Proof for the determinant of a Cauchy matrix

Question

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<j} (x_i-x_j)(y_i-y_j)}{\prod_{i,j} x_i+x_j}
\end{align}
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<j} (x_i-x_j)(y_i-y_j) }{\prod_\delta \prod_j x_{\delta j} +y_j}\\&=\frac{\prod_{i<j} (x_i-x_j)(y_i-y_j)}{\prod_{i,j} x_i+x_j}
\end{align}
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?

Answer 1

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 $\ell<k$, 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 $$c_n(\mathbf{x},\mathbf{y})=k_n\frac{\prod_{1\leq i<j\leq n}(x_i-x_j)(y_i-y_j)}{\prod_{1\leq i,j\leq n}(x_i+y_j)}$$ 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<j\leq n}(x_i-x_j)(y_i-y_j)}{\prod_{2\leq i,j\leq n}(x_i+y_j)} \end{align} 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<j\leq n}(x_i-x_j)(y_i-y_j)}{\prod_{1\leq i,j\leq n}(x_i+y_j)}\Big)\\ &=k_n\frac{\prod_{2\leq i<j\leq n}(x_i-x_j)(y_i-y_j)}{\prod_{2\leq i,j\leq n}(x_i+y_j)} \end{align} Putting things together, we have that $k_n=k_{n-1}$ Clearly $k_1=1$. The desired formula follows.