C40: If T: \(\mathbb{C}^{2}\) \(\rightarrow\) \(\mathbb{C}^{2}\) satisfies T \(\left (\begin{bmatrix} 2 \\1 \end{bmatrix} \right)\) = \(\begin{bmatrix} 3 \\4 \end{bmatrix}\) and T \(\left (\begin{bmatrix} 1 \\1 \end{bmatrix}\right )\) = \(\begin{bmatrix} -1 \\2 \end{bmatrix}\) find T \(\begin{bmatrix} 4 \\3 \end{bmatrix}\)
Because \(\begin{bmatrix} 4 \\3 \end{bmatrix}\) = \(\begin{bmatrix} 2 \\1 \end{bmatrix} + 2\begin{bmatrix} 1 \\1 \end{bmatrix}\) solving this problem is much more straight forward than I had initially thought.
So: \(T\left (\begin{bmatrix} 4 \\3 \end{bmatrix} \right) = T\left (\begin{bmatrix} 2 \\1 \end{bmatrix} +2\begin{bmatrix} 1 \\1 \end{bmatrix} \right) = T\left (\begin{bmatrix} 2 \\1 \end{bmatrix}\right) +2T\left(\begin{bmatrix} 1 \\1 \end{bmatrix}\right)\)
Which simplifies to: \(\begin{bmatrix} 3 \\4 \end{bmatrix} + 2\begin{bmatrix} -1 \\2 \end{bmatrix}\)
Coming to the conclusion that: \(T\left (\begin{bmatrix} 4 \\3 \end{bmatrix} \right) = \begin{bmatrix} 1 \\8 \end{bmatrix}\)