# Inverse of Stirling's Triangle expressed as Matrix

## Theorem

Consider Stirling's triangle of the first kind (signed) expressed as a (square) matrix $\mathbf A$, with the top left element holding $s \left({0, 0}\right)$.

$\begin{pmatrix} 1 & 0 & 0 & 0 & 0 & 0 & \cdots \\ 0 & 1 & 0 & 0 & 0 & 0 & \cdots \\ 0 & -1 & 1 & 0 & 0 & 0 & \cdots \\ 0 & 2 & -3 & 1 & 0 & 0 & \cdots \\ 0 & -6 & 11 & -6 & 1 & 0 & \cdots \\ 0 & 24 & -50 & 35 & -10 & 1 & \cdots \\ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \ddots \\ \end{pmatrix}$

Thenconsider Stirling's triangle of the second kind expressed as a (square) matrix $\mathbf B$, with the top left element holding $\displaystyle \left \{ {0 \atop 0}\right\}$.

$\begin{pmatrix} 1 & 0 & 0 & 0 & 0 & 0 & \cdots \\ 0 & 1 & 0 & 0 & 0 & 0 & \cdots \\ 0 & 1 & 1 & 0 & 0 & 0 & \cdots \\ 0 & 1 & 3 & 1 & 0 & 0 & \cdots \\ 0 & 1 & 7 & 6 & 1 & 0 & \cdots \\ 0 & 1 & 15 & 25 & 10 & 1 & \cdots \\ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \ddots \\ \end{pmatrix}$

Then:

$\mathbf A = \mathbf B^{-1}$

that is:

$\mathbf B = \mathbf A^{-1}$

## Proof

First note that from Relation between Signed and Unsigned Stirling Numbers of the First Kind:

$\displaystyle \left[{n \atop m}\right] = \left({-1}\right)^{n + m} s \left({n, m}\right)$
$\displaystyle \sum_k \left[{n \atop k}\right] \left\{ {k \atop m}\right\} \left({-1}\right)^{n - k} = \delta_{m n}$
$\displaystyle \sum_k \left\{ {n \atop k}\right\} \left[{k \atop m}\right] \left({-1}\right)^{n - k} = \delta_{m n}$

By definition of matrix multiplication, the element $a_{r n}$ of the matrix formed by multiplying the two matrices above.

As can be seen, this results in the identity matrix .

$\blacksquare$