proof of determinant of the Vandermonde matrix

To begin, note that the determinant of the $n\times n$ Vandermonde matrix (which we shall denote as ‘ $\Delta$ ’) is a homogeneous polynomial of order $n(n-1)/2$ because every term in the determinant is, up to sign, the product of a zeroth power of some variable times the first power of some other variable , $\ldots$ , the $n-1$ -st power of some variable and $0+1+\cdots+(n-1)=n(n-1)/2$ .

Next, note that if $a_{i}=a_{j}$ with $i\neq j$ , then $\Delta=0$ because two columns of the matrix would be equal. Since $\Delta$ is a polynomial, this implies that $a_{i}-a_{j}$ is a factor of $\Delta$ . Hence,

$\Delta=C\prod_{1\leq i<j\leq n}(a_{j}-a_{i})$

where C is some polynomial. However, since both $\Delta$ and the product on the right hand side have the same degree, $C$ must have degree zero, i.e. $C$ must be a constant. So all that remains is the determine the value of this constant.

One way to determine this constant is to look at the coefficient of the leading diagonal, $\prod_{n}(a_{n})^{n-1}$ . Since it equals 1 in both the determinant and the product, we conclude that $C=1$ , hence

$\Delta=\prod_{1\leq i<j\leq n}(a_{j}-a_{i}).$

Title	proof of determinant of the Vandermonde matrix
Canonical name	ProofOfDeterminantOfTheVandermondeMatrix
Date of creation	2013-03-22 15:44:50
Last modified on	2013-03-22 15:44:50
Owner	rspuzio (6075)
Last modified by	rspuzio (6075)
Numerical id	10
Author	rspuzio (6075)
Entry type	Proof
Classification	msc 15A57
Classification	msc 65F99
Classification	msc 65T50