A relationship between matrices, bernoulli polynomials, and binomial coefficients We define the following polynomials, for $n≥0$:
$$P_n(x)=(x+1)^{n+1}-x^{n+1}=\sum_{k=0}^{n}{\binom{n+1}{k}x^k}$$
For $n=0,1,2,3$ this gives us,
$$P_0(x)=1\enspace P_1(x)=2x+1\enspace P_2(x)=3x^2+3x+1\enspace P_3(x)=4x^3+6x^2+4x+1$$
We then define the set $P_{(3)}=\{P_0,P_1,P_2,P_3\}$. It can be easily shown that this set is a basis over the vector space of polynomials of degree $3$ and lower. We take $3$ for the sake of brevity. 
Taking the coefficients of these polynomials and turning them into column vectors, we can construct the matrix (coefficients from the lowest term to the highest term)
$$\large{M_{P_{(3)}}}=\begin{pmatrix}
 1 & 1 & 1 & 1 \\
 0 & 2 & 3 & 4 \\
 0 & 0 & 3 & 6 \\
 0 & 0 & 0 & 4
\end{pmatrix}$$
We'll call this matrix the pascal in the context of this post, and the above polynomials as pascal polynomials. 
The inverse of this matrix is the matrix,
$$M_{P_{(3)}}^{-1}=\begin{pmatrix}
 1 & -\frac{1}{2} & \frac{1}{6} & 0 \\
 0 & \frac{1}{2} & -\frac{1}{2} & \frac{1}{4} \\
 0 & 0 & \frac{1}{3} & -\frac{1}{2} \\
 0 & 0 & 0 & \frac{1}{4}
\end{pmatrix}$$
We'll factor this matrix into two matrices as follows:
$$M_{P_{(3)}}^{-1}=\begin{pmatrix}
 1 & 0 & 0 & 0 \\
 0 & \frac{1}{2} & 0 & 0 \\
 0 & 0 & \frac{1}{3} & 0 \\
 0 & 0 & 0 & \frac{1}{4}
\end{pmatrix}×\begin{pmatrix}
 1 & -\frac{1}{2} & \frac{1}{6} & 0 \\
 0 & 1 & -1 & \frac{1}{2} \\
 0 & 0 & 1 & -\frac{3}{2} \\
 0 & 0 & 0 & 1
\end{pmatrix}$$
We can see the Bernoulli numbers in the first row of the matrix. Every column is a coefficient vector of a Bernoulli polynomial 
The following are extended versions of these matrices:  


*

*Pascal Matrix

*Inverse Pascal Matrix

*Last, Bernoulli Matrix
Are there accepted names for these matrices and polynomials? What is the meaning of these relationships? 
In particular, is there some treatment of using these matrices as change of basis transformations between representations of polynomials? E.g. from a linear combination of pascal polynomials to a linear combination of monomial terms.

Regarding the interpretation of the matrices: 
Let 
$$X = 
\left(\begin{array}{c}
1 \\ 
x \\
\vdots \\
x^n 
\end{array}\right) 
\qquad \textrm{and}\qquad  
A =
\left(\begin{array}{c}
a_0 \\ 
a_1 \\ 
\vdots \\ 
a_n
\end{array}\right)$$ 
so 
$$X^T A = \sum_{k=0}^n a_k x^k.$$
(I've written $X^T A$ instead of the preferred $A^T X$ to agree with your notation and avoid things like $(M^{-1})^T$.)
Then 
$$\begin{eqnarray*}
X^T A &=& X^T M M^{-1} A \\
&=& {X'}^T A'
\end{eqnarray*}$$
where 
$X' = M^T X$
is the new basis and $A' = M^{-1} A$ are the coefficients in the new basis.
This interpretation holds for any invertible transformation $M$.
The formalism can be generalized in a straightforward way to transformations between any two bases of polynomials.
For your transformation we find the coefficients of 
$\sum_{k=0}^3 a_k x^k$ in the "Pascal basis" are 
$$M_{P_{(3)}}^{-1}A = \left(
\begin{array}{c}
a_0 - \frac{1}{2} a_1 + \frac{1}{6} a_2 \\
\frac{1}{2}a_1 - \frac{1}{2} a_2 + \frac{1}{4} a_3 \\
\frac{1}{3} a_2 - \frac{1}{2} a_3 \\
\frac{1}{4} a_3
\end{array}
\right).$$
Thus, 
$$\begin{eqnarray*}
\sum_{k=0}^3 a_k x^k &=& 
\left(a_0 - \frac{1}{2} a_1 + \frac{1}{6} a_2\right) P_0(x) \\
&&+ \left(\frac{1}{3} a_2 - \frac{1}{2} a_3\right) P_1(x) \\
&&+ \left(\frac{1}{3} a_2 - \frac{1}{2} a_3\right) P_2(x) \\
&&+ \frac{1}{4} a_3 P_3(x).
\end{eqnarray*}$$
