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H
A
≈
(
I
+
X
1
)
,
≈
X
2
.
Substituting these approximations into (8.10) and discarding all terms quadratic or
higher order in (
X
1
,
X
2
) yields
X
1
X
2
=
−
X
1
X
2
k
H
I
3
I
3
−
.
(8.28)
k
A
I
3
0
Since
k
H
,
0, the linearized error system is exponentially stable. This proves the
local exponential stability of the equilibrium (
I
k
A
>
,
0).
Lastly, let us prove part
iii)
. First, we remark that the function
V
is constant and
strictly positive on the set
E
u
. This can be easily verified from (8.8) and the defini-
tion of
E
u
, using the fact that on this set
A
= 0,
H
T
H
=
H
2
=
2
I
+(
1
λ
)
vv
T
,and
λ
−
λ
2
tr(
vv
T
)=1since
= 1. We denote by
V
u
the value of
V
on
E
u
. The fact that
V
u
is
strictly positive readily implies (in accordance with part
ii)
)that
E
s
is an asymptoti-
cally stable equilibrium, since
V
is nonincreasing along the system's solutions, and
each of them converges to
E
s
∪
v
E
u
. Using the same arguments, the proof of part
iii)
reduces to showing that for any point (
H
0
,
0)
∈
E
u
, and any neighborhood
U
of this
point, one can find (
H
1
,
A
1
)
∈U
such that
V
(
H
1
,
A
1
)
<
V
u
.
(8.29)
˙
Let
H
(
H
=
HC
with
C
a constant
.
) denote a smooth curve on
SL
(3), solution of
(3) that will be specified latter on. We also assume that (
H
(0)
element of
sl
,
0)
∈
E
u
.
−
H
(
t
)
2
Let
f
(
t
)=
I
/
2 so that, by (8.8),
f
(0)=
V
u
. The first derivative of
f
is
given by
−
H
(
t
))
T
˙
f
(
t
)=
H
(
t
))
−
tr((
I
−
H
(
t
))
T
H
(
t
)
C
)
=
−
tr((
I
(
H
T
(
t
)(
I
−
H
(
t
)))
=
−
P
,
C
.
For all elements (
H
0
,
(
H
0
(
I
−
H
0
)) = 0, so that
f
(0)=0. We now
0)
∈
E
u
, one has
P
calculate the second order derivative of
f
:
f
(
t
)=tr(
˙
H
(
t
)
T
˙
−
H
(
t
))
T
¨
H
(
t
))
H
(
t
))
−
tr((
I
tr
(
I
H
(
t
)
C
= tr(
˙
H
(
t
)
T
˙
−
H
(
t
))
T
˙
H
(
t
))
−
where we have used the fact that
C
is constant. Evaluating the above expression at
t
= 0 and replacing
˙
H
(0) by its value
H
(0)
C
yields
tr
(
I
H
(0)
C
2
f
(0)=
H
(0)
C
−
H
(0))
T
2
−
.
(8.30)
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