The Derivative and the Integral as Infinite Matrice

[Sauraj]

Hello,
if i have polynomials
$\displaystyle p(x) = c_0 + c_1x +c_2x^2 + c_3x^3$
$\displaystyle p'(x) = 0 + c_1 + 2c_2x + 3c_3x^2$
$\displaystyle P(x) = c_0x + 1/2c_1x^2 + 1/3c_2x^3 + 1/4c3x^4$
p(x) is polynomial, p'(x) is derivation of p(x) and P(x) is integral of p(x) (c0 ,c1,c2,c3 are coefficients)

Now I have to find matrices $\displaystyle A_{3} \in \mathbb{R} ^{3x4}, B_{2} \in \mathbb{R} ^{4x3}$ so that
$\displaystyle c' = A_{3} \cdot c$ where c' is vector of coefficients of p'(x) and c is vector of coefficients of p(x) (in this case c is: (c0,c1,c2,c3)^T)
$\displaystyle c^{*} = B_{2} \cdot c$ where c* is vector of coefficients of P(x) and c is vector of coefficients of p(x) (in this case c is: (c0,c1,c2,c3)^T)
I have to find matrices $\displaystyle A_3 (derivation), B_2(integral)$
My solution:
$\displaystyle A_3 = \begin{pmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & 3 \end{pmatrix}$ because
$\displaystyle \begin{pmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & 3 \end{pmatrix} \cdot \begin{pmatrix} c_{0} \\ c_{1} \\ c_{2} \\ c_{3} \end{pmatrix}$= $\displaystyle (c_1, 2 \cdot c_2, 3 \cdot c_3)^T$

Now how to find $\displaystyle B_2$?
I have
$\displaystyle B_2 = \begin{pmatrix} 0 & 0 & 0 \\ 1/2 & 0 & 0 \\ 0 & 1/3 & 0 \\ 0 & 0 & 1/4 \end{pmatrix} \cdot \begin{pmatrix} c_{1} \\ c_{2} \\ c_{3} \end{pmatrix} =$$\displaystyle (0, 1/2c_1, 1/3c_2, 1/4c_3)^T$

The product of A*B should be identity (from here https://demonstrations.wolfram.com/TheDerivativeAndTheIntegralAsInfiniteMatrices/#more) matrix but in my case it is not. So I assume B is not correct.
Can somebody help please, it should be easy but I dont get it..

 

[Romsek]

you should always include all coefficients of the initial polynomial in the coefficient vector.

[MATH]B_2 = \begin{pmatrix} 0 & 0 & 0 \\ 1 & 0 & 0 \\ 0 & \frac{1}{2} & 0 \\ 0 & 0 & \frac{1}{3} \\ \end{pmatrix}[/MATH]
and this would be multiplied by [MATH]\begin{pmatrix}c_0\\c_1\\c_2\end{pmatrix}[/MATH]
[MATH]A_3 B_2 = I_3[/MATH]