Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $x_1, \ldots, x_n$ be a collection of distinct elements of a field $\mathbb{F}$ and let $B \in M_n(\mathbb{F})$ be the matrix with $a_{ij} = x^{j-1}_i$. Prove that $\operatorname{det}B= \prod_{i<j}(x_j - x_i)$. And find the matrix $B^{-1}$.

So I know that the matrix has the form

$\begin{pmatrix} 1 & x_1 & x_1^2 &\dots &x_1^{n-1}\\ 1 & x_2 & x_2^2 &\dots &x_2^{n-1}\\ \vdots &&&&\vdots\\ 1 & x_n & x_n^2 &\dots &x_n^{n-1}\\ \end{pmatrix}$

Using row operations this matrix can become

$\begin{pmatrix} 1 & x_1 & x_1^2 &\dots &x_1^{n-1}\\ 0 & x_2-x_1 & x_2^2-x_1^2 &\dots &x_2^n-x_1^{n-1}\\ \vdots &&&&\vdots\\ 0 & x_n-x_{n-1} & x_n^2-x_{n-1}^2 &\dots &x_n^n-x_{n-1}^{n-1}\\ \end{pmatrix}$

Not completely sure what to do from here. And I don't know how to find $B^{-1}$.

share|improve this question
2  
This might be of help. –  Git Gud Feb 12 '13 at 7:29
2  
Your last column should have $x_{i}^{n-1}$ to make the matrix square. –  Andreas Caranti Feb 12 '13 at 7:40
1  
I'd also recommend this Inverse of Vandermonde's Matrix –  Kaster Feb 12 '13 at 7:55
    
@user60504 Do you still need help with this? –  Git Gud Feb 12 '13 at 19:54
    
No. I think I got it. Thanks for all your help! –  user4593 Feb 12 '13 at 20:15

1 Answer 1

A way to calculate the determinant is to see it as a polynomial in $x = x_{n}$. Clearly it has roots $x_{1}, \dots , x_{n-1}$ (a matrix with two equal rows has determinant zero), Thus the determinant is $$ A \cdot (x-x_{1}) \cdot \dots \cdot (x-x_{n-1}), $$ where $A$ is a polynomial in $x_{1}, \dots , x_{n-1}$. Now notice that $A$ is the coefficient of $x^{n-1}$ in this polynomial, and expanding the determinant with respect to the last row you see that this is the same (Vandermonde) determinant of size one less.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.