Vandermonde Matrix Identity for Cauchy Matrix/Examples/3x3
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Example of Vandermonde Matrix Identity for Cauchy Matrix
Illustrate $3 \times 3$ case for Vandermonde Matrix Identity for Cauchy Matrix 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:
| \(\ds C\) | \(=\) | \(\ds -P V_x^{-1} V_y Q^{-1}\) | Vandermonde Matrix Identity for Cauchy Matrix | |||||||||||
| \(\ds \map \det C\) | \(=\) | \(\ds \paren {-1}^3 \dfrac {\paren {x_3 - x_1} \paren {x_3 - x_2} \paren {x_2 - x_1} \paren {y_3 - y_1} \paren {y_3 - y_2} \paren {y_2 - y_1} } {\paren {x_1 - y_1} \paren {x_1 - y_2} \paren {x_1 - y_3} \paren {x_2 - y_1} \paren {x_2 - y_2} \paren {x_2 - y_3} \paren {x_3 - y_1} \paren {x_3 - y_2} \paren {x_3 - y_3} }\) | Determinant of Matrix Product |
Proof
Define Vandermonde matrices
| \(\ds V_x\) | \(=\) | \(\ds \begin {pmatrix} 1 & 1 & 1 \\ x_1 & x_2 & x_3 \\ x_1^2 & x_2^2 & x_3^2 \end {pmatrix}\) | ||||||||||||
| \(\ds V_y\) | \(=\) | \(\ds \begin {pmatrix} 1 & 1 & 1 \\ y_1 & y_2 & y_3 \\ y_1^2 & y_2^2 & y_3^2 \end {pmatrix}\) |
Define polynomials:
| \(\ds \map p x\) | \(=\) | \(\ds \paren {x - x_1} \paren {x - x_2} \paren {x - x_3}\) | ||||||||||||
| \(\ds \map {p_1} x\) | \(=\) | \(\ds \paren {x - x_2} \paren {x - x_3}\) | ||||||||||||
| \(\ds \map {p_2} x\) | \(=\) | \(\ds \paren {x - x_1} \paren {x - x_3}\) | ||||||||||||
| \(\ds \map {p_3} x\) | \(=\) | \(\ds \paren {x - x_1} \paren {x - x_2}\) |
Define invertible diagonal matrices:
| \(\ds P\) | \(=\) | \(\ds \begin {pmatrix} \map {p_1} {x_1} & 0 & 0 \\ 0 & \map {p_2} {x_2} & 0 \\ 0 & 0 & \map {p_3} {x_3} \end {pmatrix}\) | ||||||||||||
| \(\ds Q\) | \(=\) | \(\ds \begin {pmatrix} \map p {y_1} & 0 & 0 \\ 0 & \map p {y_2} & 0 \\ 0 & 0 & \map p {y_3} \end {pmatrix}\) |
Then:
| \(\ds P\) | \(=\) | \(\ds \begin {pmatrix} \paren {x_1 - x_2} \paren {x_1 - x_3} & 0 & 0 \\ 0 & \paren {x_2 - x_1} \paren {x_2 - x_3} & 0 \\ 0 & 0 & \paren {x_3 - x_1} \paren {x_3 - x_2} \end {pmatrix}\) | ||||||||||||
| \(\ds Q\) | \(=\) | \(\ds \begin {pmatrix} \paren {y_1 - x_1} \paren {y_1 - x_2} \paren {y_1 - x_3} & 0 & 0 \\ 0 & \paren {y_2 - x_1} \paren {y_2 - x_2} \paren {y_2 - x_3} & 0 \\ 0 & 0 & \paren {y_3 - x_1} \paren {y_3 - x_2} \paren {y_3 - x_3} \end {pmatrix}\) |
Determinant of Diagonal Matrix gives:
| \(\ds \map \det P\) | \(=\) | \(\ds \paren {x_1 - x_2} \paren {x_1 - x_3} \paren {x_2 - x_1} \paren {x_2 - x_3} \paren {x_3 - x_1} \paren {x_3 - x_2}\) | ||||||||||||
| \(\ds Q\) | \(=\) | \(\ds \paren {y_1 - x_1} \paren {y_1 - x_2} \paren {y_1 - x_3} \paren {y_2 - x_1} \paren {y_2 - x_2} \paren {y_2 - x_3} \paren {y_3 - x_1} \paren {y_3 - x_2} \paren {y_3 - x_3}\) |
Value of Vandermonde Determinant gives:
| \(\ds \map \det {V_x}\) | \(=\) | \(\ds \paren {x_3 - x_2} \paren {x_3 - x_1} \paren {x_2 - x_1}\) | ||||||||||||
| \(\ds \map \det {V_y}\) | \(=\) | \(\ds \paren {y_3 - y_2} \paren {y_3 - y_1} \paren {y_2 - y_1}\) |
Determinant of Matrix Product and Definition:Inverse Matrix give:
| \(\ds \map \det {V_x^{-1} }\) | \(=\) | \(\ds \dfrac 1 {\map \det {V_x} }\) | ||||||||||||
| \(\ds \map \det {Q^{-1} }\) | \(=\) | \(\ds \dfrac 1 {\map \det Q}\) |
Then:
| \(\ds \map \det C\) | \(=\) | \(\ds \map \det {-I} \map \det P \map \det {V_x^{-1} } \map \det {V_y} \map \det {Q^{-1} }\) | Determinant of Matrix Product | |||||||||||
| \(\ds \) | \(=\) | \(\ds \paren {-1}^3 \dfrac {\map \det P \map \det {V_y} } {\map \det {V_x} \map \det Q}\) | from $\det \begin {pmatrix} -1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & -1 \end {pmatrix} = \paren {-1}^3$ |
Insert the four determinant equations and simplify to obtain the equation for $3 \times 3$ $\map \det C$.
$\blacksquare$