A Vandermonde matrix of order n has the form

1 1 1 . . . 1
x_{1} x_{2} x_{3} . . . x_{n}
x_{1}^{2} x_{2}^{2} x_{3}^{2} . . . x_{n}^{2}
. . . . . . .
. . . . . . .
. . . . . . .
x_{1}^{n-1} x_{2}^{n-1} x_{3}^{n-1} . . . x_{n}^{n-1}

Although the transpose of this may be used as the definition; and it may also be referred to as an *alternant* matrix. Whilst solving an equation with an nXn Vandermonde matrix requires O(n^{2}) operations, it turns out that the determinant is very easy to calculate, using the formula:

_____
| |
| | (x_{j}-x_{i})
| |
1≤i<j≤n

For example, the matrix

1 1 1
x y z
x^{2} y^{2} z^{2}

Or its transpose, has determinant (y-x)(z-x)(z-y). This could be proved by multiplying out this expression and checking that it gives the same result as a

row expansion of the above matrix, but a more

elegant solution (which also illustrates why the general nXn result holds) is to use

elementary column operations to obtain a

lower-triangular matrix, for which the determinant is simply the product of the

diagonal.

|1 1 1 | |1 0 0 |
|x y z | = |x y-x z-x |
|x^{2} y^{2} z^{2} | |x^{2} y^{2}-x^{2} z^{2}-x^{2} |

By subtracting column 1 from each of columns 2 and 3. Then

factorising the bottom row we get

|1 0 0 |
|x y-x z-x |
|x^{2} (y-x)(y+x) (z-x)(z+x) |

So we can pull out

factors of (y-x) and (z-x) by the

multilinearity of the

determinant function:

|1 0 0 |
(y-x)(z-x) |x 1 1 |
|x^{2} (y+x) (z+x) |

Now we perform a further

column operation, subtracting the second column from the third, which cancels the 1 and x:

|1 0 0 |
(y-x)(z-x) |x 1 0 |
|x^{2} (y+x) (z-y) |

Again pull a factor forward:

|1 0 0 |
(y-x)(z-x)(z-y) |x 1 0 |
|x^{2} (y+x) 1 |

Observe that this is a

triangular matrix, so its determinant is the product of the diagonal =1*1*1 =1. So we are left with simply the factors (y-x)(z-x)(z-y) as desired.