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.


Ben Lynn blynn@cs.stanford.edu 💡