Matrix inequalities can be dilated in order to obtain a larger matrix inequality. This can be a useful technique to separate design variables in a BMI (bi-linear matrix inequality), as the dilation often introduces additional design variables.
A common technique of LMI dilation involves using the projection lemma in reverse, or the "reciprocal projection lemma." For instance, consider the matrix inequality
[
P
A
+
A
T
P
−
P
P
∗
−
P
]
<
0
,
{\displaystyle {\begin{bmatrix}{\mathbf {PA}}+{\mathbf {A^{T}P}}-{\mathbf {P}}&{\mathbf {P}}\\*&{\mathbf {-P}}\end{bmatrix}}<0,}
where
P
∈
§
n
×
m
{\displaystyle {\mathbf {P}}\in \S ^{n\times m}}
,
A
∈
R
n
×
n
{\displaystyle {\mathbf {A}}\in \mathbb {R} ^{n\times n}}
, with
P
>
0.
{\displaystyle {\mathbf {P}}>0.}
This can be rewritten as
[
A
T
1
0
1
0
1
]
[
0
P
0
∗
−
P
0
∗
∗
−
P
]
[
A
1
1
0
0
1
]
<
0.
{\displaystyle {\begin{bmatrix}{\mathbf {A^{T}}}&{\mathbf {1}}&{\mathbf {0}}\\{\mathbf {1}}&{\mathbf {0}}&{\mathbf {1}}\end{bmatrix}}{\begin{bmatrix}{\mathbf {0}}&{\mathbf {P}}&{\mathbf {0}}\\*&{\mathbf {-P}}&{\mathbf {0}}\\*&*&{\mathbf {-P}}\end{bmatrix}}{\begin{bmatrix}{\mathbf {A}}&{\mathbf {1}}\\{\mathbf {1}}&{\mathbf {0}}\\{\mathbf {0}}&{\mathbf {1}}\end{bmatrix}}<0.}
(1)
Then since
P
>
0
,
{\displaystyle {\mathbf {P}}>0,}
[
−
P
0
∗
−
P
]
<
0
,
{\displaystyle {\begin{bmatrix}{\mathbf {-P}}&{\mathbf {0}}\\*&{\mathbf {-P}}\end{bmatrix}}<0,}
which is equivalent to
[
0
1
0
0
0
1
]
[
0
P
0
∗
−
P
0
∗
∗
−
P
]
[
0
0
1
0
0
1
]
<
0.
{\displaystyle {\begin{bmatrix}{\mathbf {0}}&{\mathbf {1}}&{\mathbf {0}}\\{\mathbf {0}}&{\mathbf {0}}&{\mathbf {1}}\end{bmatrix}}{\begin{bmatrix}{\mathbf {0}}&{\mathbf {P}}&{\mathbf {0}}\\*&{\mathbf {-P}}&{\mathbf {0}}\\*&*&{\mathbf {-P}}\end{bmatrix}}{\begin{bmatrix}{\mathbf {0}}&{\mathbf {0}}\\{\mathbf {1}}&{\mathbf {0}}\\{\mathbf {0}}&{\mathbf {1}}\end{bmatrix}}<0.}
(2)
These expanded inequalities (1) and (2) are now in the form of the strict projection lemma, meaning they are equivalent to
Φ
(
P
)
+
G
(
A
)
V
H
T
+
H
V
T
G
T
(
A
)
,
{\displaystyle {\mathbf {\Phi }}({\mathbf {P}})+{\mathbf {G}}({\mathbf {A}}){\mathbf {VH^{T}}}+{\mathbf {HV^{T}G^{T}}}({\mathbf {A}}),}
(3)
where
N
(
G
T
(
A
)
)
=
R
(
N
G
(
A
)
)
,
N
(
H
T
)
=
R
(
N
H
)
,
{\displaystyle N({\mathbf {G^{T}}}({\mathbf {A}}))=R({\mathbf {N}}_{G}({\mathbf {A}})),N({\mathbf {H^{T}}})=R({\mathbf {N}}_{H}),}
and
V
∈
R
n
×
n
.
{\displaystyle V\in \mathbb {R} ^{n\times n}.}
By choosing
G
(
A
)
=
[
−
1
A
T
1
]
,
H
=
[
1
0
0
]
,
{\displaystyle {\mathbf {G}}({\mathbf {A}})={\begin{bmatrix}{\mathbf {-1}}\\{\mathbf {A^{T}}}\\{\mathbf {1}}\end{bmatrix}},{\mathbf {H}}={\begin{bmatrix}{\mathbf {1}}\\{\mathbf {0}}\\{\mathbf {0}}\end{bmatrix}},}
we can now rewrite the inequality (3) as
[
−
(
V
+
V
T
)
V
T
A
+
P
V
T
∗
−
P
0
∗
∗
−
P
]
<
0
,
{\displaystyle {\begin{bmatrix}-({\mathbf {V}}+{\mathbf {V^{T}}})&{\mathbf {V^{T}A}}+{\mathbf {P}}&{\mathbf {V^{T}}}\\*&{\mathbf {-P}}&{\mathbf {0}}\\*&*&{\mathbf {-P}}\end{bmatrix}}<0,}
which is the new dilated inequality.
