Determinants and permanents of an arbitrary Hadamard degree of a Cauchy matrix and a proof of a generalization of a conjecture of R.F.Scott (1881)

Abstract

In this paper we give the absolutely new proof of a conjecture of R.F.Scott(1881) on the permanent of a Cauchy matrix $\ls \frac{1}{x_i-y_j} \rs_{1 \leqslant i,j \leqslant n},$ where $x_1, ..., x_n$ and $y_1, ..., y_n$ are the distinct roots of the polynomials $x^n-1$ and $y^n +1,$ respectively. The simple formula is given for the permanent of the Cauchy matrix $A= \ls \frac{1}{x_i-y_j} \rs_{1 \leqslant i,j \leqslant n},$ where $x_1, ..., x_n$ and $y_1, ..., y_n$ are the distinct roots of the polynomials $x^n+a$ and $y^n +b$, respectively: \begin{gather*} \per (A) =\frac{n}{(b-a)^n} \prod_{k=1}^{n-1}[nb-k(b-a)] = =\begin{cases} %\begin{eqnarray} (-1)^{\frac{n-1}{2}} \cfrac{n}{(b-a)^n} \prod\limits_{k=1}^{\frac{n-1}{2}}[-na-k(b-a)][nb-k(b-a)], \mbox{if $n \equiv 1 (\mod 2)$,} \cfrac{n}{2} \cdot \cfrac{n(a+b)}{(b-a)^n} \prod\limits_{k=1}^{\frac{n}{2}-1}[na+k(b-a)][nb+k(a-b)], \mbox{if $n\equiv 0(\mod 2)$}. %\end{eqnarray} \end{cases} \end{gather*} from which the corrected formula of R.F.Scott follows instantly. Proof follows from obtained by the author a formula for the determinant of an arbitrary of the Hadamard degree $m$ of a Cauchy matrix $A$ and Borchard's theorem.