Vandermonde Matrix Identity for Cauchy Matrix


This article needs to be tidied.
In particular: Rewrite so as to make sense
Please fix formatting and $\LaTeX$ errors and inconsistencies. It may also need to be brought up to our standard house style.
To discuss this page in more detail, feel free to use the talk page.
When this work has been completed, you may remove this instance of {{Tidy}} from the code.


Theorem

Assume values $\set {x_1, \ldots, x_n, y_1, \ldots, y_n}$ are distinct in matrix

\(\ds C\) \(=\) \(\ds \begin {pmatrix} \dfrac 1 {x_1 - y_1} & \dfrac 1 {x_1 - y_2} & \cdots & \dfrac 1 {x_1 - y_n} \\ \dfrac 1 {x_2 - y_1} & \dfrac 1 {x_2 - y_2} & \cdots & \dfrac 1 {x_2 - y_n} \\ \vdots & \vdots & \cdots & \vdots \\ \dfrac 1 {x_n - y_1} & \dfrac 1 {x_n - y_2} & \cdots & \dfrac 1 {x_n - y_n} \\ \end {pmatrix}\) Cauchy matrix of order $n$

Then:

\(\ds C\) \(=\) \(\ds -P V_x^{-1} V_y Q^{-1}\) Vandermonde matrix identity for a Cauchy matrix


Definitions of Vandermonde matrices $V_x$, $V_y$ and diagonal matrices $P$, $Q$:

$V_x = \begin {pmatrix} 1 & 1 & \cdots & 1 \\ x_1 & x_2 & \cdots & x_n \\ \vdots & \vdots & \ddots & \vdots \\ {x_1}^{n - 1} & {x_2}^{n - 1} & \cdots & {x_n}^{n - 1} \\ \end {pmatrix}, \quad V_y = \begin {pmatrix} 1 & 1 & \cdots & 1 \\ y_1 & y_2 & \cdots & y_n \\ \vdots & \vdots & \ddots & \vdots \\ {y_1}^{n - 1} & {y_2}^{n - 1} & \cdots & {y_n}^{n - 1} \\ \end {pmatrix}$ Vandermonde matrices
$P = \begin {pmatrix} \map {p_1} {x_1} & \cdots & 0 \\ \vdots & \ddots & \vdots \\ 0 & \cdots & \map {p_n} {x_n} \\ \end {pmatrix}, \quad Q = \begin {pmatrix} \map p {y_1} & \cdots & 0 \\ \vdots & \ddots & \vdots \\ 0 & \cdots & \map p {y_n} \\ \end {pmatrix}$ Diagonal matrices

Definitions of polynomials $p, p_1, \ldots, p_n$:

$\ds \map p x = \prod_{i \mathop = 1}^n \paren {x - x_i}$
$\ds \map {p_k} x = \dfrac {\map p x} {x - x_k} = \prod_{i \mathop = 1, i \mathop \ne k}^n \paren {x - x_i}$, $1 \mathop \le k \mathop \le n$


Proof

Matrices $P$ and $Q$ are invertible because all diagonal elements are nonzero.

For $1 \le i \le n$ express polynomial $p_i$ as:

$\ds \map {p_i} x = \sum_{k \mathop = 1}^n a_{i k} x^{k - 1}$

Then:

\(\ds \paren {\map {p_i} {x_j} }\) \(=\) \(\ds \paren {a_{i j} } V_x\) Definition of Matrix Product (Conventional)
\(\ds P\) \(=\) \(\ds \paren {a_{i j} } V_x\) as $\map {p_i} {x_j} = 0$ for $i \ne j$.
\(\ds \paren {a_{i j} }\) \(=\) \(\ds P V_x^{-1}\) solving for matrix $paren {a_{i j} }$
\(\ds \paren {\map {p_i} {y_j} }\) \(=\) \(\ds \paren {a_{i j} } V_y\) Definition of Matrix Product (Conventional)
\(\ds \paren {\map {p_i} {y_j} }\) \(=\) \(\ds P V_x^{-1} V_y\) substituting $paren {a_{i j} } = P V_x^{-1}$


Use second equation $\map {p_i} {y_j} = \dfrac {\map p {y_j} } {y_j - x_i}$:

\(\ds \paren {\map {p_i} {y_j} }\) \(=\) \(\ds -C Q\) Definition of Matrix Product (Conventional)
\(\ds -C Q\) \(=\) \(\ds P V_x^{-1} V_y\) equating competing equations for $\paren {\map {p_i} {y_j} }$
\(\ds C\) \(=\) \(\ds -P V_x^{-1} V_y Q^{-1}\) solving for $C$

$\blacksquare$


Examples

$3 \times 3$ Matrix

Illustrate $3 \times 3$ case for and Value of Cauchy Determinant.

