go from an extended polynomial to a Bezier polynomial

Question

What is the analytical expression to do a change of basis from the extended polynomial form i.e. $$ \mathbf {B} (t)=\sum _{j=0}^{n}{t^{j}\mathbf {C} _{j}} $$ to a Bezier form $$ \mathbf{B} (t) =\sum _{i=0}^{n}{n \choose i}(1-t)^{n-i}t^{i} \mathbf{P} _{i}$$ where $\mathbf{P} _{i}$ are the control points and we know the relation $$\mathbf{C} _{j}={\frac {n!}{(n-j)!}}\sum _{l=0}^{j}{\frac {(-1)^{l+j}\mathbf {P} _{l}}{l!(j-l)!}}$$ so what has to be done is to solve the following system for $\mathbf{P}$ \begin{gather} \begin{bmatrix} \mathbf{C}_0 \\ \vdots \\ \mathbf{C}_j \\ \vdots \\ \mathbf{C}_n \end{bmatrix} = AB \begin{bmatrix} \mathbf{P}_0 \\ \vdots \\ \mathbf{P}_j \\ \vdots \\ \mathbf{P}_n \end{bmatrix} \end{gather} where $A\in \mathcal{M}_{n\times n}(\mathbb{R})$ is defined by
\begin{align} A_{ij} =& \frac{n!}{(n-i)!} &\mbox{ if }i=j \\ & 0 &\mbox{ sinon} \end{align} and $B \in \mathcal{M}_{n\times n}(\mathbb{R})$ is defined by
\begin{align} B_{ij} =& \frac{(-1)^{i+j}}{i!(j-i)!} &\mbox{ if }i\geq j \\ & 0 &\mbox{ sinon} \end{align} an then $(AB)^{-1}$ will be the matrix to change the basis, anyone knows the analytical value for $(AB)^{-1}$?

Answer 1

See section 7 of these notes:

https://www.uio.no/studier/emner/matnat/math/MAT-INF4160/h15/undervisningsmateriale/bezier1.pdf

Or section 3.3 of these notes:

http://hdl.lib.byu.edu/1877/2822