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3.6 The Jacobi Identity and Variants 45
(n+1) (n)
But A = A . Hence, A n P = 0. The result follows.
i,n+1;j,n+1 ij
Three particular cases of (B) are required for the proof of the next
theorem.
Put (i, p, q)=(r, r, n), (n − 1,r,n), (n, r, n) in turn:
(n)
(n) (n+1)
A rr A rn − A n A =0, (B 1 )
(n+1)
A A
(n+1) r,n+1;rn
n+1,r n+1,n
(n) (n)
A A (n+1)
n−1,r n−1,n − A n A =0, (B 2 )
(n+1) (n+1) n−1,n+1;rn
A A
n+1,r n+1,n
(n)
(n) (n+1)
− A n A =0. (B 3 )
A nr
A nn
(n+1)
A A (n+1) n,n+1;rn
n+1,r n+1,n
Theorem 3.7.
(n+1) (n) (n)
A
r,n+1;rn A rr A rn
(n+1) (n) (n)
A A A =0.
n−1,n+1;rn n−1,r n−1,n
(n+1) (n)
A
(n)
n,n+1;rn A nr A nn
Proof. Denote the determinant by Q. Then,
(n) (n)
A A
n−1,r n−1,n
Q 11 = (n) (n)
A nr A nn
(n)
= A n A
n−1,n;rn
(n−1)
= A n A ,
n−1,r
Q 21 = −A n A (n−1) ,
rr
(n)
Q 31 = A n A . (3.6.17)
r,n−1;rn
Hence, expanding Q by the elements in column 1 and applying (B 1 )–(B 3 ),
(n+1) (n−1) (n+1) (n−1)
Q = A n A A − A A
r,n+1;rn n−1,r n−1,n+1;rn rr
(n+1) (n)
+ A A (3.6.18)
n,n+1;rn r,n−1;rn
(n) (n)
(n)
(n−1) (n) (n−1) A A
= A A rr A rn − A n−1,r n−1,n
(n+1)
(n+1)
(n+1)
A A (n+1) rr A A
n−1,r
n+1,r n+1,n
n+1,r n+1,n
(n)
(n) (n)
+ A A nr A nn
(n+1)
A A (n+1)
r,n−1;rn
n+1,r n+1,n
(n−1)
(n+1) (n) (n) (n)
= A A A − A rr A rr
(n−1)
(n)
A A
nr
n+1,n r,n−1;rn
n−1,r n−1,r
(n−1)
(n+1) (n) (n)
− A A n−1 A − A rr A rn . (3.6.19)
(n−1)
(n)
A A
n+1,r r,n−1;rn
n−1,r n−1,n