16
1 From Riemann manifolds to Riemann manifolds
A sketch of a proof is presented in the following. Note that there are four pairs of {F 11 , F 22 }
dependent on the sign choice {+, +}, {+, −},{−, +}, and {−, −}. In Lemma 1.6, we have chosen
the solution sign {F 11 , F 22 } = {+, +}. Furthermore, note that the proof for representing the right
eigencolumns or right eigendirections runs analogeously. The dimension four of the solution space of
eigencolumns or eigendirections has already been documented by J. M. Gere and W. Weaver (1965),
for instance.
Proof (1st and 2nd left eigencolumns).
1st left eigencolumn, Λ 1 :
c 11 − Λ
2
1 G 11 c 12 − Λ
2
1 G 12
c 12 − Λ
2
1 G 12 c 22 − Λ
2
1 G 22
F 11
F 21
=
0
0
,
(1.62)
2nd identity:
c 12 − Λ
2
1 G 12
F 11 +
c 22 − Λ
2
1 G 22
F 21 = 0 =⇒
=⇒ F 21 = −
c 12 − Λ
2
1 G 12
c 22 − Λ 2
1 G 22
F 11 ⇐⇒
⎡
⎣
F 11
F 21
⎤
⎦ = F 11
⎡
⎣
1
−
c 12 −Λ
2
1 G 12
c 22 −Λ 2
1 G 22
⎤
⎦ .
(1.63)
2nd left eigencolumn, Λ 2 :
c 11 − Λ
2
2 G 11 c 12 − Λ
2
2 G 12
c 12 − Λ
2
2 G 12 c 22 − Λ
2
2 G 22
F 21
F 22
=
0
0
,
(1.64)
1st identity:
c 11 − Λ
2
2 G 11
F 21 +
c 12 − Λ
2
2 G 12
F 22 = 0 =⇒
=⇒ F 12 = −
c 12 − Λ
2
2 G 12
c 11 − Λ 2
2 G 11
F 22 ⇐⇒
⎡
⎣
F 12
F 22
⎤
⎦ = F 22
⎡
⎣
−
c 12 −Λ
2
2 G 12
c 11 −Λ 2
2 G 11
1
⎤
⎦ .
(1.65)
Left conditions:
F
T
l G l F l = I 2 ⇐⇒
F 11 F 21
F 12 F 22
G 11 G 12
G 12 G 22
F 11 F 12
F 21 F 22
=
1 0
0 1
.
(1.66)
1st and 2nd partitioning:
F 11 , F 21
G 11 G 12
G 12 G 22
F 11
F 21
= 1 ,
F 12 , F 22
G 11 G 12
G 12 G 22
F 12
F 22
= 1 .
(1.67)
1 From Riemann manifolds to Riemann manifolds
A sketch of a proof is presented in the following. Note that there are four pairs of {F 11 , F 22 }
dependent on the sign choice {+, +}, {+, −},{−, +}, and {−, −}. In Lemma 1.6, we have chosen
the solution sign {F 11 , F 22 } = {+, +}. Furthermore, note that the proof for representing the right
eigencolumns or right eigendirections runs analogeously. The dimension four of the solution space of
eigencolumns or eigendirections has already been documented by J. M. Gere and W. Weaver (1965),
for instance.
Proof (1st and 2nd left eigencolumns).
1st left eigencolumn, Λ 1 :
c 11 − Λ
2
1 G 11 c 12 − Λ
2
1 G 12
c 12 − Λ
2
1 G 12 c 22 − Λ
2
1 G 22
F 11
F 21
=
0
0
,
(1.62)
2nd identity:
c 12 − Λ
2
1 G 12
F 11 +
c 22 − Λ
2
1 G 22
F 21 = 0 =⇒
=⇒ F 21 = −
c 12 − Λ
2
1 G 12
c 22 − Λ 2
1 G 22
F 11 ⇐⇒
⎡
⎣
F 11
F 21
⎤
⎦ = F 11
⎡
⎣
1
−
c 12 −Λ
2
1 G 12
c 22 −Λ 2
1 G 22
⎤
⎦ .
(1.63)
2nd left eigencolumn, Λ 2 :
c 11 − Λ
2
2 G 11 c 12 − Λ
2
2 G 12
c 12 − Λ
2
2 G 12 c 22 − Λ
2
2 G 22
F 21
F 22
=
0
0
,
(1.64)
1st identity:
c 11 − Λ
2
2 G 11
F 21 +
c 12 − Λ
2
2 G 12
F 22 = 0 =⇒
=⇒ F 12 = −
c 12 − Λ
2
2 G 12
c 11 − Λ 2
2 G 11
F 22 ⇐⇒
⎡
⎣
F 12
F 22
⎤
⎦ = F 22
⎡
⎣
−
c 12 −Λ
2
2 G 12
c 11 −Λ 2
2 G 11
1
⎤
⎦ .
(1.65)
Left conditions:
F
T
l G l F l = I 2 ⇐⇒
F 11 F 21
F 12 F 22
G 11 G 12
G 12 G 22
F 11 F 12
F 21 F 22
=
1 0
0 1
.
(1.66)
1st and 2nd partitioning:
F 11 , F 21
G 11 G 12
G 12 G 22
F 11
F 21
= 1 ,
F 12 , F 22
G 11 G 12
G 12 G 22
F 12
F 22
= 1 .
(1.67)
