Vandermonde Matrix Identity for Hilbert Matrix/Examples/3x3

Example of Vandermonde Matrix Identity for Hilbert Matrix
Define polynomial root sets $\set {1, 2, 3}$ and $\set {0, -1, -2}$ for Definition:Cauchy Matrix because Hilbert Matrix is Cauchy Matrix.

Illustrate $3\times 3$ case for Vandermonde Matrix Identity for Hilbert Matrix and value of Hilbert matrix determinant:

Then:

Details:

Define Vandermonde matrices


 * $\ds V_x = \begin{pmatrix}

1 & 1 & 1 \\ 1 & 2 & 3 \\ 1^2 & 2^2 & 3^2 \\ \end{pmatrix}, \quad V_y = \begin {pmatrix} 1 & 1 & 1 \\ 0 & -1 & -2 \\ 0 & \paren {-1}^2 & \paren {-2}^2 \\ \end {pmatrix}$

Define polynomials:


 * $\ds \map p x = \paren {x - 1} \paren {x - 2} \paren {x - 3}$


 * $\ds \map {p_1} x = \paren {x - 2} \paren {x - 3},

\quad \map {p_2} x = \paren {x - 1} \paren {x - 3}, \quad \map {p_3} x = \paren {x - 1} \paren {x - 2}$ Define invertible diagonal matrices:


 * $\ds P = \begin {pmatrix}

\map {p_1} 1 & 0       & 0 \\ 0       & \map {p_2} 2 & 0 \\ 0       & 0        & \map {p_3} 3 \\ \end {pmatrix}, \quad Q = \begin {pmatrix} \map p 0 & 0       & 0 \\ 0       & \map p {-1} & 0 \\ 0       & 0        & \map p {-2} \\ \end {pmatrix}$

Then:
 * $\ds P = \begin {pmatrix}

\paren {1 - 2} \paren {1 - 3} & 0                            & 0 \\ 0                            & \paren {2 - 1} \paren {2 - 3} & 0 \\ 0                            & 0                             & \paren {3 - 1} \paren {3 - 2} \\ \end {pmatrix}, \quad Q = \begin {pmatrix} \paren {0 - 1} \paren {0 - 2} \paren {0 - 3} & 0       & 0 \\ 0       &  \paren {-1 - 1} \paren {-1 - 2} \paren {-1 - 3} & 0 \\ 0       & 0        &  \paren {-2 - 1} \paren {-2 - 2} \paren {-2 - 3} \\ \end {pmatrix}$

Determinant of Diagonal Matrix implies


 * $\ds \map \det P = \paren {1 - 2} \paren {1 - 3} \paren {2 - 1}\paren {2 - 3} \paren {3 - 1} \paren {3 - 2} = -4$


 * $\ds \map \det Q = \paren {0 - 1} \paren {0 - 2} \paren {0 - 3}

\paren {-1 - 1} \paren {-1 - 2} \paren {-1 - 3} \paren {-2 - 1} \paren {-2 - 2} \paren {-2 - 3} = -8640$

Vandermonde Determinant implies


 * $\ds \map \det {V_x} = \paren {3 - 2} \paren {3 - 1} \paren {2 - 1} = 2$


 * $\ds \map \det {V_y} = \paren {-2 - \paren {-1} } \paren {-2 - 0} \paren {-1 - 0} = -2$

Determinant of Matrix Product and Definition:Inverse Matrix imply


 * $\ds \map \det {V_x^{-1} } = \dfrac 1 {\map \det {V_x} }, \quad

\map \det {Q^{-1} } = \dfrac 1 {\map \det Q}$

Then: