} = \begin{bmatrix} \alpha (2u_1 - u_2 + 5u_3) \\ \alpha (-4u_1 +2u_2-10u_3) \end{bmatrix} = \alpha \begin{bmatrix} 2u_1 - u_2 + 5u_3 \\ -4u_1 + 2u_2 - 10u_3 \end{bmatrix} \]
Therefore, \(T(\alpha u) = \alpha T(u)\) and thus this is a Linear Transformation