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and, as the matrix is positive definite,
the Cholesky factor gives the SOS
z
1
)
2
,so
p
(
)
=
(
z
1
−
4
z
2
−
2
z
1
z
2
+
(
)
∈
z
.
decomposition
p
z
z
Consider now a region
defined by polynomial boundaries as follows:
={
z
:
g
1
(
z
)>
0
,...,
g
m
g
(
z
)>
0
,
h
1
(
z
)
=
0
,...,
h
m
h
(
z
)
=
0
}
(5.22)
Lemma 5.2
(Jarvis-Wloszek et al.
2005
)
If SOS polynomials s
i
(
z
)
∈
z
and arbi-
trary ones r
j
(
z
)
∈
R
z
can be found fulfilling:
m
g
m
h
p
(
z
)
−
ε(
z
)
−
s
i
(
z
)
g
i
(
z
)
+
r
j
(
z
)
h
j
(
z
)
∈
z
(5.23)
i
=
1
j
=
1
then p
(
z
)
is locally greater or equal than
ε(
z
)
in the region
.
,
i
s
i
(
is positive and
j
r
j
(
Proof
Inded, note that, in the region
z
)
g
i
(
z
)
z
)
h
i
(
z
)
)
≥
i
s
i
(
is zero, so
p
(
z
)
−
ε(
z
z
)
g
i
(
z
)
≥
0 for all
z
∈
.
are analogous to KKT ones in constrained
optimization (Bertsekas
1999
). The lemma is a simplified version of the
Positivstel-
lensatz
result (Jarvis-Wloszek et al.
2005
), in which less conservative expressions
are stated (details omitted for brevity as computational complexity increases consid-
erably).
Multiplier polynomials
s
i
(
z
)
,
r
j
(
z
)
x
3
is non-negative in the interval region of
Example 5.4
The polynomial
p
(
x
)
=
1
−
x
2
interest
.If
we execute the code included in AppendixB, we obtain that the quadratic multiplier
s
=−
0
.
5
≤
x
≤
0
.
5. Indeed, let us first describe
={
x
:
0
.
25
−
>
0
}
4076
x
2
which fulfills 1
x
3
x
2
4076
x
4
x
3
(
x
)
=
1
.
−
−
s
(
x
)(
0
.
25
−
)
=
1
.
−
−
3519
x
2
x
3
0
.
+
1
∈
x
,soweproved1
−
>
0
∀
x
∈
.
Next proposition allows checking positive-definiteness of polynomial matrices.
In this way, the linear matrix inequality framework (positive-definiteness of matrices
with linear expressions as elements (Boyd et al.
1994
)) is extended to polynomial
cases.
(
)
×
Proposition 5.1
(Prajna et al.
2004b
)
Let L
x
be an N
N symmetric polynomial
n
.MatrixL
n
matrix of degree
2
dinx
∈ R
(
x
)
is positive semi-definite for all x
∈ R
T
L
if polynomial
v
(
x
)v
∈
x
,v
(equivalently, if there exists a vector of monomials
T
L
T
Q
m
(
x
)
and a matrix Q
0
such that
v
(
x
)v
=
(v
⊗
m
(
x
))
(v
⊗
m
(
x
))
where
⊗
denotes Kronecker product).
Example 5.5
Testing if there exist coefficients
a
,
b
, such that the matrix
1
ax
2
x
2
−
0
.
1
x
+
b
L
(
x
)
=
(5.24)
1
x
3
0
.
1
x
+
b
3
−
0
.
+
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