The Vandermonde determinant

Posted: September 12, 2022 in Basic Algebra, Matrices
Tags: Vandermonde matrix

One of the determinants that appear a lot in mathematics is the so-called Vandermonde determinant.

Definition. The $n \times n$ Vandermonde matrix is defined as follows

$\displaystyle V(x_1, \cdots , x_n):=\begin{pmatrix}1 & x_1 & \cdots & x_1^{n-1} \\ 1 & x_2 & \cdots & x_2^{n-1} \\ \vdots & \vdots & \cdots & \vdots \\ 1 & x_n & \cdots & x_n^{n-1}\end{pmatrix}.$

The determinant of the matrix $V(x_1, \cdots ,x_n)$ is called the Vandermonde determinant.

Of course, the value of $\det V(x_1, \cdots , x_n)$ is quite well-known but let’s find it anyway.

Remark 1. Expanding along the first row, gives

$\det V(0,x_2, \cdots ,x_n)=x_2 \cdots x_n \det V(x_2, \cdots x_n).$

Problem 1. Show that

$\displaystyle \det V(x_1, \cdots ,x_n)=\prod_{n \ge i > j \ge 1}(x_i-x_j).$

Solution. The proof is by induction on $n.$ For $n=2$ we have $\det V(x_1,x_2)=x_2-x_1.$ For the general case, first note that if one of the $x_i$ ‘s is zero, say $x_1=0,$ then, by Remark 1,

$\det V(x_1, \cdots ,x_n)=\det V(0,x_2, \cdots ,x_n)=x_2 \cdots x_n \det V(x_2, \cdots ,x_n)$

and we are done by induction. Suppose now that $x_i \ne 0$ for all $i.$ Consider the function

$p(t):=\det V(t,x_2, \cdots ,x_n),$

which is clearly a polynomial of degree at most $n-1.$ It is also clear that $p(x_2)= \cdots p(x_n)=0.$ Thus $p(t)=c(t-x_2) \cdots (t-x_n),$ for some constant $c.$ Now, using Remark 1,

$(-1)^{n-1}cx_2 \cdots x_n=p(0)=\det V(0, x_2, \cdots , x_n)= x_2 \cdots x_n \det V(x_2, \cdots , x_n),$

which gives $c=(-1)^{n-1}\det V(x_2, \cdots , x_n).$ Hence

$\begin{aligned}p(t)=(-1)^{n-1}(t-x_2) \cdots (t-x_n)\det V(x_2, \cdots , x_n)=(x_2-t) \cdots (x_n-t)\det V(x_2, \cdots , x_n)\end{aligned}$

and so $\det V(x_1, \cdots , x_n)=p(x_1)=(x_2-x_1) \cdots (x_n-x_1)\det V(x_2, \cdots , x_n).$ We are now done by induction. $\ \Box$

Remark 2. By Problem 1, $V(x_1, \cdots , x_n)$ is invertible if and only if $x_i \ne x_j$ for all $i \ne j.$

We now use Problem 1 to find the determinant of a slightly more complicated matrix.

Problem 2. Let $A$ be the $n \times n$ matrix whose $(i,j)$ -entry is $(x_i+y_j)^{n-1}.$ Show that

$\displaystyle \det A=\prod_{k=1}^{n-1}\binom{n-1}{k}\prod_{n \ge i > j \ge 1}(x_i-x_j)(y_j-y_i).$

Solution. Since

$\displaystyle (x_i+y_j)^{n-1}=\sum_{k=0}^{n-1}\binom{n-1}{k}x_i^ky_j^{n-1-k},$

we have $A=BC,$ where

$\displaystyle B=\begin{pmatrix}1 & \binom{n-1}{1}x_1 & \cdots & \binom{n-1}{n-1}x_1^{n-1} \\ 1 & \binom{n-1}{1}x_2 & \cdots & \binom{n-1}{n-1}x_2^{n-1} \\ \vdots & \vdots & \cdots & \vdots \\ 1 & \binom{n-1}{1}x_n & \cdots & \binom{n-1}{n-1}x_n^{n-1}\end{pmatrix}, \ \ \ \ \ \ C=\begin{pmatrix}y_1^{n-1} & y_2^{n-1} & \cdots & y_n^{n-1} \\ y_1^{n-2} & y_2^{n-2} & \cdots & y_n^{n-2} \\ \vdots & \vdots & \cdots & \vdots \\ 1 & 1 & \cdots & 1\end{pmatrix}.$