Short Answer

Vandermonde determinants (introduced by Alexandre-Théophile Vandermonde). Consider distinct real numbers $a_{0}, a_{1}, . . . . ., a_{n .}$ . We define $(n + 1) \times (n + 1)$ the matrix
$A = [\begin{matrix} 1 & 1 & . . . . 1 \\ a_{0} & a_{1} & . . . . a_{n} \\ a_{0}^{2} & a_{1}^{2} & . . . . a_{1}^{2} \\ a_{0}^{n} & a_{1}^{n} & . . . . a_{n}^{n} \end{matrix}]$
Vandermonde showed that
$d e t (A) = \prod_{i > j} (a_{i} - a_{j})$
the product of all differences $(a_{i} - a_{j})$ , 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
$f (t) = \det [\begin{matrix} 1 & 1 & . . . & 1 & 1 \\ a_{0} & a_{1} & . . . & a_{n - 1} & t \\ a_{0}^{2} & a_{1}^{2} & . . . & a_{n - 1} & t^{2} \\ ⋮ & ⋮ & . . . & ⋮ & ⋮ \\ a_{0}^{n} & a_{1}^{n} & . . . & a_{n - 1}^{n} & t^{n} \end{matrix}]$
Explain why f(t) is a polynomial of $n^{t h}$ degree. Find the coefficient k of $t^{n}$ using Vandermonde's formula for $a_{0}, . . ., a_{n - 1}$ . Explain why
role="math" localid="1659522435181" $f (a_{0}) = f (a_{1}) = . . . = f (a_{n - 1}) = 0$
Conclude that
$f (t) = k (t - a_{0}) (t - a_{1}) . . . (t - a_{n - 1})$
for the scalar k you found above. Substitute $t = a_{n}$ to demonstrate Vandermonde's formula.

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

$f (t) = k \prod_{i = 0}^{n - 1} (t - a_{i})$

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Step by Step Solution

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TABLE OF CONTENTS

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

For $n = 1$ , we have a $2 \times 2$ matrix

$A = [\begin{matrix} 1 & 1 \\ a_{0} & a_{1} \end{matrix}]$

Using Vandermonde's formula, we have

$\prod_{i > j} (a_{i} - a_{j}) = a_{1} - a_{0} = d e t A$

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 $\prod_{i, j - 1 i > j}^{n - 1} (a_{i} - a_{j})$ .

For $t = a_{m}, m = 0, 1, . . ., n - 1$ , the $(m + 1)$ -th and the $(n + 1)$ -th column will be the same, thus the determinant will be 0 . So,

$f (a_{m}) = 0, \forall m = 0, 1, . . ., n$

For k being the Vandermonde's determinant for , we have exactly $f (t) = k \prod_{i = 0}^{n - 1} (t - a_{i})$ .

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

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

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Linear Algebra With Applications

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

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

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