# Inverse of a Cauchy-like matrix

Consider $$n\times n$$ symmetric Cauchy-like matrix $$M$$ with elements $$(M_{ij})_{i,j=1}^{n}$$ given by

$$M_{ij} = \frac{1}{(n-i)!(n-j)!(2n-i-j+1)} = \displaystyle\int_{0}^{1}\frac{x^{n-i}}{(n-i)!} \frac{x^{n-j}}{(n-j)!}\:{\rm{d}}x.$$

Is there a way to compute the elements of the inverse $$(M^{-1})_{ij}$$ analytically?

• I don't have a proof, but empirically it seems that $\det(M) = 1/a(n)$ where $a$ is OEIS sequence A107254, and that the matrix elements of $M^{-1}$ are all integers. – Robert Israel Aug 26 '20 at 2:49
• @RobertIsrael: I can prove the determinant formula you wrote. I am not sure if computing the adjugate is the cleanest way to proceed for inverse. – Abhishek Halder Aug 26 '20 at 4:13

I was able to figure this out by viewing $$M$$ as a scaled Cauchy matrix.
Theorem. $$\left(M^{-1}\right)_{ij} = \dfrac{(n-i)!(n-j)!}{2n-i-j+1}\dfrac{\displaystyle\prod_{r=1}^{n}(2n-i-r+1)(2n-j-r+1)}{\left(\displaystyle\prod_{\stackrel{r=1}{r\neq i}}^{n}(r-i)\right) \left(\displaystyle\prod_{\stackrel{r=1}{r\neq j}}^{n}(r-j)\right)}$$.
Proof. Define $$n\times 1$$ vector $$\alpha$$ with elements $$(\alpha)_{i} = 1/(n-i)!$$. Then $$M = \text{diag}(\alpha) N \text{diag}(\alpha)$$, where $$N_{ij} := 1/(2n-i-j+1)$$. Hence $$\left(M^{-1}\right)_{ij} = (n-i)!(n-j)!\left(N^{-1}\right)_{ij}$$.
Now, write $$N$$ as a Cauchy matrix: $$N_{ij} = 1/(a_i + b_j)$$ where $$a_i := n-i$$, $$b_j := n-j+1$$. Then using the known result [1, Sec. 1.2.3, Exercise 41] for Cauchy matrix inverse: $$\left(N^{-1}\right)_{ij} = \dfrac{1}{2n-i-j+1}\dfrac{\displaystyle\prod_{r=1}^{n}(2n-i-r+1)(2n-j-r+1)}{\left(\displaystyle\prod_{\stackrel{r=1}{r\neq i}}^{n}(r-i)\right) \left(\displaystyle\prod_{\stackrel{r=1}{r\neq j}}^{n}(r-j)\right)},$$ the result follows.