Let $C$ denote the Cauchy matrix of order $3$:

$C = \begin {pmatrix} \dfrac 1 {x_1 - y_1} & \dfrac 1 {x_1 - y_2} & \dfrac 1 {x_1 - y_3} \\ \dfrac 1 {x_2 - y_1} & \dfrac 1 {x_2 - y_2} & \dfrac 1 {x_2 - y_3} \\ \dfrac 1 {x_3 - y_1} & \dfrac 1 {x_3 - y_2} & \dfrac 1 {x_3 - y_3} \\ \end{pmatrix}$

where the values in $\set {x_1, x_2, x_3, y_1, y_2, y_3}$ are assumed to be distinct.

Then:

\(\ds C\) \(=\) \(\ds -P V_x^{-1} V_y Q^{-1}\)
\(\ds \map \det C\) \(=\) \(\ds \paren {-1}^3 \dfrac {\paren {x_3 - x_1} \paren {x_3 - x_2} \paren {x_2 - x_1} \paren {y_3 - y_1} \paren {y_3 - y_2} \paren {y_2 - y_1} } {\paren {x_1 - y_1} \paren {x_1 - y_2} \paren {x_1 - y_3} \paren {x_2 - y_1} \paren {x_2 - y_2} \paren {x_2 - y_3} \paren {x_3 - y_1} \paren {x_3 - y_2} \paren {x_3 - y_3} }\) Determinant of Matrix Product


$n \times n$ Matrix

The methods of the $3 \times 3$ example apply unchanged for the general $n \times n$ Cauchy matrix:

Assume values $\set {x_1, \ldots, x_n, y_1, \ldots, y_n}$ are distinct. Then:

$\map \det {\begin{smallmatrix} \dfrac 1 {x_1 - y_1} & \dfrac 1 {x_1 - y_2} & \cdots & \dfrac 1 {x_1 - y_n} \\ \dfrac 1 {x_2 - y_1} & \dfrac 1 {x_2 - y_2} & \cdots & \dfrac 1 {x_2 - y_n} \\ \vdots & \vdots & \cdots & \vdots \\ \dfrac 1 {x_n - y_1} & \dfrac 1 {x_n - y_2} & \cdots & \dfrac 1 {x_n - y_n} \\ \end{smallmatrix} } = \paren {-1}^n \dfrac {\ds \prod_{1 \mathop \le j \mathop < i \mathop \le n} \paren {x_i - x_j} \quad \prod_{1 \mathop \le j \mathop < i \mathop \le n} \paren {y_i - y_j} } {\ds \prod_{i \mathop = 1}^n \prod_{j \mathop = 1}^n \paren {x_i - y_j} }$ Value of Cauchy Determinant

Assume values $\set {x_1, \ldots, x_n, -y_1, \ldots, -y_n}$ are distinct, then replace in the preceding equation $y_i$ by $-y_i$, $1 \le i \le n$:

$\map \det {\begin{smallmatrix} \dfrac 1 {x_1 + y_1} & \dfrac 1 {x_1 + y_2} & \cdots & \dfrac 1 {x_1 + y_n} \\ \dfrac 1 {x_2 + y_1} & \dfrac 1 {x_2 + y_2} & \cdots & \dfrac 1 {x_2 + y_n} \\ \vdots & \vdots & \cdots & \vdots \\ \dfrac 1 {x_n + y_1} & \dfrac 1 {x_n + y_2} & \cdots & \dfrac 1 {x_n + y_n} \\ \end{smallmatrix} } = \paren {-1}^n \dfrac {\ds \prod_{1 \mathop \le j \mathop < i \mathop \le n} \paren {x_i - x_j} \quad \prod_{1 \mathop \le j \mathop < i \mathop \le n} \paren {y_j - y_i} } {\ds \prod_{i \mathop = 1}^n \prod_{j \mathop = 1}^n \paren {x_i + y_j} }$ Value of Cauchy Determinant

$\blacksquare$


Also see

  • Definition:Vandermonde Determinant


Historical Note

Roderick Gow established using interpolation polynomials and change of basis facts.


Sources

  • 1944: A.C. Aitken: Determinants and Matrices (3rd ed.): Chapter $\text{VI}$. $47$: Alternant Matrices and Determinants
  • March 1992: Roderick Gow: Cauchy's matrix, the Vandermonde matrix and polynomial interpolation (Bull. Irish Math. Soc. Vol. 28: pp. 45 – 52)
  • 2015: David C. Lay, Steven R. Lay and Judi J. McDonald: Linear Algebra and its Applications (5th ed.): Matrix Algebra Ch2 and Determinants Ch3