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:
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| \(\text {(4)}: \quad\) | \(\ds b_{ij}\) | \(=\) | \(\ds \dfrac {\ds \prod_{k \mathop = 1}^n \paren {x_k - y_i} } {\ds \prod_{k \mathop = 1,\, k \mathop \ne j}^n \paren {x_j - x_k} } \dfrac {\ds \prod_{k \mathop = 1, \, k \mathop \ne i}^n \paren {x_j - y_k} } {\ds \prod_{k \mathop = 1,\, k \mathop \ne i}^n \paren {-y_i + y_k} }\) |
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
- 1997: Donald E. Knuth: The Art of Computer Programming: Volume 1: Fundamental Algorithms (3rd ed.) ... (previous) ... (next): $\S 1.2.3$: Sums and Products: Exercise $41$