Value of Vandermonde Determinant/Formulation 1/Proof 3

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.

$\blacksquare$