Vandermonde determinant

Statement

Let $K$ be a commutative ring, let $n \ge 1$, and let $x_1, \dots, x_n \in K$. Let $V$ be the $n \times n$ Vandermonde matrix with entries $V_{ij} = x_i^{\,j-1}$. Then $$\det V = \prod_{1 \le i < j \le n} (x_j - x_i),$$ where $\det$ is the determinant.

Remarks

Over a field, $\det V \ne 0$ if and only if the $x_i$ are pairwise distinct. Then, for all $y_1, \dots, y_n$, there is exactly one polynomial $p$ of degree less than $n$ with $p(x_i) = y_i$ for all $i$. For $n = 2$, the matrix is $\begin{pmatrix} 1 & x_1 \\ 1 & x_2 \end{pmatrix}$ with determinant $x_2 - x_1$.

Explanations

L1 human

A Vandermonde matrix lists the powers $1, x, x^2, \dots$ of some numbers, one row for each number. Its determinant is the product of all differences of the numbers. So it is nonzero exactly when the numbers are distinct.