Value of Vandermonde Determinant/Formulation 1/Proof 4

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

Let $\map f x$ be any monic polynomial of degree $n - 1$:


 * $\ds \map f x = x^{n - 1} + \sum_{i \mathop = 0}^{n - 2} a_i x^i$

Apply elementary column operations to $V_n$ repeatedly to show:


 * $V_n = W$

where:


 * $ W = \begin {vmatrix}

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

Select a specific degree $n - 1$ monic polynomial:


 * $\ds \map f x = \prod_{k \mathop = 1}^{n - 1} \paren {x - x_k}$

The selected polynomial is zero at all values $x_1, \ldots, x_{n - 1}$.

Then the last column of $W$ is all zeros except the entry $\map f {x_n}$.

Expand $\map \det W$ by cofactors along the last column to prove:

For $n \ge 2$, let $\map P n$ be the statement:


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

Mathematical induction will be applied.

Basis for the Induction
By definition, determinant $V_1 = 1$.

To prove $\map P 2$ is true, use equation $(1)$ with $n = 2$:


 * $V_2 = \paren {x_2 - x_1} V_1$

This is the basis for the induction.

Induction Step
This is the induction step:

Let $\map P n$ is be assumed true.

We are to prove that $\map P {n + 1}$ is true.

As follows:

Thus $\map P {n + 1}$ has been shown to be true.

The induction is complete.