Prove determinant of a matrix $=\prod_{j<i}(a_i-a_j)$

Solution 1:

It's the determinant of a Vandermonde matrix. The standard argument to compute the determinant is to write $$f(x) = \begin{vmatrix} 1 & 1 & \cdots & 1 \\ a_0 & a_1 & \cdots & x \\ a_0^2 & a_1^2 & \cdots & x^2 \\ \vdots & \vdots & \ddots & \vdots \\ a_0^n & a_1^n & \cdots & x^n \\\end{vmatrix}.$$ This is a polynomial in $x$ of degree $n$, and by the properties of determinants (the determinant of a matrix with two columns equal is zero), you have that $a_0, a_1, \dots, a_{n-1}$ are roots. Therefore $$ f(x) = A \cdot (x-a_0) \cdot (x-a_1) \cdot\ldots \cdot (x- a_{n-1}). $$ Now looking at the matrix, you see that the coefficient $A$ of $x^{n}$ is a similar determinant of size one less. Use induction to conclude.

Solution 2:

Let $$V(a_0,\ldots,a_n)=\begin{vmatrix} 1 & 1 & \cdots & 1 \\ a_0 & a_1 & \cdots & a_n \\ a_0^2 & a_1^2 & \cdots & a_n^2 \\ \vdots & \vdots & \ddots & \vdots \\ a_0^n & a_1^n & \cdots & a_n^n \\\end{vmatrix}.$$

We have $V(a_0,a_1)=a_1-a_0.$ Suppose $n\geq3$, and we denote $l_{0},l_1,\ldots,l_n$the lines of $V$. We write $$V=\det((l_{0},l_1-a_0l_{0},l_2-a_0l_1,\ldots,l_n-a_0l_{n-1})^T)$$ we find $$V=\begin{vmatrix} 1 & 1 & \cdots & 1 \\ 0 & a_1-a_0 & \cdots & a_n-a_0 \\ 0 & a_1(a_1-a_0) & \cdots & a_n(a_n-a_0) \\ \vdots & \vdots & \ddots & \vdots \\ 0 & a_1^{n-1}(a_1-a_0) & \cdots & a_n^{n-1}(a_n-a_0) \\\end{vmatrix}.$$ So, $$V(a_0,\ldots,a_n)=V(a_1,\ldots,a_n)\prod_{k=1}^n(a_k-a_0).$$ Now, immediate induction provides: $$V(a_0,\ldots,a_n)=\prod_{0\leq i<j\leq n}(x_j-x_i).$$