Recognize or interpret this involution : $\frac{\prod_{x\neq i}(1-a_xa_j)}{\prod_{y\neq j}(a_j-a_y)}$ Let $a_1,a_2,\ldots ,a_n$ be distinct numbers. For indices $i,j$
between $1$ and $n$, put
$d_i=\prod_{j\neq i}(a_i-a_j)$ and $b_{ij}=\frac{\prod_{x\neq i}(1-a_xa_j)}{d_j}$.
Let $B$ be the $n\times n$ matrix $B=(b_{ij})$. Does anyone know
how to show that $B$ is an involution ? A proof adding some interpretation to the computations would be better, of course. Perhpas this matrix has a name already ?
 A: Let us denote
$$x_i=a_i^{-1},\quad y_j=a_j,\quad\beta_j=\prod_k (1-a_k a_j),\quad \gamma_i=a_i\prod_{k\neq i}(a_i-a_k),$$
then $B$ can be rewritten as
$$ B_{ij}=
\frac{\prod_{k\neq i}(1-a_k a_j)}{\prod_{k\neq i}(a_i-a_k)}=
\frac{\prod_k (1-a_k a_j)}{a_i\prod_{k\neq i}(a_i-a_k)}\times \frac{1}{a_i^{-1}-a_j}
=\frac{\beta_j}{\gamma_i}\frac{1}{x_i-y_j},$$
Hence up to left and right diagonal factors $B$ is a Cauchy matrix, and therefore it is very easy to compute its inverse (see e.g. (2) here):
\begin{align}B^{-1}_{ij}=&\frac{\beta_j}{\alpha_i}\times \frac{1}{y_i-x_j}\frac{\prod_k(x_j-y_k)\prod_k(y_i-x_k)}{\prod_{k\neq j}(x_j-x_k)
\prod_{k\neq i}(y_i-y_k)}=\\
=&\frac{\gamma_j}{\beta_i}\times \frac{1}{x_j-y_i}\frac{\prod_k(1-a_ja_k)\prod_k(1-a_ka_i)}{a_j^2\prod_{k\neq j}(a_j-a_k)\prod_{k\neq i}(a_i-a_k)}=\\=&
\frac{\gamma_j}{\beta_i}\times \frac{1}{x_j-y_i}\frac{\beta_j\beta_i a_i}{\gamma_j\gamma_i  a_j}=\\
=&\frac{\beta_j}{\gamma_i}\times \frac{a_i}{a_j(x_j-y_i)}=\\
=&\frac{\beta_j}{\gamma_i}\frac{1}{a_i^{-1}-a_j}=\\
=&B_{ij}.\end{align}
A: Here is an interpretation in terms of Lagrange interpolation.
Notice that $$b_{ij} = \frac{1}{a_i^{n-1}} \frac{\prod_{k \neq i} (a_j - a_k^{-1})}{\prod_{k \neq i} (a_i^{-1} - a_k^{-1})},$$ so comparing with the formulas for Langrange interpolation we get the following interpretation for $B$:

Given a vector $y = (y_i)_i$, to compute $By$, find the unique polynomial $F$ of degree at most $n-1$ such that $F(a_i^{-1}) = \frac{y_i}{a_i^{n-1}}$; then $(By)_i = F(a_i)$.

Now, given any polynomial of degree $n-1$ or lower, $F(x)=c_0 + \cdots + c_{n-1}x^{n-1}$, let $F^R(x) = c_n + \cdots + c_0 x^{n-1}$ be its "reversal" (maybe its more correct to call it its $(n-1)$-reversal, in the case $\deg F<n-1$). We have $F^R(x) = x^{n-1}F(x^{-1})$, so we can rephrase the interpretation of $B$ as follows:

Given a vector $y = (y_i)_i$, to compute $By$, find the unique polynomial $G$ of degree at most $n-1$ such that $G(a_i) = y_i$; then $(By)_i = G^R(a_i)$.

This last form makes it clear it is an involution.
