C26. Verify that the function below is a linear transformation. \(T: P_2 \rightarrow \mathbb{C}^2\), \(T(a + bx + cx^2) = \begin{bmatrix} 2a - b \\ b + c \end{bmatrix}\) For a function to be a linear transformation, it must satisfy two key properties:
1. Additivity
Let \(u_1 = a' + b'x + c'x^2\) and \(u_2 = d' + e'x + f'x^2\). Then, \[ T(u_1 + u_2) = T((a' + d') + (b' + e')x + (c' + f')x^2) \] We can observe that: \[ \begin{aligned} &a = (a' + d'), \quad b = (b' + e'), \quad c = (c' + f') \\ \end{aligned} \]
Thus, \[ \begin{aligned} &T(u_1 + u_2) = [2(a' + d') - (b' + e') \\ &(b' + e') + (c' + f')] \\ &= [(2a' - b') + (2d' - e')] \\ &+ [(b' + c') + (e' + f')] \\ &= [(2a' - b')(b' + c')] + [(2d' - e')(e' + f')] \\ &= T(u_1) + T(u_2) \end{aligned} \]
2. Homogeneity
Let \(\alpha\) be any scalar and \(u = a + bx + cx^2\). Then, \[ \begin{aligned} &T(\alpha u) = T(\alpha(a + bx + cx^2)) \\ &= T((\alpha a) + (\alpha b)x + (\alpha c)x^2) \\ &= [2(\alpha a) - (\alpha b) \\ &(\alpha b) + (\alpha c)] \\ &= \alpha [2a - b \quad b + c] \\ &= \alpha T(u) \end{aligned} \]
Both Property 1 and Property 2 of a linear transformation have been demonstrated. Therefore the given: \[ T: P_2 \rightarrow \mathbb{C}^2, \quad T(a + bx + cx^2) = \begin{bmatrix} 2a - b \\ b + c \end{bmatrix} \] is truly a linear transformation.