Page 121 - Determinants and Their Applications in Mathematical Physics

106 4. Particular Determinants

Let
A n = |φ i+j−2 | n = |φ m | n , 0 ≤ m ≤ 2n − 2,
(4.8.9)
B n = |x i+j−2 φ i+j−2 | n = |x φ m | n , 0 ≤ m ≤ 2n − 2.
m
Lemma.

a. B n = x n(n−1) A n .
(n) n(n−1)−(i+j−2) (n)
b. B = x A .
ij ij
c. B ij = x −(i+j−2) A .
ij
n n
Proof of (a). Perform the following operations on B n : Remove the factor
x i−1 from row i,1 ≤ i ≤ n, and the factor x j−1 from column j,1 ≤ j ≤ n.
The effect of these operations is to remove the factor x i+j−2 from the
element in position (i, j).
The result is
B n = x 2(1+2+3+···+n−1) A n ,

which yields the stated result. Part (b) is proved in a similar manner, and
(c), which contains scaled cofactors, follows by division.

4.8.2 Hankelians Whose Elements are Differences
The h difference operator ∆ h is defined in Appendix A.8.

Theorem.
|φ m | n = |∆ φ 0 | n ;
m
h
that is, a Hankelian remains unaltered in value if each φ m is replaced by
∆ φ 0 .
m
h
Proof. First Proof. Denote the determinant on the left by A and perform
the row operations
i−1
i − 1

R = (−h) r R i−r , i = n, n − 1,n − 2,..., 2, (4.8.10)
r
i
r=0
on A. The result is
i−1
A = ∆ φ j−1 . (4.8.11)

h
n
Now, restore symmetry by performing the same operations on the columns,
that is,
j−1
j − 1

C = (−h) r C j−r , j = n, n − 1,n − 2,..., 2. (4.8.12)
r
j
r=0
The theorem appears. Note that the values of i and j are taken in
descending order of magnitude.