4.3 Skew-Symmetric Determinants 75
In the particular case in whichaij=1,i<j, denote Pfnby pf
n
and
denote Pf
(n)
r by pf
(n)
r.
Lemma.
pf
n
=1.
The proof is by induction. Assume pf
m
=1,m<n, which implies
pf
(n)
r
= 1. Then, from (4.3.19),
pfn=
2 n− 1
∑
r=1
(−1)
r+1
=1.
Thus, if every Pfaffian of orderm<nis equal to 1, then every Pfaffian of
ordernis also equal to 1. But from (4.3.16), pf
1
= 1, hence pf
2
= 1, which
is confirmed by (4.3.17), pf
3
= 1, and so on.
The following important theorem relates Pfaffians to skew-symmetric
determinants.
Theorem.
A 2 n= [Pfn]
2
.
The proof is again by induction. Assume
A 2 m= [Pfm]
2
,m<n,
which implies
A
(2n−1)
ii
=
[
Pf
(n)
i
] 2
.
Hence, referring to (4.3.9),
[
A
(2n−1)
ij
] 2
=A
(2n−1)
ii
A
(2n−1)
jj
=
[
Pf
(n)
i Pf
(n)
j
] 2
A
(2n−1)
ij
Pf
(n)
i Pf
(n)
j
=± 1 (4.3.22)
for all elementsaijfor whichaji=−aij. Letaij=1,i<j. Then
A
(2n−1)
ij
→E
(2n−1)
ij
=(−1)
i+j
,
Pf
(n)
i →pf
(n)
i =1.
Hence,
A
(2n−1)
ij
Pf
(n)
i
Pf
(n)
j
=
E
(2n−1)
ij
pf
(n)
i
pf
(n)
j
=(−1)
i+j
, (4.3.23)