Vandermonde matrix and determinant
I failed to mention the Vandermonde matrix because I couldn't see how it fit in with the story. Let x_{1}, ..., x_{n} be distinct numbers and let
P(x)  = (x  x_{1})· ... ·(x  x_{n}) 
 = x^{n} + a_{1}x^{n1} + a_{2}x^{n2} + ... + a_{n}.

Let
G = 
  0  1  0  ...  ...   
  0  0  1  ...  ...   
  .  .  .  .  .   
  0  .  .  .  1   
  a_{n}  a_{n1}  a_{n2}  ...  a_{1}   

be the companion matrix of P(x). Then (1, x_{i}, x_{i}^{2}, ..., x_{i}^{n1})^{T} is an eigenvector of G with eigenvalue x_{i}. Since the x_{i} are distinct, these eigenvectors are linearly independent. Hence the Vandermonde matrix, whose columns are these vectors, is nonsingular.
[an error occurred while processing this directive]
Contact
Front page
Contents
Algebra
Up
Copyright © 19962018 Alexander Bogomolny
[an error occurred while processing this directive]