Geography Reference
In-Depth Information
F
12
F
22
=
1
G
22
(
e
11
− K
2
G
11
)
2
−
2
G
12
(
e
11
− K
2
G
11
)(
e
12
− K
2
G
12
)+
G
11
(
e
12
− K
2
G
12
)
2
×
−
.
(
e
12
−
K
2
G
12
)
×
e
11
−
K
2
G
11
Right eigenvalues
(the right general eigenvalue problem reduces to the right special eigenvalue problem):
|
E
r
−
κ
i
I
r
|
=0
,
tr[E
r
]
±
(tr[E
r
])
2
−
4det [E
r
]
=
κ
1
,
2
=
κ
±
=
1
2
(2.15)
E
11
+
E
22
±
(
E
11
+
E
22
)
2
+(2
E
12
)
2
.
=
1
2
Right eigencolumns:
f
11
f
21
=
E
22
− k
1
−
,
⎨
√
(
E
22
−k
1
)
2
+
E
12
1
E
12
F
r
=
f
11
f
12
(2.16)
f
21
f
22
f
12
f
22
=
⎩
−
E
12
√
(
E
11
−k
2
)
2
+
E
12
1
E
11
−
k
2
Since the right Frobenius matrix F
r
is an orthonormal matrix, it can be represented by
F
r
=
cos
φ
sin
φ
∀φ ∈
[0
,
2
π
]
,
−
sin
φ
cos
φ
(2.17)
E
12
E
11
−
2
E
12
E
11
−
tan
φ
=
,
tan2
φ
=
E
22
.
κ
−
End of Lemma.
Lemma
1.7
is the basis of the proof if we specialize G
r
=I
2
. Again, we emphasize that within the
right eigenspace analysis the right Frobenius matrix is orthonormal. As an orthonormal matrix,
i.e. F
r
2
×
2
F
r
F
r
=I
2
and det [F
r
]=+1
, it can be properly parameterized
by a rotation angle
φ
. Such an angle of rotation orientates the right eigenvectors
∈
SO(2) :=
{
F
r
∈
R
|
}
{
f
1
,
f
2
|
O}
2
=span
with respect to
. Indeed, the “ tan2
φ
identity” leads to an easy
computation of the orientation of the right eigenvectors.
{
e
1
,
e
2
|O}
,
R
{
e
1
,
e
2
}
Search WWH ::
Custom Search