Consider the Vandermonde matrix: \[ A \da \left(\begin{array}{ccc} 1 & \cdots & 1 \\ \lambda_{1} & \cdots & \lambda_{k} \\ \vdots & & \vdots \\ \lambda_{1}^{k-1} & \cdots & \lambda_{k}^{k-1} \end{array}\right) .\]
Show that \[ \det A = \prod_{i < j} (\lambda_j - \lambda_i) .\]
Solution
Regard \(D(\lambda_1,\ldots,\lambda_k)=\det A\) as a polynomial in the \(\lambda_i\).
1. For every \(i<j\), the polynomial \(\lambda_j-\lambda_i\) divides \(D\).
Interchanging \(\lambda_i\) and \(\lambda_j\) interchanges columns \(i\) and \(j\) of \(A\), so it changes the sign of \(D\). In particular, after setting \(\lambda_i=\lambda_j\), the two columns coincide and \(D=0\). Viewing \(D\) as a polynomial in \(\lambda_j\) over the coefficient ring generated by the other variables, the factor theorem gives \[ \lambda_j-\lambda_i\mid D. \]
2. Hence\[ \prod_{i<j}(\lambda_j-\lambda_i)\mid D. \]
The factors \(\lambda_j-\lambda_i\) are pairwise nonassociate irreducible polynomials in the UFD generated by the \(\lambda_i\), hence are pairwise relatively prime. By 1 each divides \(D\), so their product divides \(D\).
3. Both \(D\) and the product in 2 have total degree \(k(k-1)/2\).
Expanding the determinant, \[ D=\sum_{\sigma\in S_k}\operatorname{sgn}(\sigma) \prod_{r=1}^k \lambda_{\sigma(r)}^{r-1}. \] Every monomial in this sum has total degree \[ 0+1+\cdots+(k-1)=\frac{k(k-1)}2. \] The product \(\prod_{i<j}(\lambda_j-\lambda_i)\) has one linear factor for each of the \(\binom{k}{2}\) pairs, so it has the same total degree.
4. The quotient in 2 is the constant \(1\).
By 2 and 3, there is a constant \(c\) such that \[ D=c\prod_{i<j}(\lambda_j-\lambda_i). \] Consider the monomial \[ \lambda_2\lambda_3^2\cdots\lambda_k^{k-1}. \] In the determinant expansion of 3, this monomial occurs only for the identity permutation and has coefficient \(1\). In the product, it is obtained uniquely by choosing the term \(\lambda_j\) from every factor with second index \(j\), and again has coefficient \(1\). Hence \(c=1\).
5. Therefore\[ \det A=\prod_{i<j}(\lambda_j-\lambda_i). \]
This is 4 with \(c=1\).