Linear Transformation Matrix with polynomials

Question

A linear transformation $T : P_2 \to P_2$ has matrix with respect to $S$ given by:

$$[T]\,( S) = \begin{bmatrix} 1/2&-3&1/2\\ -1&4&-1\\ 1/2&2&1/2\\ \end{bmatrix} $$

How do you find $T(a+bx+cx^2)$?

Thank you!!

Answer 1

Every polynomial $S$ of degree $2$ s.t. $S \in P_2$ can be represented as a vector in three dimensional space: $$ S = a + bx + cx^2 \quad \iff \quad \begin{bmatrix} a \\ b \\ c \\ \end{bmatrix} \cdot \begin{bmatrix} 1 \\ x \\ x^2 \\ \end{bmatrix}, $$ therefore we can associate $S$ with a 3D vector $$ S \longleftrightarrow \begin{bmatrix} a \\ b \\ c \\ \end{bmatrix}. $$

When you have the polynomial represented as a vector, applying linear transformation $T$, given in form of a matrix, is a piece of cake:

$$ [T]\,( S) = \begin{bmatrix} 1/2&-3&1/2\\ -1&4&-1\\ 1/2&2&1/2\\ \end{bmatrix} \begin{bmatrix} a \\ b \\ c \\ \end{bmatrix} = \begin{bmatrix} \frac{1}{2}a - 3 b + \frac{1}{2}c \\ -a + 4 b -c \\ \frac{1}{2}a +2 b + \frac{1}{2}c \\ \end{bmatrix} , $$ which will correspond to the polynomial

$$ T\left( a+bx+cx^2\right) = \left(\frac{1}{2}a - 3 b + \frac{1}{2}c\right) + \left( -a + 4 b -c \right) x + \left(\frac{1}{2}a +2 b + \frac{1}{2}c \right)x^2 $$