Transformation Matrix for Derivative

Question

I have figured out how to show Part A by using properties of derivatives. For Part B, we know that $T(f)$ is $\begin{pmatrix} 4a+2b+c\\4a+b\\2a\end{pmatrix}$, so when asked to find a matrix $A$, would it have to be the case that when $A$ is multiplied by the $3*1$ matrix $[f]_B = \begin{pmatrix} 1\\x\\x^2\end{pmatrix}$?

Answer 1

Hint: Use the fact that $T$ is linear, and that if we take $B=(1,x,x^2)=(b_1,b_2,b_3)$, then any $f$ can be written as $$f= ab_1+bb_2+cb_3= \begin{bmatrix} b_1 \quad b_2 \quad b_3 \end{bmatrix}\begin{bmatrix} a \\ b \\ c \end{bmatrix} $$ Then $$ f'=\cdots =\begin{bmatrix} \cdots \quad \cdots \quad \cdots \end{bmatrix}\begin{bmatrix} a \\ b \\ c \end{bmatrix} $$ and $$f''=\cdots =\begin{bmatrix} \cdots \quad \cdots \quad \cdots \end{bmatrix}\begin{bmatrix} a \\ b \\ c \end{bmatrix} $$ Thus $A=\cdots$

Answer 2

I assume $f(x) = a x^2 + b x + c$ and $$ [f]_B = [a x^2 + b x + c]_B = (a, b, c)^t $$ and the task is finding $A$ such that $$ A [f]_B = A (a, b, c)^t \overset{!}{=} T(f) = (4a+2b+c,4a+b,2a)^t $$ Another choice could be $$ [f]_B = [a x^2 + b x + c]_B = (c, b, a)^t $$ so please check your definition of $[.]_B$.