Determinants and Their Applications in Mathematical Physics

4.3 Skew-Symmetric Determinants 71

b.

2 n

∑

k=i

Ejk=(−1)

j+1

δi,odd,i≤j

=(−1)

j+1

δi,even, i>j

c.

i− 1

∑

k=1

Ejk=(−1)

j+1

δi,even,i≤j

=(−1)

j+1

δi,odd, i>j.

Proof. Referring to Lemma 4.14(b,c),

2 n

∑

k=1

Ejk=

j− 1

∑

k=1

Ejk+Ejj+

2 n

∑

k=j+1

Ejk

=−

j− 1

∑

k=1

(−1)

j+k+1

+0+

2 n

∑

k=j+1

(−1)

j+k+1

=(−1)

j+1

(δj,even+δj,odd)

=(−1)

j+1

,

which proves (a).

Ifi≤j,

2 n

∑

k=i

Ejk=

[

2 n

∑

k=1

−

i− 1

∑

k=1

]

Ejk

=(−1)

j+1

+

i− 1

∑

k=1

(−1)

j+k+1

=(−1)

j+1

(1−δj,even)

=(−1)

j+1

δi,odd.

Ifi>j,

2 n

∑

k=i

Ejk=

2 n

∑

k=i

(−1)

j+k+1

=(−1)

j+1

δi,even,

which proves (b).

i− 1

∑

k=1

Ejk=

[

2 n

∑

k=1

−

2 n

∑

k=i

]

Ejk.

Part (c) now follows from (a) and (b).

LetEnbe a skew-symmetric determinant defined as follows:

En=|εij|n,

whereεij=1,i<j, andεji=−εij, which impliesεii=0.