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Vandermonde Determinant

Date: 02/11/2003 at 19:55:04
From: Roger
Subject: Determinant

Compute the n x n determinant:

           |  1          1       1    ...    1        |
           |  x1         x2      x3          xn       | 
           |  x1^2      x2^2     x3^2   ...  xn^2     |
           |  .          .                            |
           |  .          .                            |
           |  .          .                            |
           |  x1^(n-1) x2^(n-1) x3^(n-1) ... xn^(n-1) |

Date: 02/13/2003 at 09:11:40
From: Doctor Jacques
Subject: Re: Determinant

Hi Roger,

Let us consider a simple example, with n = 3. The determinant is:

    |  1    1    1  |
D=  |  x    y    z  |
    | x^2  y^2  z^2 |

If you expand this determinant, you will get a polynomial in x, y, z. 
Each term in this polynomial is of the form:


where a and b are taken from x, y, z. This means that each term is of 
degree 3.

Notice now that, if x = y, the determinant has two equal columns, and 
is therefore equal to 0. This means that our polynomial is divisible 
by (x-y). In a similar way, D is divisible by (y-z) and (z-x).

We conclude that:

  D = (x-y)(y-z)(z-x)*p(x,y,z)

where p(x,y,z) is an unknown polynomial in x, y, z.

However, (x-y)(y-z)(z-x) is already of degree 3, so p(x,y,z) is just 
a constant, say A.

We have thus:

  D = A(x-y)(y-z)(z-x)

It remains to find the value of A. This can be done by looking at the 
coefficient of a particular term, for example y*z^2.

You should have no trouble generalizing this to higher degrees. Note 
that the degree of the polynomial is n(n-1)/2, and this is just (by 
chance?) the number of pairs of different variables.

This determinant is called a Vandermonde determinant.

Please feel free to wrie back if you require more assistance.

- Doctor Jacques, The Math Forum 
Associated Topics:
College Linear Algebra

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