# Cauchy Determinant Formula

## Main Result

This note is about **Cauchy matrices** of the form
$$
M = M(\mathbf{a}, \mathbf{b}) = \left( \frac{1}{a_i - b_j} \right)_{1 \leq i, j \leq n}
$$
where $\mathbf{a}, \mathbf{b} \in \mathbb{C}^n$. We assume throughout that the matrix is well-defined, in the sense that $a_i \neq b_j$ for all $i,j$. The Cauchy determinant formula says that
$$
\det M = \frac{\prod_{i > j} (a_i - b_j)(b_j - a_i)}{\prod_{i,j} (a_i - b_j)}.
$$
This note explains the argument behind this result, as given in the paper On the Inversion of Certain Matrices by Samuel Schechter. Some of the argument is already on the Wikipedia page for Cauchy matrices. Schechter’s argument also contains some facts which are useful elsewhere:

- there is a formula for $M^{-1}$ involving the Lagrange interpolation polynomials for $\mathbf{a}$ and $\mathbf{b}$
- in fact $M^{-1}$ can be written as $$ M^{-1} = D_r M^{T} D_c $$ where $M^T$ is the transpose of $M$ and $D_r, D_c$ are the diagonal matrices whose entries are the row and column sums of $M^{-1}$ (respectively). These row and column sums can also be nicely expressed in terms of the Lagrange interpolation polynomials.

We also note that neither the results nor the arguments rely on $\mathbf{a}, \mathbf{b}$ being complex. In general they can come from any field.

## Lagrange Interpolation Polynomials

Recall that for $a_1, \ldots, a_n \in \mathbb{C}$ all distinct we form the Lagrange interpolation polynomials (in barycentric form) by defining
$$
A(z) = \prod_{i} (z - a_i), \quad A_i(z) = \frac{A(z)}{A'(a_i) (z - a_i)}.
$$
The polynomials $A_i(z)$ are of degree $n-1$ (since $A$ is of degree $n$) and satisfy
$$
\begin{equation}\label{eq:A_identity}
A_i(a_j) = \delta_{ij} = \mathbf{1} \{ i = j \}.
\end{equation}
$$
Thus these $n$ polynomials are linearly independent, and so form a basis of the $n$-dimensional space of degree $n-1$ polynomials. Thus for *any* polynomial $p$ of degree $n-1$ with coefficients in $\mathbb{C}$ one has
$$
\begin{equation}
p(z) = \sum_{i=1}^n p(a_i) A_i(z). \label{eq:p_general}
\end{equation}
$$
Of course both $\eqref{eq:A_identity}$ and $\eqref{eq:p_general}$ hold for $B(z)$ and $B_i(z)$ defined by
$$
B(z) = \prod_i (z - b_i), \quad B_i(z) = \frac{B(z)}{B'(b_i)(z - b_i)}.
$$

