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 ?

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.
