Miscellaneous Mathematics - Vandermonde Determinant

Vandermonde Determinant

Let $R$ be a commutative ring. Let $a_{1}, a_{2}, . . . \in R$ . Then

∣ \begin{matrix} 1 & a_{1} & . . . & a_{1}^{n - 1} \\ ⋮ & ⋮ & ⋱ & ⋮ \\ 1 & a_{n} & . . . & a_{n}^{n - 1} \end{matrix} ∣ = \prod_{1 \leq r < s \leq n} (a_{s} - a_{r})

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

∣ \begin{matrix} 1 & a_{1} & . . . & a_{1}^{n} \\ ⋮ & ⋮ & ⋱ & ⋮ \\ 1 & a_{n} & . . . & a_{n + 1}^{n} \end{matrix} ∣

This is equal to

∣ \begin{matrix} 1 & a_{1} & . . . & a_{1}^{n - 1} & f (a_{1}) \\ ⋮ & ⋮ & ⋱ & ⋮ & ⋮ \\ 1 & a_{n} & . . . & a_{n + 1}^{n - 1} & f (a_{n + 1}) \end{matrix} ∣

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.