Vandermonde Matrix Identity for Hilbert Matrix/Examples/3x3

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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:

\(\ds H\) \(=\) \(\ds {\begin{pmatrix} \frac 1 1 & \frac 1 2 & \frac 1 3 \\ \frac 1 2 & \frac 1 3 & \frac 1 4 \\ \frac 1 3 & \frac 1 4 & \frac 1 5 \\ \end{pmatrix} }\) Hilbert matrix of order $3$

Then:

\(\ds H\) \(=\) \(\ds -P V_x^{-1} V_y Q^{-1}\) Vandermonde Matrix Identity for Hilbert Matrix
\(\ds \map \det H\) \(=\) \(\ds \dfrac 1 {2140}\) Determinant of Matrix Product


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

$\ds \map \det P = \paren {1 - 2} \paren {1 - 3} \paren {2 - 1}\paren {2 - 3} \paren {3 - 1} \paren {3 - 2} = -4$
$\ds \map \det Q = \paren {0 - 1} \paren {0 - 2} \paren {0 - 3} \paren {-1 - 1} \paren {-1 - 2} \paren {-1 - 3} \paren {-2 - 1} \paren {-2 - 2} \paren {-2 - 3} = -8640$

Value of Vandermonde Determinant gives:

$\ds \map \det {V_x} = \paren {3 - 2} \paren {3 - 1} \paren {2 - 1} = 2$
$\ds \map \det {V_y} = \paren {-2 - \paren {-1} } \paren {-2 - 0} \paren {-1 - 0} = -2$

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:

\(\ds \map \det H\) \(=\) \(\ds \map \det {-I} \map \det P \map \det {V_x^{-1} } \map \det {V_y} \map \det {Q^{-1} }\) Determinant of Matrix Product
\(\ds \) \(=\) \(\ds \paren {-1}^3 \dfrac {\map \det P \map \det {V_y} } {\map \det {V_x} \map \det Q}\) $\map \det {\begin{matrix} -1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & -1 \\ \end{matrix} } = (-1)^3$
\(\ds \) \(=\) \(\ds \paren {-1}^3 \dfrac {\paren {-4} \paren {-2} } {\paren 2 \paren {-8640} }\) inserting the four determinant equations
\(\ds \) \(=\) \(\ds \dfrac 1 {2140}\) simplifying

$\blacksquare$