Page 90 - Determinants and Their Applications in Mathematical Physics

4.3 Skew-Symmetric Determinants 75

In the particular case in which a ij =1, i<j, denote Pf n by pf and
n
denote Pf (n) by pf (n) .
r r
Lemma.
pf =1.
n
The proof is by induction. Assume pf =1, m<n, which implies
m
(n)
pf = 1. Then, from (4.3.19),
r
2n−1
r+1
pf = (−1) =1.
n
r=1
Thus, if every Pfaffian of order m<n is equal to 1, then every Pfaffian of
order n is also equal to 1. But from (4.3.16), pf = 1, hence pf = 1, which
2
1
is confirmed by (4.3.17), pf = 1, and so on.
3
The following important theorem relates Pfaffians to skew-symmetric
determinants.
Theorem.
2
A 2n = [Pf n ] .
The proof is again by induction. Assume
2
A 2m = [Pf m ] , m < n,
which implies
(2n−1) (n) 2

A = Pf .
ii i
Hence, referring to (4.3.9),
(2n−1) 2 (2n−1) (2n−1)

A = A A
ij ii jj

(n) (n) 2
= Pf Pf
i j
(2n−1)
A
= ±1 (4.3.22)
ij
(n) (n)
Pf Pf
i j
for all elements a ij for which a ji = −a ij . Let a ij =1, i<j. Then
(2n−1) (2n−1)
A → E =(−1) i+j ,
ij ij
(n) (n)
Pf → pf =1.
i i
Hence,
(2n−1) (2n−1)
A E
= =(−1) , (4.3.23)
ij ij i+j
(n) (n) (n) (n)
Pf Pf pf pf
i j i j