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6.10 The Einstein and Ernst Equations 289
Proof. Multiply the rth row of A by ω −r ,1 ≤ r ≤ n and the sth column
by ω ,1 ≤ s ≤ n. The effect of these operations is to multiply A by the
s
factor 1 and to multiply the element a rs by ω s−r . Hence, by (6.10.6), A is
transformed into B and the lemma is proved.
Unlike A, which is real, the cofactors of A are not all real. An example
is given in the following lemma.
Lemma 6.18.
2
A 1n = ω n−1 B 1n (ω = −1).
Proof.
A 1n =(−1) n+1 |e rs | n−1 ,
where
e rs = a r+1,s
= ω |r−s+1| u |r−s+1|
= a r,s−1
and
B 1n =(−1) n+1 |β rs | n−1 ,
where
β rs = b r+1,s
= b r,s−1 ,
that is,
β rs = ω s−r−1 e rs .
(n) −r−1
Multiply the rth row of A by ω ,1 ≤ r ≤ n − 1 and the sth column
1n
(n)
by ω ,1 ≤ s ≤ n − 1. The effect of these operations is to multiply A by
s
1n
the factor
ω −(2+3+···+n)+(1+2+3+···+n−1) = ω 1−n
and to multiply the element e rs by ω s−r−1 . The lemma follows.
Both A and B are persymmetric (Hankel) about their secondary diag-
onals. However, A is also symmetric about its principal diagonal, whereas
B is neither symmetric nor skew-symmetric about its principal diagonal.
In the analysis which follows, advantage has been taken of the fact that A
with its complex elements possesses a higher degree of symmetry than B
with its real elements. The expected complicated analysis has been avoided
by replacing B and its cofactors by A and its cofactors.