Verify that the function below is a linear transformation:

\[T: P_2 \rightarrow \mathbb{C}^2,\quad T(a+bx+cx^2) = \begin{bmatrix} 2a - b \\ b + c \\ \end{bmatrix}\]

Solution:

We need to make sure that \(T(u_1+u2) = T(u_1) + T(u2) \textrm{ for all } u_1,u_2 \in U\)

So:

\[\begin{align*} & T(u+v) = T((a+bx+cx^2) + (d+ex+fx^2)) \\ &= T((a+d) + x(b+e) + x^2(c+f)) \\ &=\begin{bmatrix} 2(a+d) - (b+e)\\ (b+e)+(c+f) \end{bmatrix} \\ &=\begin{bmatrix} (2a -bd) + (2d-e)\\ (b+c)+(e+f)\\ \end{bmatrix} \\ &=\begin{bmatrix} 2a-b\\ b+c \end{bmatrix}+ \begin{bmatrix} 2d-e \\ e+f \end{bmatrix} \\ &=T(u)+T(v) \end{align*}\]

And we also need to make sure that \(T(\alpha u) = \alpha T(u) \textrm{ for all } u \in U \textrm{ and } \alpha \in \mathbb{C}\)

So:

\[\begin{align*} &T(\alpha u) = T(\alpha(a+bx+cx^2))\\ &= T((\alpha a)+x(\alpha b) + x^2(\alpha c))\\ &=\begin{bmatrix} 2(\alpha a) + (\alpha b)\\ (\alpha b) + (\alpha c)\\ \end{bmatrix} \\ &=\begin{bmatrix} \alpha(2a -b)\\ \alpha(b+c) \end{bmatrix}\\ &=\alpha \begin{bmatrix} 2a-b \\ b+c \end{bmatrix} \\ &=\alpha(u) \end{align*}\]