I'm interested in pairs $A=(a_{i,j})_{i,j=0,1,\ldots}$ and $B=(b_{i,j})_{i,j=0,1,\ldots}$ of infinite matrices for which:
- They are uni-lower-triangular, i.e., $a_{i,i}=1$ for all $i$ and $a_{i,j}=0$ for all $j>i$.
- They are inverses, i.e. their product $AB$ is the identity.
- The generating function of the $k$th column of $A$ is the reciprocal of the generating function of the $(k+1)$st row of $B$, that is,
$$ \sum_{j=k}^{\infty}a_{j,k} \; x^{j-k}= \big(\sum_{i=0}^{k+1}b_{k+1,k+1-i} \; x^{i}\big)^{-1}.$$
Notice how for the generating function of columns we start at the $1$ on the diagonal and go down, while for the generating function of rows we start at the $1$ on the diagonal and go left (and all g.f.'s are normalized to start with the constant term).
I have a handful of examples of this occurring with matrices of combinatorial significance:
Example 1. We let $a_{i,j}=\binom{i}{j}$ and $b_{i,j}=(-1)^{i-j}\binom{i}{j}$. Note the generating function of the $k$th row of $B$ is $(1-x)^{k}$, and the generating function of the $(k-1)$th column of $A$ is $1/(1-x)^{k}$.
Example 2. We let $a_{i,j}=S(i+1,j+1)$ and $b_{i,j} = s(i+1,j+1)$, where $S(n,k)$ and $s(n,k)$ are the Stirling numbers of the 2nd and 1st kind, respectively (the shift by one is just to match my convention that indexing of rows/columns starts at $0$). This looks like
$$ A = \begin{pmatrix} 1 & 0 & 0 & 0 & \cdots \\ 1 & 1 & 0 & 0 \\ 1 & 3 & 1 & 0 \\ 1 & 7 & 6 & 1 \\ \vdots & & & & \ddots \end{pmatrix} \qquad B = \begin{pmatrix} 1 & 0 & 0 & 0 & \cdots \\ -1 & 1 & 0 & 0 \\ 2 & -3 & 1 & 0 \\ -6 & 11 & -6 & 1 \\ \vdots & & & & \ddots \end{pmatrix}$$
Note that the generating function of the $k$th row of $B$ (and the reciprocal of the generating function of the $(k-1)$th column of $A$) is $(1-x)(1-2x)\cdots(1-kx)$.
Example 3. The previous examples were over $\mathbb{Z}$, this example is over $\mathbb{Z}[q]$; actually it is a $q$-analog of Example 1. We let $a_{i,j} = \binom{i}{j}_q$ be the usual $q$-binomial $\binom{i}{j}_q = \frac{(1-q^i)(1-q^{i-1})\ldots(1-q^{i-j+1})}{(1-q^j)(1-q^{j-1})\ldots(1-q)}$, and we let $b_{i,j} = (-1)^{i-j} \; q^{\binom{i-j}{2}} \; \binom{i}{j}_q$. This looks like:
$$ A = \begin{pmatrix} 1 & 0 & 0 & 0 & \cdots \\ 1 & 1 & 0 & 0 \\ 1 & q+1 & 1 & 0 \\ 1 & q^{2}+q+1 & q^{2}+q+1 & 1 \\ \vdots & & & & \ddots \end{pmatrix} \qquad B = \begin{pmatrix} 1 & 0 & 0 & 0 & \cdots \\ -1 & 1 & 0 & 0 \\ q & -(q+1) & 1 & 0 \\ -q^3 & q^3+q^2+q & -(q^2+q+1) & 1 \\ \vdots & & & & \ddots \end{pmatrix}$$
Note that the generating function of the $k$th row of $B$ (and the reciprocal of the generating function of the $(k-1)$th column of $A$) is $(1-x)(1-qx)\cdots(1-q^{k-1}x)$.
Question 1. What is going on here? Why are these pairs of matrices of combinatorial sequences inverse "in two ways"? How is being inverse in one way related to being inverse in the other way?
Note that all of these examples come from sequences of uniform posets: they are the Whitney numbers of the 2nd and 1st kind of these posets. Hence, this question is related to my previous question. (But it's easy to come up with sequences of uniform posets whose matrices are not "inverse in the 2nd way.")
Also note that in all the examples, the generating function of the $(k+1)$st column of $A$ is obtained from the generating function of the $k$th column by multiplying by a simple factor; but (except for Example 1), it is not exactly the same factor every time, so these matrices are not quite Riordan arrays.
Question 2. Can you find some more examples of matrices of combinatorially significant sequences with these properties?
EDIT: One further observation is that you can take any such pair $A=(a_{i,j})$, $B=(b_{i,j})$ of matrices and get another one $A'$, $B'$ by choosing some constant $\kappa$ and setting $a'_{i,j}=\kappa^{i-j} \; a_{i,j}$, $b'_{i,j}=\kappa^{i-j} \; b_{i,j}$. So for example from the Pascal's triangle example you can (basically) get the $f$-vectors of the cross polytope this way.
Best Answer
Following up on David's nice answer, there is a different parametrization that makes the pattern much more obvious. Namely, let $s_1,s_2,\dots$ be arbitrary, then you can write $$A=\Big(h_{i-j}(s_1,s_2,\dots, s_{j+1})\Big)_{i,j=0}^{\infty}$$ $$B=\Big((-1)^{i-j}e_{i-j}(s_1,s_2,\dots, s_{i})\Big)_{i,j=0}^{\infty}$$ And checking that the matrices and the corresponding generating functions are inverses becomes a little more clear. The $h_i, e_i$ denote the complete homogeneous and elementary symmetric functions respectively, with the convention $h_0, e_0 =1$.