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


 * $\displaystyle 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:


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


 * $\displaystyle \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:


 * $\displaystyle P = \paren {\begin{smallmatrix}

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

Then:
 * $\displaystyle P = \paren {\begin{smallmatrix}

\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{smallmatrix} }, \quad Q = \paren {\begin{smallmatrix} \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{smallmatrix} }$

Determinant of Diagonal Matrix implies


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


 * $\displaystyle \det\paren { 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


 * $\displaystyle \det\paren {V_x} =

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


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

Determinant of Matrix Product and Definition:Inverse Matrix imply


 * $\displaystyle \det\paren {V_x^{-1} } =\dfrac{1}{\det\paren {V_x} },\quad

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

Then: