<s>[INST] 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.
 [/INST] This is a very fascinating problem, but in the $P_{(4)}$ case, we have
$$\tag{1}
M_{P_{(4)}}^{-1}=\begin{pmatrix}
1 & -1/2 &  1/6 &   0  & -1/30\\
  &  1/2 & -1/2 &  1/4 &   0\\
  &      &  1/3 & -1/2 &  1/3\\
  &      &      &  1/4 & -1/2\\
  &      &      &      &  1/5 
\end{pmatrix}
$$
This factors out as
$$
M_{P_{(4)}}^{-1}=\begin{pmatrix}
1 & -1 &  1 &   0  & -1/6\\
  &  1 & -3/2 &  1 &   0\\
  &      &  1 & -2 &  5/3\\
  &      &      &  1 & -5/2\\
  &      &      &      &  1
\end{pmatrix}
\begin{pmatrix}
1 &  &   &     & \\
  &  1/2 &  &   &   \\
  &      &  1/3 & &  \\
  &      &      &  1/4 & \\
  &      &      &      &  1/5 
\end{pmatrix}
$$
Unfortunately, as we can read off, the columns in the matrix on the left do not give us Bernoulli polynomials :(
Although, please note the first row of the matrix in (1) gives us Bernoulli numbers. Perhaps the first row of the inverse matrix does give you the Bernoulli numbers, but you do not obtain the Bernoulli polynomials.
Again, read this with caution and suspicion: the good reader should compute this and verify it for him or herself!
Addendum
Write out
$$
M_{P_{(n)}} = DU
$$
where $D$ is a diagonal matrix, and $U=I+X$ is a unit upper triangular matrix (here $X$ is the strictly upper triangular part of $U$). The claim is that
$$(DU)^{-1}=(I+X)^{-1}D^{-1}=(I-X+X^{2}-X^{3}+\dots+(-1)^{n}X^{n})D^{-1}$$
produces the coefficients of Bernoulli polynomials in the columns, and the first row is the Bernoulli numbers. I claim the first (Bernoulli polynomial data) implies the latter (the first row consists of Bernoulli numbers), and claim this is obvious.
Conjecture
It seems that the OP found the matrix version for the Bernoulli polynomials described on Wikipedia using forward differences.
I would have urged the OP to change notation, and use lower triangular matrices. Why? Because then you could write
$$
M_{P_{(n)}}\begin{bmatrix}0\\ 0\\\vdots\\0\\ x^{n}\end{bmatrix} \cong (1+x)^{n+1}-x^{n+1}
$$
in vector form. Current notation demands we use row vectors for polynomials.
I'm about to go to bed, so I do not believe I have time to prove my conjecture. If no one has proven it by tomorrow, I'll try writing up a proof.
</s>