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and plugging this equation into (8.12), the homography for the second solution
(
λ
1
=
λ
2
) can be expressed as follows:
H
=
λ
−
3
1)(
Ue
3
)(
Ue
3
)
)
−
.
λ
(
I
+(
Setting
v
=
Ue
3
, we finally find that
H
must have the following form:
H
=
λ
−
3
1)
vv
)
λ
(
I
+(
−
where
v
is a unitary vector:
v
= 1and
λ
is the unique real constant value that
2
+ 1 = 0.
It remains to show that
A
= 0. The inverse of
H
is
3
verifies the equation
λ
−
λ
H
−
1
=
λ
−
1
(
I
+(
3
1)
vv
)
λ
−
.
The derivative of
H
is
˙
H
=
λ
−
3
1)(
vv
+
v v
)
λ
(
−
so that
A
=
H
−
1
˙
H
=(
λ
−
3
3
1)
vv
)(
vv
+
v v
)
−
1)(
I
+(
λ
−
.
Knowing that
v
v
= 0, this equation becomes:
A
=(
λ
−
3
1)(
vv
+
v v
+(
3
1)
v v
)
−
λ
−
3
=
2
and knowing that
λ
λ
−
1, we obtain:
A
=(
λ
−
3
1)(
vv
+
v v
+(
2
2)
v v
)
−
λ
−
(8.26)
λ
−
3
1)(
vv
−
v v
+
2
v v
)
=(
−
λ
.
(8.27)
Since
vv
−
v v
=[
v
1
, we finally obtain
×
v
]
×
=[[
v
]
×
v
]
×
A
=(
λ
−
3
2
v v
)
−
1)([[
v
]
×
v
]
×
+
λ
.
Since [[
v
]
×
v
]
×
is a skew-symmetric matrix, the diagonal elements of
A
are
a
ii
=
(
2
v
i
v
i
. Knowing that each
a
ii
is constant we have two possible cases. The
first one is
a
ii
= 0 for each
i
.Then
v
is constant so that
H
is also constant and
A
= 0.
If there exists
i
such that
a
ii
λ
−
3
−
1)
λ
= 0, then there exists
i
such that
a
ii
<
0. This is due
to the fact that
A
∑
i
a
ii
= 0. In this case, the corresponding
v
i
diverges to infinity because
v
i
v
i
is a strictly positive constant. This contradicts the
fact that
∈
sl
(3) and therefore
= 1. This concludes the proof of part
i)
of the theorem.
Let us prove part
ii)
. We compute the linearization of system (8.10) at
E
s
=(
I
v
,
0).
Let us define
X
1
and
X
2
as elements of
sl(3) corresponding to the first order approx-
imations of
H
and
A
around (
I
,
0):
1
[
v
]
×
represents the skew-symmetric matrix associated with the cross-product by
v
∈
R
3
,
3
.
i.e.
[
v
]
×
y
=
v
×
y
, ∀
y
∈
R
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