Determinant of a Cauchy matrix

Throughout this posting, the expressions $$ \sum_{\mathit{vars} \mathop{:} \mathit{conds}} f(\mathit{vars}) \qquad\text{and}\qquad \prod_{\mathit{vars} \mathop{:} \mathit{conds}} f(\mathit{vars}) $$ will refer to the sum and product of $f(\mathit{vars})$ over all possible values of $\mathit{vars}$ subject to the constraints $\mathit{conds}$, respectively.

Theorem. (Determinant of a Cauchy matrix) Let $A$ be an $n\times n$ matrix with elements $a_{ij}$ in the form $$ a_{ij} = \frac{1}{x_i + y_j}. $$ (Think of $A$ as an matrix over the field of rational functions in variables $x_i$'s and $y_j$'s.) Then $$ \det A = \frac{\prod_{i, j \mathbin{:} i \lt j} (x_i - x_j)(y_i - y_j)}{\prod_{i, j} (x_i + y_j)}. $$

Proof. We prove the claim by induction on $n$. The base case is clearly true, so it suffices to establish the inductive step.

Suppose that the claim holds for all $(n-1)\times(n-1)$ matrices. Also, recall that if $p(z)$ is a polynomial of degree less than $n$ and if $\alpha_1, \ldots, \alpha_n$ are distinct, then we have the partial fraction decomposition $$ \frac{p(z)}{\prod_{j} (z - \alpha_j)} = \sum_{k} \frac{p(\alpha_k)}{(z - \alpha_k) \prod_{j \mathbin{:} j \neq k} (\alpha_k - \alpha_j)}. $$ So, by regarding $x_n$ as the variable and applying the partial fraction decomposition, \begin{align*} &\frac{\prod_{i \mathbin{:} i \lt n} (x_i - x_n)}{\prod_{j} (x_n + y_j)} \\ &= \sum_{k} \frac{\prod_{i \mathbin{:} i \lt n} (x_i + y_k)}{(x_n + y_k) \prod_{j \mathbin{:} j \neq k} (y_j - y_k)} \\ &= \sum_{k} (-1)^{n-k} \frac{\prod_{i \mathbin{:} i \lt n} (x_i + y_k)}{(x_n + y_k) \bigl[ \prod_{j \mathbin{:} j \lt k} (y_j - y_k) \bigr]\bigl[ \prod_{j \mathbin{:} k \lt j} (y_k - y_j) \bigr]} \end{align*} From this, it follows that \begin{align*} &\frac{\prod_{i, j \mathbin{:} i \lt j} (x_i - x_j)(y_i - y_j)}{\prod_{i, j} (x_i + y_j)} \\ &= \frac{\bigl[ \prod_{i, j \mathbin{:} i \lt j \lt n} (x_i - x_j) \bigr]\bigl[ \prod_{i, j \mathbin{:} i \lt j} (y_i - y_j) \bigr]}{\prod_{i, j \mathbin{:} i \lt n} (x_i + y_j)} \cdot \frac{\prod_{i \mathbin{:} i \lt n} (x_i - x_n)}{\prod_{j} (x_n + y_j)} \\ &= \sum_{k} \frac{(-1)^{n-k}}{x_n + y_k} \cdot \frac{\bigl[ \prod_{i, j \mathbin{:} i \lt j \lt n} (x_i - x_j) \bigr]\bigl[ \prod_{i, j \mathbin{:} i \lt j; i \neq k; j \neq k} (y_i - y_j) \bigr]}{\prod_{i, j \mathbin{:} i \lt n; j\neq k} (x_i + y_j)} \end{align*} However, the last line is precisely the cofactor expansion of $\det A$ along the $n$th row, where the determinants of the submatrices are computed using the inductive hypothesis. Therefore the inductive step is established and the proof is done. $\square$