I am given a question as follows:
Suppose $T: P_{2} \rightarrow M_{2,2}$ is a linear transformation whose action on a basis for $P_{2}$ is
$$T(2x^2+2x+2)=\begin{bmatrix} 2&4 \\ 2&8 \end{bmatrix} \\T(4x^2+2x+2)=\begin{bmatrix}4&6 \\ -2&10 \end{bmatrix} \\T(1)=\begin{bmatrix} 1&1 \\ 2&1 \end{bmatrix}.$$Find the action of $T$ on a general polynomial with $a,b,c$ as constants.
Assuming that $T$ is a linear transformation, then for any vectors $\vec{v_{1}}, …, \vec{v_{n}}$ and any scalars $a_{1}, …, a_{n}$, $T(a_{1}\vec{v_{1}}+…+a_{n}\vec{v_{n}})=a_{1}T(\vec{v_{1}})+…+a_{n}T(\vec{v_{n}})$.
The answer is supposed to be in matrix form.
What I have tried doing so far is taking the basis polynomials and putting them into a matrix, which I then multiplied by a constant vector consisting of $a,b,c$:
$$\begin{bmatrix} 2&4&0 \\ 2&2&0 \\ 2&2&1\end{bmatrix} \begin{bmatrix} a \\ b \\c \end{bmatrix}= \begin{bmatrix} 2a+4b \\ 2a+2b \\ 2a+2b+c \end{bmatrix}$$
Given what I have done so far, I am not sure if this is the right approach, and I am not sure where to proceed from here. Any help would be appreciated!
Best Answer
First determine the matrix representation relative to the basis, $T(1), T(x),T(x^2)$.$$T(x^2)=\frac{T(4x^2+2x+2)-T(2x^2+2x+2)}{2}\\=(\begin{bmatrix}4&6 \\ -2&10 \end{bmatrix}-\begin{bmatrix} 2&4 \\ 2&8 \end{bmatrix})/2 =\begin{bmatrix}1&1 \\ -2&1 \end{bmatrix}$$ $$T(x)=\frac{T(2x^2+2x+2)-2T(x^2)-2T(1)}{2}=(\begin{bmatrix} 2&4 \\ 2&8 \end{bmatrix}-2\begin{bmatrix}1&1 \\ -2&1 \end{bmatrix}-2\begin{bmatrix} 1&1 \\ 2&1 \end{bmatrix})/2=\begin{bmatrix} -1&0 \\ 1&2 \end{bmatrix}$$ $$T(1)=\begin{bmatrix} 1&1 \\ 2&1 \end{bmatrix}$$Therefore,$$T(ax^2+bx+c)=aT(x^2)+bT(x)+cT(1)=\begin{bmatrix} a-b+c&a+c \\ -2a+b+2c&a+2b+c \end{bmatrix}$$