Value of Vandermonde Determinant/Formulation 1/Proof 4
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:
- $V_n = \begin {vmatrix}
1 & x_1 & {x_1}^2 & \cdots & {x_1}^{n - 2} & {x_1}^{n - 1} \\ 1 & x_2 & {x_2}^2 & \cdots & {x_2}^{n - 2} & {x_2}^{n - 1} \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 1 & x_n & {x_n}^2 & \cdots & {x_n}^{n - 2} & {x_n}^{n - 1} \end {vmatrix}$
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:
\(\text {(1)}: \quad\) | \(\ds V_n\) | \(=\) | \(\ds \map f {x_n} V_{n - 1}\) | Expansion Theorem for Determinants for columns | ||||||||||
\(\ds \) | \(=\) | \(\ds V_{n - 1} \prod_{k \mathop = 1}^{n - 1} \paren {x_n - x_k}\) |
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:
\(\ds V_{n + 1}\) | \(=\) | \(\ds V_n \prod_{k \mathop = 1}^n \paren {x_{n + 1} - x_k}\) | from $(1)$, setting $n \to n + 1$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \prod_{1 \mathop \le m \mathop < k \mathop \le n} \paren {x_k - x_m} \prod_{k \mathop = 1}^n \paren {x_{n + 1} - x_k}\) | Induction hypothesis with new indexing symbols: $i \to m, j \to k$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \prod_{1 \mathop \le i \mathop < j \mathop \le n + 1} \paren {x_j - x_i}\) | simplifying |
Thus $\map P {n + 1}$ has been shown to be true.
The induction is complete.
$\blacksquare$