Vandermonde determinants

enescu

Vandermonde determinants

Let $a_{1},a_{2},\ldots a_{n}$ be complex numbers and let
$D_{n}=\left| \begin{array} [c]{cccc} 1 & 1 & \ldots & 1\\ a_{1} & a_{2} & \ldots & a_{n}\\ a_{1}^{2}...$
$D_{n}$ 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
$P\left( x\right) =\left| \begin{array} [c]{cccc} 1 & 1 & \ldots & 1\\ a_{1} & a_{2} & \ldots & x\\...$
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
$c=\left| \begin{array} [c]{cccc} 1 & 1 & \ldots & 1\\ a_{1} & a_{2} & \ldots & a_{n-1}\\ a_{1}^{2} &...$
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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$\left(\int_{I}f(t)dt\right)^{k}= \int_{I^{k}}f(t_{1})\cdot\ldots\cdot f(t_{k}) dt_{1}\ldots dt_{k}$

darij grinberg

Total nitpicking, but I just want to keep stuff in Theorems & Formulas correct:

Enescu's proof requires $a_1$ , $a_2$ , ..., $a_{n - 1}$ to be distinct (otherwise, we cannot conclude that $P(x) = c(x - a_{1})\ldots(x - a_{n - 1})$ from $P(a_{i}) = 0$ for $1\leq i\leq n - 1$ ). But the other case is even easier (both $D_n$ and $\prod_{1\leq i < j\leq n}\left(a_j - a_i\right)$ are zero), so the proof is not in danger.

Unfortunately, this trick doesn't work for arbitrary commutative rings. Instead one has to prove Vandermonde over $\mathbb{Z}$ and conclude that it holds over an arbitrary commutative ring. Or proceed in a different way (I know of at least one good proof which works over any arbitrary commutative ring with $1$ ; can you find it?).

darij

**Posted:** Tue Oct 06, 2009 10:05 am

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mavropnevma

darij, the proof works over $\mathbb{Z}$ , since Euclidian algorithm works for monic divisors like $x-a_i$ , and $\mathbb{Z}$ is an integral domain. The problem with other commutative rings $R$ , even with unit, I think is that the factorization in $R[x]$ may no more be unique, e.g. in $\mathbb{Z}_4[x]$ we have $x^2 =(x-2)^2 \neq x(x-2)$ .

**Posted:** Tue Oct 06, 2009 10:37 am

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darij grinberg

Yes, that's what I meant. For $\mathbb{Z}$ it follows from $\mathbb{Q}$ .

darij

**Posted:** Tue Oct 06, 2009 10:46 am

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Now the die is cast, the first step taken, a glimmer of hope lights up our lives
Visions of the past, dreams forsaken forming right under our eyes
We are alive...