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 $3 \times 3$ $\map \det C$.

$\blacksquare$

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