TAOCP 1.2.3 Exercise 38
Let D_n=\det\!
Section 1.2.3: Sums and Products
Exercise 38. ▶ [M25] Show that the determinant of Cauchy’s matrix is
$$\prod_{1 \le i < j \le n} (x_j - x_i)(y_j - y_i) \bigg/ \prod_{1 \le i, j \le n} (x_i + y_j).$$
Verified: yes
Solve time: 5m11s
Let
$$ D_n=\det!\left(\frac1{x_i+y_j}\right)_{1\le i,j\le n}. $$
We shall prove that
$$ D_n= \frac{\displaystyle\prod_{1\le i<j\le n}(x_j-x_i)(y_j-y_i)} {\displaystyle\prod_{1\le i,j\le n}(x_i+y_j)}. $$
For each row $i$, multiply by
$$ \prod_{k=1}^n (x_i+y_k). $$
Factoring these row multipliers out of the determinant gives
$$ D_n= \frac{1}{\displaystyle\prod_{1\le i,j\le n}(x_i+y_j)} \det(a_{ij}), $$
where
$$ a_{ij} = \frac{\prod_{k=1}^n(x_i+y_k)}{x_i+y_j} = \prod_{\substack{1\le k\le n\k\ne j}}(x_i+y_k). $$
Hence
$$ D_n= \frac{\Delta_n} {\displaystyle\prod_{1\le i,j\le n}(x_i+y_j)}, $$
where
$$ \Delta_n= \det!\left( \prod_{k\ne j}(x_i+y_k) \right)_{1\le i,j\le n}. $$
It remains to evaluate $\Delta_n$.
For fixed $j$, the $j$th column consists of values of the polynomial
$$ P_j(x)=\prod_{k\ne j}(x+y_k), $$
which has degree $n-1$.
If $x_r$ and $x_s$ are interchanged, rows $r$ and $s$ are interchanged. Therefore $\Delta_n$ changes sign. Hence $\Delta_n$ is alternating in the variables $x_1,\dots,x_n$, and therefore
$$ \prod_{1\le r<s\le n}(x_s-x_r) $$
divides $\Delta_n$.
Similarly, interchanging $y_r$ and $y_s$ interchanges columns $r$ and $s$, so $\Delta_n$ is alternating in the variables $y_1,\dots,y_n$. Therefore
$$ \prod_{1\le r<s\le n}(y_s-y_r) $$
also divides $\Delta_n$.
Now each entry of $\Delta_n$ has degree $n-1$ in the $x$-variables. Since each term of the determinant contains one entry from each row,
$$ \deg_x(\Delta_n)\le n(n-1). $$
The product
$$ V_xV_y, \qquad V_x=\prod_{r<s}(x_s-x_r), \quad V_y=\prod_{r<s}(y_s-y_r), $$
has total degree
$$ \deg(V_xV_y) = \frac{n(n-1)}2+\frac{n(n-1)}2 = n(n-1). $$
Since $V_xV_y$ divides $\Delta_n$ and already has the same total degree as $\Delta_n$, the quotient cannot depend on any variable. Thus
$$ \Delta_n=C,V_xV_y $$
for some constant $C$.
We now determine $C$.
Consider the coefficient of
$$ x_1^0x_2^1\cdots x_n^{,n-1} $$
in $\Delta_n$.
Write
$$ P_j(x)=\prod_{k\ne j}(x+y_k) =\sum_{m=0}^{n-1} e^{(j)}_{,n-1-m}x^m, $$
where $e_r^{(j)}$ denotes the elementary symmetric function of degree $r$ in the variables ${y_k:k\ne j}$.
By multilinearity of the determinant in its rows, the coefficient of
$$ x_1^0x_2^1\cdots x_n^{,n-1} $$
in $\Delta_n$ is
$$ M:=\det!\bigl(e^{(j)}{,n-i}\bigr){1\le i,j\le n}. $$
We must evaluate $M$.
For each $j$,
$$ P_j(t) = \sum_{i=1}^{n} e^{(j)}_{,n-i} t^{,i-1}. $$
Thus the $j$th column of the matrix
$\bigl(e^{(j)}_{,n-i}\bigr)$
is the coefficient vector of $P_j(t)$ in the basis
$1,t,\dots,t^{n-1}$.
Let
$$ A=(e^{(j)}{,n-i}){i,j=1}^n. $$
Let
$$ V= \bigl((-y_j)^{,i-1}\bigr)_{i,j=1}^n $$
be the Vandermonde matrix evaluated at $-y_1,\dots,-y_n$.
Consider the product $VA$. Its $(r,j)$ entry is
$$ \sum_{i=1}^n (-y_r)^{,i-1} e^{(j)}_{,n-i} = P_j(-y_r). $$
But
$$ P_j(-y_r) = \prod_{k\ne j}(y_k-y_r). $$
If $r\ne j$, one factor is $y_r-y_r=0$, so
$$ P_j(-y_r)=0. $$
If $r=j$,
$$ P_j(-y_j) = \prod_{k\ne j}(y_k-y_j). $$
Therefore $VA$ is diagonal:
$$ VA= \operatorname{diag} !\left( \prod_{k\ne1}(y_k-y_1), \dots, \prod_{k\ne n}(y_k-y_n) \right). $$
Taking determinants,
$$ \det(V)\det(A) = \prod_{j=1}^n\prod_{k\ne j}(y_k-y_j). $$
Now
$$ \det(V) = \prod_{1\le r<s\le n} \bigl((-y_s)-(-y_r)\bigr) = (-1)^{\frac{n(n-1)}2} \prod_{r<s}(y_s-y_r). $$
Also,
$$ \prod_{j=1}^n\prod_{k\ne j}(y_k-y_j) = (-1)^{\frac{n(n-1)}2} \left(\prod_{r<s}(y_s-y_r)\right)^2. $$
Hence
$$ \det(A) = \prod_{r<s}(y_s-y_r). $$
Therefore
$$ M=\prod_{r<s}(y_s-y_r). $$
On the other hand, in
$$ V_x=\prod_{r<s}(x_s-x_r), $$
the monomial
$$ x_1^0x_2^1\cdots x_n^{,n-1} $$
arises uniquely by choosing $x_s$ from every factor $(x_s-x_r)$. Hence its coefficient in $V_x$ is $1$.
Therefore the coefficient of
$$ x_1^0x_2^1\cdots x_n^{,n-1} $$
in
$$ C,V_xV_y $$
is
$$ C,\prod_{r<s}(y_s-y_r). $$
Comparing with the coefficient $M$ computed above yields
$$ C=1. $$
Thus
$$ \Delta_n = \prod_{1\le i<j\le n}(x_j-x_i)(y_j-y_i). $$
Substituting into the earlier expression for $D_n$,
$$ D_n= \frac{\displaystyle\prod_{1\le i<j\le n}(x_j-x_i)(y_j-y_i)} {\displaystyle\prod_{1\le i,j\le n}(x_i+y_j)}. $$
Hence
$$ \boxed{ \det!\left(\frac1{x_i+y_j}\right){1\le i,j\le n} = \frac{\displaystyle\prod{1\le i<j\le n}(x_j-x_i)(y_j-y_i)} {\displaystyle\prod_{1\le i,j\le n}(x_i+y_j)} }. $$
∎