Page 76 - Determinants and Their Applications in Mathematical Physics

4.1 Alternants 61

n

b. V n = V (x 2 ,x 3 ,...,x n ) (x r − x 1 ).
r=2
n

c. V (x t ,x t+1 ,...,x n )= V (x t+1 ,x t+2 ,...,x n ) (x r − x t ).
r=t+1
(n) n+1 (−1) n+1
d. V =(−1) V (x 2 ,x 3 ,...,x n )= V n .
1n n
(x r − x 1 )
r=2
(n)
e. V =(−1) n+i V (x 1 ,...,x i−1 ,x i+1 ,...,x n )
in
(−1) n+i
= V n , i > 1
i−1
(x i − x r ) (x r − x i )
n
r=1 r=i+1
f. If {j 1 j 2 ··· j n } is a permutation of {12 ...n}, then

1 2 ··· n

= sgn V (x 1 ,x 2 ,...,x n ).
V x j 1 ,x j 2 ,...,x j n
j 1 j 2 ··· j n
The proofs of (a) and (b) follow from the difference–product formula
in Section 4.1.2 and are elementary. A proof of (c) can be constructed as
follows. In (b), put n = m−t+1, then put x r = y r+t−1 , r =1, 2, 3,..., and
change the dummy variable in the product from r to s using the formula
s = r + t − 1. The resut is (c) expressed in different symbols. When t =1,
(c) reverts to (b). The proofs of (d) and (e) are elementary. The proof
of (f) follows from Property (c) in Section 2.3.1 and Appendix A.2 on
permutations and their signs.
Let the minors of V n be denoted by M ij . Then,

M i = M in = V (x 1 ,...,x i−1 ,x i+1 ,...,x n ),

M n = M nn = V n−1 .

Theorems.

m m−1
V (x m+1 ,x m+2 ,...,x n )V
a. M r = n , 1 ≤ m ≤ n − 1
V (x 1 ,x 2 ,...,x m )
r=1
n
n−2
b. M r = V
n
r=1
m m−1
V (x k m+1 ,x k m+2 ,...,x k n
)V
c. M k r = n
)
V (x k 1 ,x k 2 ,...,x k m
r=1
Proof. Use the method of induction to prove (a), which is clearly valid
when m = 1. Assume it is valid when m = s. Then, from Lemma (e) and