Page 96 - Determinants and Their Applications in Mathematical Physics

4.4 Circulants 81

Hence,

A = z r W r C 2 C 3 ··· C n . (4.4.10)

It follows that each z r ,0 ≤ r ≤ n − 1, is a factor of A n . Hence,
n−1

A n = K z r , (4.4.11)
r=0
but since A n and the product are homogeneous polynomials of degree n
in the a r , the factor K must be purely numerical. It is clear by comparing
the coefficients of a on each side that K = 1. The theorem follows from
n
1
(4.4.7).
3
Illustration. When n =3, ω r = exp(2riπ/3), ω =1.
r
ω 0 =1,
ω = ω 1 = exp(2iπ/3),
2
2
ω 2 = exp(4iπ/3) = ω = ω =¯ω,
1
2
ω = ω 1 = ω.
2
Hence,

a 1 a 2

a 3
A 3 = a 3 a 1

a 2
a 2 a 3 a 1

2
2
=(a 1 + a 2 + a 3 )(a 1 + ω 1 a 2 + ω a 3 )(a 1 + ω 2 a 2 + ω a 3 )
1 2
2
2
=(a 1 + a 2 + a 3 )(a 1 + ωa 2 + ω a 3 )(a 1 + ω a 2 + ωa 3 ). (4.4.12)
Exercise. Factorize A 4 .
4.4.3 The Generalized Hyperbolic Functions
Define a matrix W as follows:
(r−1)(s−1)
W = ω (ω = ω 1 )
n
1 1 1 1 ··· 1
 
 1 ω ω 2 ω 3 ··· ω n−1 
 1 ω 2 ω 4 ω 6 ··· ω 2n−2 


 1 ω ω ω ··· ω 
=  3 6 9 3n−3  . (4.4.13)
 
··· ··· ··· ··· ··· ···
2
1 ω n−1 ω 2n−2 ω 3n−3 ··· ω (n−1)
n
Lemma 4.18.
1
−1
W = W.
n
Proof.
−(r−1)(s−1)
W = ω .
n