Inverse of Cauchy Matrix

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Theorem

Let $C_n$ be the square Cauchy matrix of order $n$:

$C_n = \begin{bmatrix} \dfrac 1 {x_1 + y_1} & \dfrac 1 {x_1 + y_2} & \cdots & \dfrac 1 {x_1 + y_n} \\ \dfrac 1 {x_2 + y_1} & \dfrac 1 {x_2 + y_2} & \cdots & \dfrac 1 {x_2 + y_n} \\ \vdots & \vdots & \ddots & \vdots \\ \dfrac 1 {x_n + y_1} & \dfrac 1 {x_n + y_2} & \cdots & \dfrac 1 {x_n + y_n} \\ \end{bmatrix}$


Then its inverse $C_n^{-1} = \sqbrk b_n$ can be specified as:

$\begin{bmatrix} b_{ij} \end{bmatrix} = \begin{bmatrix} \dfrac {\ds \prod_{k \mathop = 1}^n \paren {x_j + y_k} \paren {x_k + y_i} } {\ds \paren {x_j + y_i} \paren {\prod_{\substack {1 \mathop \le k \mathop \le n \\ k \mathop \ne j} } \paren {x_j - x_k} } \paren {\prod_{\substack {1 \mathop \le k \mathop \le n \\ k \mathop \ne i} } \paren {y_i - y_k} } } \end{bmatrix}$


Proof

Preliminaries:

Vandermonde Matrix Identity for Cauchy Matrix supplies matrix equation

$(1): \quad -C = PV_x^{-1} V_y Q^{-1}$
Definitions of symbols:
$V_x = \paren {\begin{smallmatrix} 1 & 1 & \cdots & 1 \\ x_1 & x_2 & \cdots & x_n \\ \vdots & \vdots & \ddots & \vdots \\ x_1^{n-1} & x_2^{n-1} & \cdots & x_n^{n-1} \\ \end{smallmatrix} },\quad V_y = \paren {\begin{smallmatrix} 1 & 1 & \cdots & 1 \\ y_1 & y_2 & \cdots & y_n \\ \vdots & \vdots & \ddots & \vdots \\ y_1^{n-1} & y_2^{n-1} & \cdots & y_n^{n-1} \\ \end{smallmatrix} }$ Vandermonde matrices
$P = \paren {\begin{smallmatrix} p_1(x_1) & \cdots & 0 \\ \vdots & \ddots & \vdots \\ 0 & \cdots & p_n(x_n) \\ \end{smallmatrix} }, \quad Q = \paren {\begin{smallmatrix} p(y_1) & \cdots & 0 \\ \vdots & \ddots & \vdots \\ 0 & \cdots & p(y_n) \\ \end{smallmatrix} }$ Diagonal matrices
$\ds \map p x = \prod_{i \mathop = 1}^n \paren {x - x_i}, \quad \map {p_k} x = \prod_{i \mathop = 1, i \mathop \ne k}^n \, \paren {x - x_i}, \quad 1 \mathop \le k \mathop \le n$ Polynomials

Compute the inverse $C^{-1}$ for:

\(\ds C\) \(=\) \(\ds \begin{bmatrix} \dfrac 1 {x_1 - y_1} & \dfrac 1 {x_1 - y_2} & \cdots & \dfrac 1 {x_1 - y_n} \\ \dfrac 1 {x_2 - y_1} & \dfrac 1 {x_2 - y_2} & \cdots & \dfrac 1 {x_2 - y_n} \\ \vdots & \vdots & \ddots & \vdots \\ \dfrac 1 {x_n - y_1} & \dfrac 1 {x_n - y_2} & \cdots & \dfrac 1 {x_n - y_n} \\ \end{bmatrix}\) Assume $\set {x_1,\ldots,x_n,y_1,\ldots,y_n}$ has distinct elements.

Replacement $y_k \to -y_k$ then gives the inverse $C_n^{-1}$ in the theorem.

Inverse of Matrix Product applied to equation (1) gives:

$ C^{-1} = -Q V_y^{-1} V_x P^{-1}$

Let ${\vec K}_1,\ldots,{\vec K}_n$ denote the columns of the $n\times n$ identity matrix.

Then $n\times n$ matrix $B = C^{-1}$ has entries $b_{ij} = {\vec K}_i^T C^{-1} {\vec K}_j$.

Hold fixed until the end of the proof the row and column index symbols $i$ and $j$.

Define column vectors $\vec A$, $\vec B$ so that $b_{ij} = {\vec A}^T \vec B$:

$\vec A = \paren { {\vec K}_i^T \, Q \, V_y^{-1} }^T, \quad \vec B = -V_x \, P^{-1} \, {\vec K}_j$

Define $u = x_j$ and simplify:

\(\ds \vec A\) \(=\) \(\ds {\map p {y_i} } \, \paren { V_y^{-1} }^T \, {\vec K}_i\) Transpose of Matrix Product
\(\ds \) \(=\) \(\ds \dfrac{ \map p {y_i} }{\det \paren {V_y} } \paren { \adj {V_y} }^T {\vec K}_i\) Matrix Product with Adjugate Matrix
\(\ds \) \(=\) \(\ds \dfrac{ \map p {y_i} }{\det \paren {V_y} } \begin{bmatrix} {\mathbf {Cofactor} } \paren {V_y,1,i } \\ \vdots \\ {\mathbf {Cofactor} } \paren {V_y,n,i } \\ \end{bmatrix}\) ${\mathbf {Cofactor} } \paren {M,r,s}$ denotes cofactor $M_{rs}$
\(\ds \vec B\) \(=\) \(\ds -\dfrac 1 {\map {p_j} {x_j} } \, V_x \, {\vec K}_j\)
\(\ds \) \(=\) \(\ds -\dfrac 1 {\map {p_j} {x_j} } \, \begin{bmatrix} 1 \\ u \\ \vdots \\ u^{n-1} \\ \end{bmatrix}\) Symbol $u = x_j$.

