Chapter D, Problem 1P

Program Plan Intro

To prove that the determinant of the Vandermonde matrix V is $\det \left(V\left(x_0,x_1,\mathrm{...},x_n\right)\right)={\displaystyle \prod _{0\le j<k\le n-1}\left(x_k-x_j\right)}$

Explanation of Solution

Given information:

Vandermonde matrix V is given by $V\left(x_0,x_1,\mathrm{...}{x}_{n-1}\right)=\left(\begin{array}{cccc}1& x_0& x_0{}^2& x_0{}^{n-1}\\ 1& x_1& x_1{}^2& x_1{}^{n-1}\\ \begin{array}{l}.\\ .\\ .\end{array}& \begin{array}{l}.\\ .\\ .\end{array}& \begin{array}{l}.\\ .\\ .\end{array}& \begin{array}{l}.\\ .\\ .\end{array}\\ 1& x_{n-1}& x_{n-1}{}^2& x_{n-1}{}^{n-1}\end{array}\right)$

Explanation:

Start with induction on n .

For n = 1, the Vandermonde matrix is of size $1\times 1$ with only one entry i.e. 1.It is easy to note that this matrix’s determinant is 1. Also product is one as no distinct i and j can be picked in the product on the right hand side.

Now, assume that the given condition is true for $n\times n$ matrix. Using this let’s show that it is true for $\left(n+1\right)\times \left(n+1\right)$ matrices.

Start from second last column on right side that is $n-1$ and go towards left adding $\left(-x_0\right)$ times that column to the one to its right.

The determinant will not be changed from this operation. All entries in the top row, except the leftmost one, have become zero as $x_0{}^{j-1}-x_0{}^{j-2}\cdot x_0=0$ . This means that expansion by minors along the top row can be done in order to compute the determinant. The top left column will be the only non-zero term in this operation. Thus determinant remains the same as the determinant of that one minor. Also, the factor of every item in i^th of that minor is $\left(x{i}_{}-x_0\right)$ . This implies that taking all those factors out will give expression ${\displaystyle {\prod }_{i=1}^{n-1}\left(x_i-x_0\right)}$ i.e. all the factors in the formula that include $x_0$ .

After taking factors out,a $n\times n$ Vandermonde matrix on the variables $x_1,\mathrm{...},x_{n-1}$ is left.This will only contain all the other factors that are needed to get the right hand side.

Thus, this completes the induction.