Page 158 - Determinants and Their Applications in Mathematical Physics

4.11 Hankelians 4 143

1 (−1) 1! 2! 3! ··· (n − 2)!
n(n−1)/2

b. = .
(i + j − 2)! n!(n + 1)! ··· (2n − 2)!
The second determinant is Hankelian.
Proof. Denote the first determinant by A n . Every element in the last
row of A n is equal to 1. Perform the column operations

C = C j − C j−1 , j = n, n − 1,n − 2,..., 2, (4.11.30)
j
which remove all the elements in the last row except the one in position
(n, 1). After applying the binomial identity

n n − 1 n − 1
− = ,
r r r − 1
the result is

n + j − 2
A n =(−1) n+1 . (4.11.31)
n − i − 1
n−1
Once again, every element in the last row is equal to 1. Repeat the column
operations with j = n − 1,n − 2,..., 2 and apply the binomial identity
again. The result is

n + j − 2
A n = − . (4.11.32)
n − i − 2
n−2
Continuing in this way,

n + j − 2

A n =+
n − i − 4

n−4

n + j − 2
= −
n − i − 6
n−6

n + j − 2

=+
n − i − 8

n−8

n + j − 2

= ±
2 − i

2
= ±1, (4.11.33)

1 when n =4m, 4m +1
sign(A n )= (4.11.34)
−1 when n =4m − 2, 4m − 1,
which proves (a).
Denote the second determinant by B n . Divide R i by (n−i)!, 1 ≤ i ≤ n−1,
and multiply C j by (n + j − 2)!, 1 ≤ j ≤ n. The result is

(n − 1)! n!(n + 1)! ··· (2n − 2)! (n + j − 2)!
B n =
(n − 1)! (n − 2)! (n − 3)! ··· 1! (n − i)! (i + j − 2)!
n