Value of Vandermonde Determinant/Formulation 1/Proof 3

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $V_n$ be the Vandermonde determinant of order $n$ defined as the following formulation:

$V_n = \begin {vmatrix}

1 & x_1 & {x_1}^2 & \cdots & {x_1}^{n - 2} & {x_1}^{n - 1} \\ 1 & x_2 & {x_2}^2 & \cdots & {x_2}^{n - 2} & {x_2}^{n - 1} \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 1 & x_n & {x_n}^2 & \cdots & {x_n}^{n - 2} & {x_n}^{n - 1} \end {vmatrix}$


Its value is given by:

$\ds V_n = \prod_{1 \mathop \le i \mathop < j \mathop \le n} \paren {x_j - x_i}$


Proof

Let:

$V_n = \begin {vmatrix}
 1 & x_1 & {x_1}^2 & \cdots & {x_1}^{n - 2} & {x_1}^{n - 1} \\
 1 & x_2 & {x_2}^2 & \cdots & {x_2}^{n - 2} & {x_2}^{n - 1} \\
 1 & x_3 & {x_3}^2 & \cdots & {x_3}^{n - 2} & {x_3}^{n - 1} \\

\vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 

 1 & x_{n - 1} & {x_{n - 1} }^2 & \cdots & {x_{n - 1} }^{n - 2} & {x_{n - 1} }^{n - 1} \\
 1 & x_n & {x_n}^2 & \cdots & {x_n}^{n - 2} & {x_n}^{n - 1}

\end {vmatrix}$


Start by replacing number $x_n$ in $V_n$ with the unknown $x$.

Thus $V_n$ is made into a function of $x$.

$\map P x = \begin{vmatrix}
 1 & x_1 & {x_1}^2 & \cdots & {x_1}^{n - 2} & {x_1}^{n - 1} \\
 1 & x_2 & {x_2}^2 & \cdots & {x_2}^{n - 2} & {x_2}^{n - 1} \\
 1 & x_3 & {x_3}^2 & \cdots & {x_3}^{n - 2} & {x_3}^{n - 1} \\

\vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 

 1 & x_{n - 1} & {x_{n - 1} }^2 & \cdots & {x_{n - 1} }^{n - 2} & {x_{n - 1} }^{n - 1} \\
 1 & x & x^2 & \cdots & x^{n - 2} & x^{n - 1}

\end{vmatrix}$.

Let $x$ equal a value from the set $\set {x_1,\ldots,x_{n-1} }$.

Then determinant $\map P x$ has equal rows, giving:

\(\ds \map P x\) \(=\) \(\ds 0\) for $x = x_1, \ldots, x_{n - 1}$

Perform row expansion by the last row.

Then $\map P x$ is seen to be a polynomial of degree $n-1$:

$\map P x = \begin {vmatrix}
 x_1 & {x_1}^2 & \cdots & {x_1}^{n - 2} & {x_1}^{n - 1} \\
 x_2 & {x_2}^2 & \cdots & {x_2}^{n - 2} & {x_2}^{n - 1} \\
 x_3 & {x_3}^2 & \cdots & {x_3}^{n - 2} & {x_3}^{n - 1} \\

\vdots & \vdots & \vdots & \ddots & \vdots \\ 

 x_{n - 1} & {x_{n - 1} }^2 & \cdots & {x_{n - 1} }^{n - 2} & {x_{n - 1} }^{n - 1}

\end{vmatrix} + \begin{vmatrix} 

 1 & {x_1}^2 & \cdots & {x_1}^{n - 2} & {x_1}^{n - 1} \\
 1 & {x_2}^2 & \cdots & {x_2}^{n - 2} & {x_2}^{n - 1} \\
 1 & {x_3}^2 & \cdots & {x_3}^{n - 2} & {x_3}^{n - 1} \\

\vdots & \vdots & \vdots & \ddots & \vdots \\ 

 1 & {x_{n - 1} }^2 & \cdots & {x_{n - 1} }^{n - 2} & {x_{n - 1} }^{n - 1}

\end {vmatrix} x + \cdots + \begin {vmatrix} 

 1 & x_1 & {x_1}^2 & \cdots & {x_1}^{n - 2} \\
 1 & x_2 & {x_2}^2 & \cdots & {x_2}^{n - 2} \\
 1 & x_3 & {x_3}^2 & \cdots & {x_3}^{n - 2} \\

\vdots & \vdots & \vdots & \ddots & \vdots \\ 

 1 & x_{n - 1} & {x_{n - 1} }^2 & \cdots & {x_{n - 1} }^{n - 2}

\end{vmatrix} x^{n - 1}$

By the Polynomial Factor Theorem:

$\map P x = \map C {x - x_1} \paren {x - x_2} \dotsm \paren {x - x_{n - 1} }$

where $C$ is the leading coefficient (with $x^{n - 1}$ power).


Thus:

$\map P x = V_{n - 1} \paren {x - x_1} \paren {x - x_2} \dotsm \paren {x - x_{n - 1} }$

which by evaluating at $x = x_n$ gives:

$V_n = V_{n - 1} \paren {x_n - x_1} \paren {x_n - x_2} \dotsm \paren {x_n - x_{n - 1} }$


Repeating the process:

\(\ds V_n\) \(=\) \(\ds \prod_{1 \mathop \le i \mathop < n} \paren {x_n - x_i} V_{n - 1}\)
\(\ds \) \(=\) \(\ds \prod_{1 \mathop \le i \mathop < n} \paren {x_n - x_i} \prod_{1 \mathop \le i \mathop < n - 1} \paren {x_{n - 1} - x_i} V_{n - 2}\)
\(\ds \) \(=\) \(\ds \dotsm\)
\(\ds \) \(=\) \(\ds \prod_{1 \mathop \le i \mathop < j \mathop \le n} \paren {x_j - x_i}\)


which establishes the solution.

$\blacksquare$

Retrieved from "https://proofwiki.org/w/index.php?title=Value_of_Vandermonde_Determinant/Formulation_1/Proof_3&oldid=535529"