Data605_Week4_discussion

To solve Exercise C40 in chapter LT:

Problem: 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 \(T\left(\begin{bmatrix} 1 \\ 1 \end{bmatrix}\right) = \begin{bmatrix} -1 \\ \frac{1}{2} \end{bmatrix}\), find \(T\left(\begin{bmatrix} 4 \\ 3 \end{bmatrix}\right)\).

Solution: By applying the transformation \(T\) to this linear combination and using the linearity property \[ T(av + bw) = aT(v) + bT(w), \] where \(v\) and \(w\) are vectors and \(a\) and \(b\) are scalars, we get:

Since \[ \begin{bmatrix} 4 \\ 3 \end{bmatrix} = \begin{bmatrix} 2 \\ 1 \end{bmatrix} + 2 \begin{bmatrix} 1 \\ 1 \end{bmatrix}, \] we have

\[ T\left( \begin{bmatrix} 4 \\ 3 \end{bmatrix} \right) = T\left( \begin{bmatrix} 2 \\ 1 \end{bmatrix} \right) + 2T\left( \begin{bmatrix} 1 \\ 1 \end{bmatrix} \right) = \begin{bmatrix} 3 \\ 4 \end{bmatrix} + 2 \begin{bmatrix} -1 \\ 2 \end{bmatrix} = \begin{bmatrix} 1 \\ 8 \end{bmatrix}. \]