PrrofOfDeterminantOfTheVandermondeMatrix 2006-03-08 11:54:13 2006-11-03 11:49:59 Proof determinant of the Vandermonde matrix 0 % this is the default PlanetMath preamble. as your knowledge % of TeX increases, you will probably want to edit this, but % it should be fine as is for beginners. % almost certainly you want these \usepackage{amssymb} \usepackage{amsmath} \usepackage{amsfonts} % used for TeXing text within eps files %\usepackage{psfrag} % need this for including graphics (\includegraphics) %\usepackage{graphicx} % for neatly defining theorems and propositions %\usepackage{amsthm} % making logically defined graphics %\usepackage{xypic} % there are many more packages, add them here as you need them % define commands here 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 \[ \Delta = \prod_{1 \leq i < j \leq n}(a_j - a_i). \]