VIII More about determinants and matrices
Vandermonde determinants have the form
Its values is
We will prove this by induction with respect to n. Obviously, (35.1) is correct for n = 2. Assume that it has been proved for determinants of order (n - 1), that is,
Now consider V(x1, ··· , xn-1, xn) as a polynomial in xn with x1, ··· , xn-1 held fixed. As you can see from the expanded determinant above, the highest power is xnn-1, and indeed it is multiplied by the adjoint minor determinant of xnn-1, that is, by V(x1, ··· , xn-1), whence
V(x1, ··· , xn-1, xn) = xnn-1V(x1, ··· , xn-1) + xnn-2 ···. (35.3)
Obviously, V(x1, ··· , xn-1, xn) vanishes for xn = x1, ··· , , xn = xn-1. whence V(x1, ··· , xn-1, xn) can be divided by the product
(xn-x1)(xn-x2) ··· (xn-xn-1).
Since the degree of this product in xn is already n - 1, that is, equal to to the total degree of V(x1, ··· , xn-1, xn) in xn, you must have
V(x1, ··· , xn-1, xn) = (xn-x1)(xn-x2) ··· (xn-xn-1)·c, (35.4)
where c does not depend on x. A comparison of the coefficients of xnn-1 in (35.3) and (25.4) yields
c = V(x1, ··· , xn-1),
that is
V(x1, ··· , xn-1, xn) = (xn-x1) ··· (xn-xn-1)·V(x1, ··· , xn-1).
Combined with (35.2) you now obtain (35.1).