## Vandermonde Determinant

Let $R$ be a commutative ring. Let $a_1,a_2,... \in R$. Then

$\begin{vmatrix} 1 & a_1 & ... & a_1^{n-1} \\ \vdots & \vdots & \ddots & \vdots \\ 1 & a_n & ... & a_n^{n-1} \end{vmatrix} = { \prod_{1\le r\lt s\le n}(a_s - a_r) }$

Proof: Assume the result holds for $n$. Consider

$\begin{vmatrix} 1 & a_1 & ... & a_1^n \\ \vdots & \vdots & \ddots & \vdots \\ 1 & a_n & ... & a_{n+1}^n \end{vmatrix}$

This is equal to

$\begin{vmatrix} 1 & a_1 & ... & a_1^{n-1} & f(a_1) \\ \vdots & \vdots & \ddots & \vdots & \vdots \\ 1 & a_n & ... & a_{n+1}^{n-1} & f(a_{n+1}) \end{vmatrix}$

for any monic $f\in R[x]$ of degree $n$, because this matrix can be obtained from the previous one via elementary column operations. In particular, if we set $f = {\prod_{i=1}^{n}(x-a_i)}$ then $f(a_1) = ... = f(a_n) = 0$ and $f(a_{n+1}) = {\prod_{i=1}^n(a_{n+1}-a_i)}$. The result follows after using the inductive hypothesis.