## Formulas for the Inverse

The quantities $(a_i - b_j)^{-1}$ come in by observing that from $\eqref{eq:p_general}$ (more precisely from the version of it for $B$) we have $$ \frac{p(z)}{B(z)} = \sum_{j=1}^n \frac{p(b_j)}{B'(b_j)} \frac{1}{z - b_j}. $$ This holds for all $p$ of degree $n-1$, in particular for $A_1, \ldots, A_n$, so $$ \frac{A_k(z)}{B(z)} = \sum_{j=1}^n \frac{A_k(b_j)}{B'(b_j)} \frac{1}{z - b_j}, ,, k=1,\ldots,n. $$ Now insert $z = a_i$ into the above to obtain $$ \delta_{ik} = B(a_k) \sum_{j=1}^n \frac{A_k(b_j)}{B'(b_j)} \frac{1}{a_i - b_j}, $$ since we could replace $B(a_i)$ with $B(a_k)$ because of the $\delta_{ik}$ term. This holds for $1 \leq i,k \leq n$ so these equations precisely say that $M^{-1}$ has entries $$ (M^{-1})_{jk} = \frac{A_k(b_j)}{B'(b_j)} B(a_k) = A_k(b_j) B_j(a_k) (a_k - b_j). $$ The last equality is just the definition of $B_j$. It is also useful to use the definition of $A_k$ and $B_j$ to rewrite the above as $$ \begin{equation}\label{eq:M_inv_entries} (M^{-1})_{jk} = \frac{A(b_j)}{A'(a_k)(b_j - a_k)} \frac{B(a_k)}{B'(b_j)(a_k - b_j)} (a_k - b_j) = - \frac{1}{a_k - b_j} \frac{A(b_j)}{B'(b_j)} \frac{B(a_k)}{A'(a_k)}. \end{equation} $$ In fact this already shows that $M^{-1}$ has the form $$ \begin{equation}\label{eq:inv_trans_relation} M^{-1} = D_1 M^{T} D_2, \end{equation} $$ where $D_1$ and $D_2$ are the diagonal matrices with respective diagonal entries $$ -\frac{A(b_i)}{B'(b_i)}, \quad \frac{B(a_i)}{A'(a_i)}, \quad 1 \leq i \leq n. $$ The last two equations already tell us the determinant of the Cauchy matrix, namely that $$ \det(M)^2 = \det(D_1)^{-1} \det(D_2)^{-1} = (-1)^n \prod_{i=1}^n \frac{A'(a_i)}{B(a_i)} \frac{B'(b_i)}{A(b_i)}. $$ The definitions of $A$ and $B$ imply that $$ \prod_{i=1}^n B(a_i) A(b_i) = \prod_{i=1}^n \prod_{j=1}^n (a_i - b_j) (b_i - a_j) = (-1)^{n^2} \prod_{i,j} (a_i - b_j)^2. $$ For the other terms we have $$ \prod_{i=1}^n A'(a_i) = \prod_{i=1}^n \prod_{k \neq i} (a_i - a_k) = (-1)^{n(n-1)/2} \prod_{i > k} (a_i - a_k)^2, $$ with a similar formula for the product of $B'(b_i)$. Consequently the negatives all cancel, leaving us with $$ \det(M)^2 = \frac{\prod_{i > k} (a_i - a_k)^2 (b_k - b_i)^2}{\prod_{i,j} (a_i - b_j)^2}. $$ It remains to be checked that the positive square root of the right hand side is the correct choice. Simply check it for $n=2,3,\ldots$ and proceed by induction.

## Row and Column Sums of $M^{-1}$

Equation $\eqref{eq:M_inv_entries}$ for the entries of $M^{-1}$ also gives a nice way of computing the row and column sums of $M^{-1}$. The formulas are $$ \sum_{k=1}^n (M^{-1})_{jk} = -\frac{A(b_j)}{B'(b_j)}, \quad \sum_{j=1}^n (M^{-1})_{jk} = \frac{B(a_j)}{A'(a_j)}. $$ Here is the calculation for the first formula (the one for the second is similar): by $\eqref{eq:M_inv_entries}$ $$ \sum_{k=1}^n (M^{-1})_{jk} = -\frac{A(b_j)}{B'(b_j)} \sum_{k=1}^n \frac{B(a_k)}{(a_k - b_j)A'(a_k)}. $$ We claim that the summation term is identically one. To see this first observe that $$ \frac{B(a_k)}{(a_k - b_j) A'(a_k)} = \operatorname{Res}\left( \frac{B(z)}{(z - b_j)A(z)} ; a_k \right). $$ The claim that the summation is one follows because $A(z)/B(z) \sim 1$ along a sufficiently large contour $C$ around the origin (recall that both $A$ and $B$ are monic of the same degree), and therefore $$ 1 = \frac{1}{2 \pi i} \oint_C \frac{B(z)}{(z - b_j)A(z)} = \sum_{k=1}^n \operatorname{Res}\left( \frac{B(z)}{(z - b_j)A(z)} ; a_k \right). $$