# Proof for the determinant of a Cauchy matrix

I want to proof the formula for the determinant of a Cauchy Matrix without recurring to matrix manipulation, but by directly applying the definition of the determinant. That is, given two sequences of numbers of length n, $$x_1,...,x_n$$ and $$y_1,...,y_n$$, where $$x_i \neq -y_i$$, I want to show that the determinant of the matrix A, whose entries in the ith row and jth column are given by \begin{align} \frac{1}{x_i+y_j} \end{align}, is given by: \begin{align} \det(A) = \frac{\prod_{i Now my proof looks like this: \begin{align} \det(A) &= \sum_{\delta \in perm(N)} sign(\delta)\prod_j \frac{1}{x_{\delta j}+y_j}\\ & = \frac{\sum_{\delta} -sign(\delta)\prod_j x_{\delta j}+y_j}{\prod_\delta \prod_j x_{\delta j} +y_j} \\&= \quad ...\\ &= \frac{\left (\frac {\prod_\delta \prod_j x_{\delta j} +y_j}{\prod_{i,j} x_i+x_j} \right ) \prod_{i Where $$N$$ is the set of the first $$n$$ natural numbers. What I have yet to understand, is the jump from equation 2 to 4. For the case of $$n = 2$$ it obviously holds. For the case of $$n>2$$ my intuition says that for every permutation there is a permutation with opposite sign such that $$\prod_j x_{\delta j}+y_j$$ has $$n-2$$ common factors and that these can be factored out such that they cancel with the $$x_i+x_j$$ appearing more than once in $$\prod_\delta \prod_j x_{\delta j} +y_j$$. Can you help me to formalize that intuition or, if its wrong, lead me towards the right path?

• Yeah, that's quite a jump, though I'm not even sure what those $\prod_\delta$ are supposed to mean (you definitely don't want to multiply over all permutations). I don't know of any direct proof by considering all permutations. I know proofs by induction and LU-decomposition and even by combinatorics (via the RSK algorithm and Lindström-Gessel-Viennot -- it's quite a detour). Feb 8, 2019 at 5:08

Here is a proof that goes along the lines of the OP's argument. The method is for numeric matrices, that is $$x_i,y_j\in\mathbb{C}$$ since it uses simple Calculus methods.
Notice that $$c_n(x_1,\ldots,x_n,y_1,\ldots,y_n):=\operatorname{det}\big(\frac{1}{x_i+y_j}\big)$$ is homogenous of over $$-n$$, that is $$c_n(\lambda x_1,\ldots,\lambda x_n,\lambda y_1,\ldots,\lambda y_n)=\lambda^{-n}c_n(x_1,\ldots, x_n,y_1,\ldots,y_n)$$ For simplicity, set $$\mathbf{x}=(x_1,\ldots, x_n)$$ (similarly for $$\mathbf{y}$$). From the definition of determinant it follows that $$c_n(\mathbf{x},\mathbf{y})=\sum_{\sigma\in S_n}(-1)^{\sigma}\frac{1}{x_1+y_{\sigma(1)}}\cdot\ldots\cdot\frac{1}{x_n+y_{\sigma(n)}}=\frac{P(\mathbf{x},\mathbf{y})}{\prod_{1\leq i,j\leq n}(x_i+y_i)}$$ where $$P$$ is a polynomial on $$\mathbf{x}$$ and $$\mathbf{y}$$ which is homogeneous of order $$n^2-n$$ (observe that $$\prod_{1\leq i,j\leq n}(x_i+y_i)$$ is the common denominator of all the rational expressions in the sum that defines $$c$$). Notice that if $$x_\ell=x_k$$ (or $$y_\ell=y_k$$) for some $$\ell, then the matrix $$C_n=\big(\frac{1}{x_i+y_j}\big)$$ would have rows (resp. columns) $$\ell$$ and $$k$$ identical. It follows that $$c_n(\mathbf{x},\mathbf{y})=k_n\frac{\prod_{1\leq i for some constant $$k_n$$.
I tried to find a particular choice of $$\mathbf{x}$$ and $$\mathbf{y}$$ that yields a matrix which a determinant easy to compute and which allow us to estimate $$k_n$$, but did not go very far. Then I turn to some simple Calculus: the function $$\mathbf{x}\mapsto x_1 c(\mathbf{x},\mathbf{y})$$ is the determinant of the matrix obtained from $$C$$ by multiplying the first row of $$C$$ by $$x_1$$ and leaving the other ones the same. The continuity of the determinant function yields \begin{align} \lim_{y_1\rightarrow\infty}\Big(\lim_{x_1\rightarrow\infty}x_1c_n(\mathbf{x},\mathbf{y})\Big)&=c_{n-1}(x_2,\ldots,x_n,y_2,\ldots,y_n)\\ &=k_{n-1}\frac{\prod_{2\leq i On the other hand, \begin{align} \lim_{y\rightarrow\infty}\Big(\lim_{x_1\rightarrow\infty} x_1c_n(\mathbf{x},\mathbf{y})\Big)&=\lim_{y_1\rightarrow\infty}\Big(\lim_{x\rightarrow\infty}k_n\frac{x_1\prod_{1\leq i Putting things together, we have that $$k_n=k_{n-1}$$ Clearly $$k_1=1$$. The desired formula follows.