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
,