Page 104 - Determinants and Their Applications in Mathematical Physics

4.5 Centrosymmetric Determinants 89

The element t 2n−1 does not appear in T n but appears in the bottom right-
hand corner of B n . Hence,

= R n−1 ,
∂R n
∂t 2n−1
= −S n−1 . (4.5.14)
∂S n
∂t 2n−1
The same element appears in positions (1, 2n) and (2n, 1) in T 2n . Hence,
referring to the second line of (4.5.12),
1 ∂T (2n)
(2n)
T =
1,2n
2 ∂t 2n−1
∂
=2 (R n S n )
∂t 2n−1
=2(R n−1 S n − R n S n−1 ). (4.5.15)
Replacing n by 2n in (4.5.13),
(2n) 2
2

T 2n−1 = T 2n T 2n−2 + T 1,2n
2
=4 4R n S n R n−1 S n−1 +(R n−1 S n − R n S n−1 )
2
=4(R n−1 S n + R n S n−1 ) .
The sign of T 2n−1 is decided by putting t 0 = 1 and t r =0, r> 0. In that
1
case, T n = I n , B n = O n , R n = S n = . Hence, the sign is positive:
2
T 2n−1 =2(R n−1 S n + R n S n−1 ). (4.5.16)
Part (a) of the theorem follows from (4.5.11).
The element t 2n appears in the bottom right-hand corner of E n but does
not appear in either T n or A n . Hence, referring to (4.5.10),

= −P n−1 ,
∂P n
∂t 2n
= Q n−1 . (4.5.17)
∂Q n
∂t 2n
(2n+1) 1 ∂T 2n+1
T 1,2n+1 =
2 ∂t 2n
∂
= (P n Q n+1 )
∂t 2n
= P n Q n − P n−1 Q n+1 . (4.5.18)
Return to (4.5.13), replace n by 2n + 1, and refer to (4.5.12):

T 2 = T 2n+1 T 2n−1 + T (2n+1) 2
2n 1,2n+1
=4P n Q n+1 P n−1 Q n +(P n Q n − P n−1 Q n+1 ) 2
2
=(P n Q n + P n−1 Q n+1 ) . (4.5.19)