Vandermonde determinants

enescu

Vandermonde determinants

Let $a_{1},a_{2},\ldots a_{n}$ be complex numbers and let
$D_{n}=\left| \begin{array} [c]{cccc} 1 & 1 & \ldots & 1\\ a_{1} & a_{2} & \ldots & a_{n}\\ a_{1}^{2} & a_{2}^{2} & \ldots & a_{n}^{2}\\ \vdots & & & \\ a_{1}^{n-1} & a_{2}^{n-1} & \ldots & a_{n}^{n-1} \end{array} \right| .$
$D_{n}$ is called Vandermonde determinant and its value equals
$D_{n}=\underset{1\leq i<j\leq n}{\prod}\left( a_{j}-a_{i}\right) .$
The proof goes by induction. The base case is obvious, so assume the asertion true for $n-1.$ Consider the polynomial
$P\left( x\right) =\left| \begin{array} [c]{cccc} 1 & 1 & \ldots & 1\\ a_{1} & a_{2} & \ldots & x\\ a_{1}^{2} & a_{2}^{2} & \ldots & x^{2}\\ \vdots & & & \\ a_{1}^{n-1} & a_{2}^{n-1} & \ldots & x^{n-1} \end{array} \right| .$
Clearly, $\deg P=n-1$ and $P(a_{i})=0$ for $1\leq i\leq n-1.$ Thus, we have
$P(x)=c(x-a_{1})\ldots(x-a_{n-1}),$
for some constant $c.$ Expanding the determinant after the last column, we find that
$c=\left| \begin{array} [c]{cccc} 1 & 1 & \ldots & 1\\ a_{1} & a_{2} & \ldots & a_{n-1}\\ a_{1}^{2} & a_{2}^{2} & \ldots & a_{n-1}^{2}\\ \vdots & & & \\ a_{1}^{n-2} & a_{2}^{n-2} & \ldots & a_{n-1}^{n-2} \end{array} \right| =D_{n-1}.$
Finally, setting $x=a_{n}$ gives the result.

**Posted:** Tue Sep 06, 2005 10:39 am

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Bogdan Enescu

-oo-

Nice proof. Of course, the $a_i$ don't have to be complex numbers. They may be elements of an arbitrary commutative unitary ring.

**Posted:** Tue Apr 18, 2006 7:48 pm

conceive

**Posted:** Mon Jun 12, 2006 11:52 am

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[will be away for a long long time..]

Rzeszut

**Posted:** Tue Jul 25, 2006 8:59 am

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$\left(\int_{I}f(t)dt\right)^{k}= \int_{I^{k}}f(t_{1})\cdot\ldots\cdot f(t_{k}) dt_{1}\ldots dt_{k}$