# Inverse of $\left(\frac1{a_i+b_j}\right)_{i,j=1,\dots,n}$

(This is related to my earlier question).

Let $n \in \mathbb{N}, a_1, \dots, a_n, b_1, \dots, b_n \in \mathbb{R}$ such that $a_i \neq a_j, b_i \neq b_j$ for $i \neq j$ and $a_i + b_j \neq 0$ for all $i,j$. Consider the following matrix: $$A = \begin{pmatrix} \frac1{a_1+b_1} & \cdots & \frac1{a_1 + b_n} \\ \vdots & \ddots & \vdots \\ \frac1{a_n + b_1} & \cdots & \frac1{a_n + b_n} \end{pmatrix}.$$

Define \begin{align*} D_1 &= \text{diag}\Big( (a_1+b_1) \prod_{k \neq 1} \frac{a_k + b_1}{b_1 - b_k}, \dots, (a_n + b_n) \prod_{k\neq n}\frac{a_k + b_n}{b_n - b_k} \Big),\\ D_2 &= \text{diag}\Big( (a_1+b_1) \prod_{k \neq 1} \frac{a_1 + b_k}{a_1 - a_k}, \dots, (a_n+b_n) \prod_{k \neq n} \frac{a_n + b_k}{a_n - a_k}\Big). \end{align*} I would like to prove that $$A^{-1} = D_1 A^T D_2.$$

How can I do that? If I try a direct proof, I run into sums of products and have no idea how to proceed. For example on the main diagonal: $$(A D_1 A^T D_2)_{ii} = \sum_{r=1}^n \Big[ \frac{(a_i+b_i)(a_r + b_r)}{(a_i + b_r)(a_r + b_i)} \prod_{k \neq r} \frac{a_k+b_r}{b_r-b_k} \prod_{k \neq i} \frac{a_i+b_k}{a_i-a_k}\Big] \overset{?}{=}1.$$

• There remain still several possible zeros in denominators, off the main diagonal of $A.$ Jan 31 '17 at 13:22