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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 ?

Ewan Delanoy
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2 Answers2

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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}

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  • I find your answer quite satisfying, but I wonder if there is a deeper explanation of why $B$ is an involution. The fact that the computation of $B^{-1}$ eventually leads to $B$ looks a little like magic here – Ewan Delanoy Mar 19 '15 at 12:25
  • @EwanDelanoy I agree. My feeling is that this shoud somehow be related to Lagrange interpolation. – Start wearing purple Mar 19 '15 at 12:26
  • Yes, the $\prod_{j\neq i}(a_j-a_i)$ reminds one of Lagrange interpolation, but on the other hand the $1-a_ia_j$ reminds me of Householder transformations – Ewan Delanoy Mar 19 '15 at 12:29
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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.

Omar Antolín-Camarena
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