Page 196 - Determinants and Their Applications in Mathematical Physics

5.2 The Generalized Cusick Identities 181

s 2 = a i+1,j + φ i ψ j ,
s 3 = a i+1,j+1 .
The lemma follows.

Let
∗
A ∗ = |a | 2n ,
2n ij
1/2
∗

Pf = A ∗ . (5.2.15)
n 2n
Lemma 5.3.
2n−1
i+1 (n) 2n−i−1
(−1) Pf x =Pf ∗ n−1 .
i
i=1
Proof. Denote the sum by F n . Then, referring to (5.2.13) and Section 3.7
on bordered determinants,
2n−1 2n−1
2
x
F = (−1) i+j Pf (n) Pf (n) 4n−i−j−2
n i j
i=1 j=1
2n−1 2n−1
(2n−1) 4n−i−j−2
= A x
ij
i=1 j=1
x 2n−2
a 11 a 12 ··· a 1,2n−1

x 2n−3
a 21 a 22 ··· a 2,2n−1

. (5.2.16)

= − .........................................
1
a 2n−1,1 a 2n−1,2 ··· a 2n−1,2n−1
x x ··· 1 •
2n−2 2n−3
2n
(It is not necessary to put a ii = 0, etc., in order to prove the lemma.)
Eliminate the x’s from the last column and row by means of the row and
column operations

R = R i − xR i+1 , 1 ≤ i ≤ 2n − 2,
i

C = C j − xC j+1 , 1 ≤ j ≤ 2n − 2. (5.2.17)
j
The result is
a a ··· a
∗ ∗ ∗
11 12 1,2n−1 •
a a ··· a
∗ ∗ ∗
21 22 2,2n−1 •
2
F = − .....................................
n
a ∗ 2n−1,1 a ∗ 2n−1,2 ··· a ∗ 2n−1,2n−1 1

• • ··· 1 •
2n
∗
=+|a | 2n−2
ij
= A ∗ .
2n−2
The lemma follows by taking the square root of each side.