Then:

\(\ds A^T\, B\) \(=\) \(\ds \dfrac {-\map p {y_i} } {\map {p_j} {x_j} \det \paren {V_y} } \sum_{k \mathop = 1}^n {\mathbf {Cofactor} } \paren {V_y, k,i } \, u^{k - 1}\) Cofactor Expansion along column $i$.
\(\ds \) \(=\) \(\ds \dfrac {-\map p {y_i} } {\map {p_j} {x_j} \det \paren {V_y} } \det \paren {\begin {matrix} 1 & \cdots & 1 & 1 & 1 & \cdots & 1 \\ y_1 & \cdots & y_{i-1} & u & y_{i+1} & \cdots & y_n \\ \vdots & \cdots & \vdots &\vdots & \vdots & \cdots & \vdots \\ y_1^{n-1} & \cdots & y_{i-1}^{n-1} & u^{n-1} & y_{i+1}^{n-1} & \cdots & y_n^{n-1} \\ \end {matrix} }\)
\(\ds \) \(=\) \(\ds \dfrac {-\map p {y_i} } {\map {p_j} {x_j} } \ \dfrac {\bigvalueat {\paren {\det \paren {V_y} } } {y_i \mathop = u} } {\det \paren {V_y} }\)

Simplify the fraction on the right:

\(\text {(2)}: \quad\) \(\ds \dfrac {\bigvalueat {\paren {\det \paren {V_y} } } {y_i \mathop = u} } {\det \paren {V_y} }\) \(=\) \(\ds \dfrac {\ds \valueat {\paren {\prod_{1 \mathop \le m \mathop < k \mathop \le n} \paren {y_k - y_m} } } {y_i \mathop = u} } {\ds \prod_{1 \mathop \le m \mathop \lt k \mathop \le n} \paren {y_k - y_m} }\) Vandermonde Determinant

Define sets $D,A,B,C$ with disjoint decomposition $D = A \cup B \cup C$:

\(\ds D\) \(=\) \(\ds \set {\tuple {m, k} : 1 \le m < k \le n }\)
\(\ds A\) \(=\) \(\ds \set {\tuple {m, k} \in D : m \ne i \mbox{ and } k \ne i}\)
\(\ds B\) \(=\) \(\ds \set {\tuple {m, k} \in D: m = i}\)
\(\ds C\) \(=\) \(\ds \set {\tuple {m, k} \in D: k = i}\)

Use $\prod_D = \prod_A \prod_B \prod_C$ to convert the numerator and denominator in the right hand side of $(2)$.

Then:

\(\ds \dfrac {\bigvalueat {\paren {\det \paren {V_y} } } {y_i \mathop = u} } {\det \paren {V_y} }\) \(=\) \(\ds \dfrac {\ds \prod_{k \mathop = 1, \, k \mathop \ne i}^n \paren {u - y_k} } {\ds \prod_{k \mathop = 1, \, k \mathop \ne i}^n \paren {y_i - y_k} }\) cancelling common factors in $(2)$

Replacement of ${ \map p {y_i} }$ and ${ \map {p_j} {x_j} }$ by products gives:

\(\text {(3)}: \quad\) \(\ds b_{ij}\) \(=\) \(\ds \paren {-1} \, \dfrac {\ds \prod_{k \mathop = 1}^n \paren {y_i - x_k } } {\ds \prod_{k \mathop = 1,\, k \mathop \ne j}^n \paren { x_j - x_k } } \ \prod_{k \mathop = 1,\, k \mathop \ne i}^n \paren {\dfrac {x_j - y_k} {y_i - y_k} }\) substituting $u = x_j$


After substituting $y_k \to -y_k$ and simplifying, Knuth's original identity (1997) becomes:



\(\text {(4)}: \quad\) \(\ds b_{ij}\) \(=\) \(\ds \dfrac {\ds \prod_{k \mathop = 1}^n \paren {x_k - y_i} } {\ds \prod_{k \mathop = 1,\, k \mathop \ne j}^n \paren {x_j - x_k} } \dfrac {\ds \prod_{k \mathop = 1, \, k \mathop \ne i}^n \paren {x_j - y_k} } {\ds \prod_{k \mathop = 1,\, k \mathop \ne i}^n \paren {-y_i + y_k} }\)

In $(3)$, factor $\paren {-1}^{n + 1}$ from the numerator and $\paren {-1}^{n - 1}$ from the denominator.

Then $\paren {-1}^{n + 1} = \paren {-1}^{n - 1}$ verifies that $(3)$ matches $(4)$.

$\blacksquare$


Sources