Determinants and Their Applications in Mathematical Physics

70 4. Particular Determinants

In detail,

Bn=

∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣

− 1 • 11 ··· 11

− 1 − 1 • 1 ··· 11

− 1 − 1 − 1 • ··· 11

.................................

− 1 − 1 − 1 − 1 ··· − 1 •

− 1 − 1 − 1 − 1 ··· − 1 − 1

∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ n

.

Lemma 4.13.

Bn=(−1)

n

.

Proof. Perform the column operation

C

′

2

=C 2 −C 1

and then expand the resulting determinant by elements from the newC 2.

The result is

Bn=−Bn− 1 =Bn− 2 =···=(−1)

n− 1

B 1.

ButB 1 =−1. The result follows.

Lemma 4.14.

a.

2 n

∑

k=1

(−1)

j+k+1

=0,

b.

i− 1

∑

k=1

(−1)

j+k+1

=(−1)

j

δi,even,

c.

2 n

∑

k=i

(−1)

j+k+1

=(−1)

j+1

δi,even,

where theδfunctions are defined in Appendix A.1. All three identities follow

from the elementary identity

q

∑

k=p

(−1)

k

=(−1)

p

δq−p,even. 

Define the functionEijas follows:

Eij=







(−1)

i+j+1

,i<j

0 ,i=j

−(−1)

i+j+1

, i>j.

Lemma 4.15.

a.

2 n

∑

k=1

Ejk=(−1)

j+1

,