History of the Cauchy matrix

Question

Cauchy matrix $C$ is defined by $$C_{i,j}=\frac{1}{a_i + b_j}$$ where $a_i$ and $b_j$ are any numbers so long $a_i + b_j \neq 0$. Why did Cauchy introduce this matrix? Did he use it in the context of another problem or application?

Answer 1

It (or, rather, its determinant) appears in the second volume of Cauchy's "Exercices d'analyse et de physique mathematique", as part of "Memoir on alternating functions and alternating sums" (https://books.google.se/books?id=DRg-AQAAIAAJ, pages 151-159, see particularly formula (10).) Reading just that section, the calculation of the determinant is presented simply "to present an application of formula (4)" (on the general decomposition of alternating sum obtained from a rational function).

Here is a summary, trying to stay reasonably close to the original:

Roughly, Cauchy defines alternating multivariable functions (the usual way), and observes that any such vanishes whenever one sets any two inputs equal to each other; thus if such a function is "entire" it must be algebraically divisible by (linear polynomial which is) the difference of any two input variables, and hence by product of these differences, $P$. In a separate observation, an alternating rational function which has symmetric nominator must have alternating denominator.

Now, given any function $f$ one can define an alternating function $s$ by taking signed sum of values of $f$ on all permutations of inputs (this is the modern antisymmetrizer, but without the $n!$ normalization). If $f$ was entire, then so is $s$; hence $s$ is divisible by $P$. If $f$ is rational, then so is $s$, and in $s=\frac{U}{V}$ one can take for $V$ the product of all denominators which appear in the alternating sum defining $s$ -- or any symmetric function divisible by this product -- thereby making $U$ alternating, and thus divisible by $P$, so $U=PW$. Hence $s=P \frac{W}{V}$. Here $W$ and $V$ are symmetric entire functions.

Now one applies this to $f(x, y, \ldots, )=\frac{1}{(x-a)(y-b)\ldots }$. Observe that we can take for $V$ the product of all the differences $x-a$, $x-b$, $\ldots$, $y-a$, $y-b$, $\ldots$ $\ldots$. Observe that $U$ will be divisible not only by $P$ but also by $\mathscr{P}$ which is a product of all $(a-b)$, $(a-c) \ldots $, $(b-c)\ldots $. Overall, $s=k\frac{P \mathscr{P}}{V}$. Counting total degree (in all variables, $x,y, \ldots, a, b, \ldots$) one sees that degree of $k$ is zero, so it's a constant, and, clearing $s=k\frac{P \mathscr{P}}{V}$ of denominators and plugging in $x=a$, $y=b, \ldots$ we get $n$ choose $2$ minus signs, so $k=(-1)^{n(n-1)/2}$, and $s=(-1)^{n(n-1)/2} \frac{P \mathscr{P}}{V}$. (Note that this is precisely what we would these days call the determinant of the Cauchy matrix).