Page 86 - Determinants and Their Applications in Mathematical Physics

4.3 Skew-Symmetric Determinants 71

2n
j+1
b. E jk =(−1) δ i,odd , i ≤ j
k=i j+1
=(−1) δ i,even , i>j
i−1
j+1
c. E jk =(−1) δ i,even , i ≤ j
k=1
=(−1) j+1 δ i,odd , i>j.
Proof. Referring to Lemma 4.14(b,c),
j−1
2n 2n

E jk = E jk + E jj + E jk
k=1 k=1 k=j+1
j−1 2n
j+k+1 j+k+1
= − (−1) +0+ (−1)
k=1 k=j+1
=(−1) j+1 (δ j,even + δ j,odd )
=(−1) j+1 ,
which proves (a).
If i ≤ j,

i−1

2n 2n

E jk = − E jk
k=1 k=1
k=i
i−1
=(−1) j+1 + (−1) j+k+1
k=1
=(−1) j+1 (1 − δ j,even )
=(−1) j+1 δ i,odd .
If i>j,

2n 2n
j+k+1
E jk = (−1)
k=i k=i
j+1
=(−1) δ i,even ,
which proves (b).
i−1
2n
2n

E jk = − E jk .
k=1 k=1 k=i
Part (c) now follows from (a) and (b).
Let E n be a skew-symmetric determinant defined as follows:
E n = |ε ij | n ,
where ε ij =1, i<j, and ε ji = −ε ij , which implies ε ii =0.