Determinants and Their Applications in Mathematical Physics

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)