Page 98 - Determinants and Their Applications in Mathematical Physics

4.4 Circulants 83
n−1

n

= exp ω (r−1)t x t
r=1 t=1
n−1 n

= exp x t ω (r−1)t
t=1 r=1
= exp(0).
The lemma follows.
Theorem.
A = A(H 1 ,H 2 ,H 3 ,...,H n )=1.

Proof. The definition (4.4.16) implies that
 
H 1 H 2 H 3 ··· H n
H 1 H 2 ··· H n−1 
 H n
 
A(H 1 ,H 2 ,H 3 ,...,H n )=  H n−1 H n H 1 ··· H n−2 
 
··· ··· ··· ··· ···
H 2 H 3 H 4 ··· H 1
n

= W −1 diag E 1 E 2 E 3 ...E n W. (4.4.18)
Taking determinants,
n

−1
A(H 1 ,H 2 ,H 3 ,...,H n )= W W E r .

r=1
The theorem follows from Lemma 4.19.
Illustrations

When n =2, ω = exp(iπ)= −1.

1 1
W = ,
1 −1
1
W −1 = W,
2
E r = exp[(−1) r−1 x 1 ], r =1, 2.
Let x 1 → x; then,

E 1 = e ,
x
E 2 = e −x ,

H 1 1 1 1 e x
= ,
H 2 2 1 −1 e −x
H 1 =ch x,
H 2 =sh x,