C26 † Page 443

Solution

For this, we need to verify the two defining conditions of a linear transformation; that is:

• 1. \(T(x + y) = T(x) + T(y)\)
• 2. \(T( \alpha x) = \alpha T(x)\)

1. First let’s work on (1) \(T(x + y) = T(x) + T(y)\).

\[T(x + y) = T \left((a_1 + b_1x + c_1x^2) + (a_2 + b_2x + c_2x^2)\right)\]

\[T(x + y) = T \left((a_1 + a_2) + (b_1 + b_2)x + (c_1+ c_2)x^2\right)\]

Let’s define as follows:

\(a = a_1 + a_2\)

\(b = b_1 + b_2\)

\(c = c_1 + c_2\)

By making a substitution; we have as follows:

\[T(x + y) = T \left((a) + (b)x + (c)x^2\right)\] And by our given problem, we have as follows:

\[T \left( a + bx + cx^2 \right) = \begin{pmatrix}2a - b\\ b + c\end{pmatrix}\]

And by substituting back we have:

\[T(x + y) =\begin{pmatrix}2(a_1 + a_2) - ( b_1 + b_2)\\ ( b_1 + b_2) + (c_1 + c_2)\end{pmatrix}\]

\[T(x + y) =\begin{pmatrix}(2a_1 + 2a_2) - ( b_1 + b_2)\\ ( b_1 + b_2) + (c_1 + c_2)\end{pmatrix}\]

\[T(x + y) =\begin{pmatrix}(2a_1 - b_1) + (2a_2 - b_2)\\ ( b_1 + c_1) + (b_2 + c_2)\end{pmatrix}\]

\[T(x + y) =\begin{pmatrix}(2a_1 - b_1) \\ ( b_1 + c_1) \end{pmatrix} + \begin{pmatrix} (2a_2 - b_2) \\ (b_2 + c_2)\end{pmatrix}\]

\[T(x + y) =\begin{pmatrix}2a_1 - b_1 \\ b_1 + c_1 \end{pmatrix} + \begin{pmatrix} 2a_2 - b_2 \\ b_2 + c_2\end{pmatrix}\]

\[T(x + y) = T \left( a_1 + b_1x + c_1x^2 \right) + T \left( a_2 + b_2x + c_2x^2 \right)\]

\[T(x + y) = T(x) + T(y)\]

This proof the first part of the definition is met.

2. Now, I will focus on (2) \(T( \alpha x) = \alpha T(x)\).

\[T(\alpha x) = T \left( \alpha (a_1 + b_1x + c_1x^2) \right) \]

\[T(\alpha x) = T \left( \alpha a_1 + \alpha b_1x + \alpha c_1x^2 \right) \]

Let’s define as follows:

\(a = \alpha a_1\)

\(b = \alpha b_1\)

\(c = \alpha c_1\)

And by making a substitution we get:

\[T(\alpha x) = T \left( a + bx + cx^2 \right) \]

Given our original equation, we have as follows:

\[T(\alpha x) = \begin{pmatrix}2a - b\\ b + c\end{pmatrix}\]

By replacing back our previous definitions, we get as follows:

\[T(\alpha x) = \begin{pmatrix}2 \alpha a_1 - \alpha b_1\\ \alpha b_1 +\alpha c_1\end{pmatrix}\]

\[T(\alpha x) = \begin{pmatrix} \alpha (2a_1 - b_1)\\ \alpha (b_1 + c_1)\end{pmatrix}\]

\[T(\alpha x) = \alpha \begin{pmatrix} 2a_1 - b_1\\ b_1 + c_1\end{pmatrix}\]

\[T(\alpha x) = \alpha T \left( a + bx + cx^2 \right)\]

\[T(\alpha x) = \alpha T(x)\]

This demonstrates that the second equation is met.

Conclusion:

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

The above function is a linear transformation since the two defining equations are met.