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Linear Algebra With Applications
Found in: Page 290
Linear Algebra With Applications

Linear Algebra With Applications

Book edition 5th
Author(s) Otto Bretscher
Pages 442 pages
ISBN 9780321796974

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Short Answer

Vandermonde determinants (introduced by Alexandre-Théophile Vandermonde). Consider distinct real numbers a0,a1,.....,an. . We define (n+1)×(n+1) the matrix


Vandermonde showed that


the product of all differences (ai-aj) , where exceeds j . a. Verify this formula in the case of n=1 . b. Suppose the Vandermonde formula holds for n=1 . You are asked to demonstrate it for n. Consider the function


Explain why f(t) is a polynomial of nthdegree. Find the coefficient k of tn using Vandermonde's formula for a0,...,an-1. Explain why

role="math" localid="1659522435181" f(a0)=f(a1)=...=f(an-1)=0

Conclude that


for the scalar k you found above. Substitute t=an to demonstrate Vandermonde's formula.

Therefore, the being the Vandermonde's determinant for , we have exactly .


See the step by step solution

Step by Step Solution

Step 1: (a) By using Vandermonde’s Formula. 

For n=1, we have a 2×2 matrix


Using Vandermonde's formula, we have


Step 2: (b) To  Find the coefficient k of tn using Vandermonde's formula.

By the Laplace expansion along the n+1 -th column, we see that f is a polynomial of n -th degree, the coefficient of being in fact the Vandermonde's determinant for n-1 , which is i,j-1i>jn-1ai-aj.

For t=am,m=0,1,...,n-1 , the m+1 -th and the n+1 -th column will be the same, thus the determinant will be 0 . So,


For k being the Vandermonde's determinant for , we have exactly ft=ki=0n-1t-ai .

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