4.4 Circulants 83
= exp
[
n
∑
r=1
n− 1
∑
t=1
ω
(r−1)t
xt
]
= exp
[
n− 1
∑
t=1
xt
n
∑
r=1
ω
(r−1)t
]
= exp(0).
The lemma follows.
Theorem.
A=A(H 1 ,H 2 ,H 3 ,...,Hn)=1.
Proof. The definition (4.4.16) implies that
A(H 1 ,H 2 ,H 3 ,...,Hn)=
H 1 H 2 H 3 ··· Hn
Hn H 1 H 2 ··· Hn− 1
Hn− 1 Hn H 1 ··· Hn− 2
··· ··· ··· ··· ···
H 2 H 3 H 4 ··· H 1
n
=W
− 1
diag
(
E 1 E 2 E 3 ...En
)
W. (4.4.18)
Taking determinants,
A(H 1 ,H 2 ,H 3 ,...,Hn)=
∣
∣
W
− 1
W
∣
∣
n
∏
r=1
Er.
The theorem follows from Lemma 4.19.
Illustrations
Whenn=2,ω= exp(iπ)=−1.
W=
[
11
1 − 1
]
,
W
− 1
=
1
2
W,
Er= exp[(−1)
r− 1
x 1 ],r=1, 2.
Letx 1 →x; then,
E 1 =e
x
,
E 2 =e
−x
,
[
H 1
H 2
]
=
1
2
[
11
1 − 1
][
e
x
e
−x
]
,
H 1 =chx,
H 2 =shx,