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1st left eigencolumn 1 :
F 11
F 21
=
1
( c 22
Λ 1 G 12 ) 2 G 22 ×
Λ 1 G 22 ) 2 G 11
2( c 12
Λ 1 G 12 )( c 22
Λ 1 G 22 ) G 12 +( c 12
c 22
,
Λ 1 G 22
×
Λ 1 G 12 )
( c 12
(1.60)
2nd left eigencolumn 2 :
F 12
F 22
=
1
( c 11
Λ 2 G 12 ) 2 G 11 ×
Λ 2 G 11 ) 2 G 22
Λ 2 G 11 )( c 12
Λ 2 G 12 ) G 12 +( c 12
2( c 11
( c 12 − Λ 2 G 12 )
c 11
,
×
Λ 2 C 11
1st right eigencolumn 1 :
f 11
f 21
=
1
( C 22
λ 1 g 12 ) 2 g 22 ×
λ 1 g 22 ) 2 g 11
2( C 12
λ 1 g 12 )( C 22
λ 1 g 22 ) g 12 +( C 12
C 22
,
λ 1 g 22
×
λ 1 g 12 )
( C 12
(1.61)
2nd right eigencolumn 2 :
f 12
f 22
=
1
( C 11 − λ 2 g 11 ) 2 g 22 2( C 11 − λ 2 g 11 )( C 12 − λ 2 g 12 ) g 12 +( C 12 − λ 2 g 12 ) 2 g 11 ×
.
λ 2 g 12 )
( C 12
×
λ 2 g 11
C 11
End of Lemma.
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 ,wehavechosen
the solution sign
. Furthermore, note that the proof for representing the right
eigencolumns or right eigendirections runs analogously. The dimension four of the solution space
of eigencolumns or eigendirections has already been documented by Gere and Weaver ( 1965 ), for
instance.
{
F 11 ,F 22
}
=
{
+ , +
}
Proof (1st and 2nd left eigencolumns).
1st left eigencolumn, Λ 1 :
c 11
F 11
F 21
= 0
,
Λ 1 G 11
Λ 1 G 12
c 12
(1.62)
Λ 1 G 12
Λ 1 G 22
c 12
c 22
0
2nd identity:
Λ 1 G 12 ) F 11 +( c 22
Λ 1 G 22 ) F 21 =0=
( c 12
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