Consider infinite matrices of the form
$$\left(
\begin{array}{ccccc}
a_0 & a_1 & a_2 & a_3 & . \\
0 & a_0 & a_1 & a_2 & . \\
0 & 0 & a_0 & a_1 & . \\
0 & 0 & 0 & a_0 & . \\
. & . & . & . & . \\
\end{array}
\right)$$
The elements on each diagonal coincide.
My questions are:
-
Do they form a commutative ring?
-
Can they be extended to form a field?
Now, let define an operation $\operatorname{reg} A=\sum_{k=0}^\infty B_k a_k,$
where $B_k$ are Bernoulli numbers.
What are the properties of this operation?
Let's define another operation $\det' A=\exp(\Re \operatorname{reg} \log A)$.
What are the properties of this operation?
Motivation part.
This is meant to be a matrix representation of divergent integrals and series.
For instance,
$\sum_{k=1}^\infty 1=
\left(
\begin{array}{ccccc}
0 & 1 & 0 & 0 & . \\
0 & 0 & 1 & 0 & . \\
0 & 0 & 0 & 1 & . \\
0 & 0 & 0 & 0 & . \\
. & . & . & . & . \\
\end{array}
\right)$
$\sum_{k=0}^\infty 1=
\left(
\begin{array}{ccccc}
1 & 1 & 0 & 0 & . \\
0 & 1 & 1 & 0 & . \\
0 & 0 & 1 & 1 & . \\
0 & 0 & 0 & 1 & . \\
. & . & . & . & . \\
\end{array}
\right)$
$\sum_{k=0}^\infty k=
\left(
\begin{array}{ccccc}
1/12 & 1/2 & 1/2 & 0 & . \\
0 & 1/12 & 1/2 & 1/2 & . \\
0 & 0 & 1/12 & 1/2 & . \\
0 & 0 & 0 & 1/12 & . \\
. & . & . & . & . \\
\end{array}
\right)$
$\int_0^\infty x dx=\int_0^\infty \frac 2{x^3}=\left(
\begin{array}{ccccc}
1/6 & 1/2 & 1/2 & 0 & . \\
0 & 1/6 & 1/2 & 1/2 & . \\
0 & 0 & 1/6 & 1/2 & . \\
0 & 0 & 0 & 1/6 & . \\
. & . & . & . & . \\
\end{array}
\right)$
There are also some expressions that include divergent integrals that can be represented this way:
$(-1)^{\int_0^\infty dx}=\left(
\begin{array}{ccccccc}
i & -\pi & -\frac{i \pi ^2}{2} & \frac{\pi ^3}{6} & \frac{i \pi ^4}{24} & -\frac{\pi
^5}{120} & . \\
0 & i & -\pi & -\frac{i \pi ^2}{2} & \frac{\pi ^3}{6} & \frac{i \pi ^4}{24} & . \\
0 & 0 & i & -\pi & -\frac{i \pi ^2}{2} & \frac{\pi ^3}{6} & . \\
0 & 0 & 0 & i & -\pi & -\frac{i \pi ^2}{2} & . \\
0 & 0 & 0 & 0 & i & -\pi & . \\
0 & 0 & 0 & 0 & 0 & i & . \\
. & . & . & . & . & . & . \\
\end{array}
\right)$
The $\operatorname{reg}$ operation gives the regularized value of the integral or series.
Best Answer
If the matrices have entries from a (unital) ring $R$ then the set of such matrices is isomorphic to $R[[x]]$, the ring of formal power series over $R$. To see this, observe that the map sending the infinite matrix with $a_0 = 0$, $a_1 = 1$ and $a_k = 0$ for $k \ge 2$ to $x$ is a ring isomorphism.
This also answers the second question: if $R$ is an integral domain then set of matrices embeds canonically in the field of fractions of $R[[x]]$ and this is the smallest field containing $R[[x]]$. In particular, if $R$ is a field then this field is $\{ \sum_{k=-m}^\infty a_k x^k : a_k \in R, m \in \mathbb{N}_0 \}$.
I'm uncertain how $\mathrm{reg}$ is (well)-defined, but certainly one can take $R$ to be the polynomial ring $\mathbb{C}[z]$ and then something like $\sum_{k=0}^\infty B_k(z) x^k$ is a well-defined element of $R[[x]] = \mathbb{C}[z][[x]]$. If, as in the correction then one wants Bernoulli numbers rather than the polynomials, just specialize to $\mathbb{C}[[x]]$ by evaluating at $z=0$.