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Right eigencolumns:
⎧
⎨
f
11
f
21
=
C
22
−
,
λ
1
√
(
C
22
−λ
1
)
2
+
C
12
1
−
C
12
F
r
=
f
11
f
12
(2.5)
f
12
f
22
=
.
f
21
f
22
⎩
C
12
C
11
− λ
2
−
√
(
C
11
−λ
2
)
2
+
C
12
1
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
(2.6)
C
12
C
11
−
2
C
12
C
11
−
tan
ϕ
=
,
tan2
ϕ
=
C
22
.
λ
2
−
End of Lemma.
The proof of Lemma
2.1
is straightforward from Lemma
1.6
as soon as we specialize G
r
=I
2
.Of
special interest is the right eigenspace analysis. Here, the right Frobenius matrix F
r
is orthonor-
mal. As an orthonormal matrix (also called “proper rotation matrix”), it can be parameterized
by a rotation angle
ϕ
. Such an angle of rotation orientates the right eigenvectors
{
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. We proceed to a short example.
{
e
1
,
e
2
|O}
,
R
{
e
1
,
e
2
}
Example 2.1 (Orthogonal projection of points of the sphere
S
2
R
+
onto the equatorial plane
2
O
P
through the origin
O
).
In Example
1.6
, we presented already to you the special map projection of the hemisphere
S
2
R
+
onto
2
O
the central equatorial plane
by computing its characteristic right Cauchy-Green deformation
tensor as well as its right eigenspace. Here, we aim at testing the right Frobenius matrix F
r
on
orthonormality. Let us transfer the right eigencolumns to build up
F
r
=
f
11
f
12
P
=
xy
y −x
.
1
x
2
+
y
2
−
(2.7)
f
21
f
22
Is this Frobenius matrix of integrating factors an orthonormal matrix? Please test F
r
F
r
=I
2
to
convince yourself. Here, we generate
F
r
cos
ϕ
sin
ϕ
=
xy
y
,
1
x
2
+
y
2
−
(2.8)
−
sin
ϕ
cos
ϕ
−
x
y
x
,
tan 2
ϕ
=
2tan
α
1
−
tan
2
α
=
2
xy
x
2
tan
ϕ
=
−
−
− y
2
,
(2.9)
x
2
y
2
xy
−
C
12
=
−
(
x
2
+
y
2
)
, C
11
− C
22
=
−
(
x
2
+
y
2
)
,
(2.10)
R
2
R
2
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