Some useful examples of dilated matrix inequalities are presented here.
Example 1
Consider matrices
A
,
G
∈
R
n
×
n
,
Δ
∈
R
m
×
m
,
P
∈
§
n
,
δ
1
,
δ
2
,
a
,
b
∈
R
>
0
,
{\displaystyle {\mathbf {A,G}}\in \mathbb {R} ^{n\times n},{\mathbf {\Delta }}\in \mathbb {R} ^{m\times m},{\mathbf {P}}\in \S ^{n},\delta _{1},\delta _{2},a,b\in \mathbb {R} _{>0},}
where
P
>
0
{\displaystyle {\mathbf {P}}>0}
and
b
=
a
−
1
.
{\displaystyle b=a^{-1}.}
The following matrix inequalities are equivalent:
A
P
+
P
A
T
+
δ
1
P
+
δ
2
A
P
A
T
+
P
Δ
T
Δ
P
<
0
;
{\displaystyle {\mathbf {AP}}+{\mathbf {PA^{T}}}+\delta _{1}{\mathbf {P}}+\delta _{2}{\mathbf {APA^{T}}}+{\mathbf {P\Delta ^{T}\Delta P}}<0;}
[
0
−
P
P
0
P
Δ
T
∗
0
0
−
P
0
∗
∗
−
δ
1
−
1
P
0
0
∗
∗
∗
−
δ
2
−
1
P
0
∗
∗
∗
∗
−
1
]
+
H
e
(
[
A
1
0
0
0
]
G
[
1
−
b
1
b
1
1
b
Δ
T
]
)
<
0.
{\displaystyle {\begin{bmatrix}{\mathbf {0}}&{\mathbf {-P}}&{\mathbf {P}}&{\mathbf {0}}&{\mathbf {P\Delta ^{T}}}\\*&{\mathbf {0}}&{\mathbf {0}}&{\mathbf {-P}}&{\mathbf {0}}\\*&*&-\delta _{1}^{-1}{\mathbf {P}}&{\mathbf {0}}&{\mathbf {0}}\\*&*&*&-\delta _{2}^{-1}{\mathbf {P}}&{\mathbf {0}}\\*&*&*&*&{\mathbf {-1}}\\\end{bmatrix}}+He({\begin{bmatrix}{\mathbf {A}}\\{\mathbf {1}}\\{\mathbf {0}}\\{\mathbf {0}}\\{\mathbf {0}}\\\end{bmatrix}}{\mathbf {G}}{\begin{bmatrix}{\mathbf {1}}&-b{\mathbf {1}}&b{\mathbf {1}}&{\mathbf {1}}&b{\mathbf {\Delta }}^{T}\end{bmatrix}})<0.}
Example 2
Consider matrices
A
,
V
∈
R
n
×
n
,
P
,
X
∈
§
n
,
B
∈
R
n
×
m
,
C
∈
R
p
×
n
,
D
∈
R
p
×
m
,
R
∈
§
m
,
{\displaystyle {\mathbf {A,V}}\in \mathbb {R} ^{n\times n},{\mathbf {P,X}}\in \S ^{n},{\mathbf {B}}\in \mathbb {R} ^{n\times m},C\in \mathbb {R} ^{p\times n},{\mathbf {D}}\in \mathbb {R} ^{p\times m},{\mathbf {R}}\in \S ^{m},}
and
S
∈
§
p
,
{\displaystyle {\mathbf {S}}\in \S ^{p},}
where
P
,
R
,
S
,
X
>
0.
{\displaystyle {\mathbf {P,R,S,X}}>0.}
The matrix inequality
[
−
V
−
V
T
V
A
+
P
V
B
0
V
∗
−
2
P
+
X
0
C
T
0
∗
∗
−
R
D
T
0
∗
∗
∗
−
S
0
∗
∗
∗
∗
−
X
]
<
0
{\displaystyle {\begin{bmatrix}-{\mathbf {V}}-{\mathbf {V^{T}}}&{\mathbf {VA}}+{\mathbf {P}}&{\mathbf {VB}}&{\mathbf {0}}&{\mathbf {V}}\\*&-2{\mathbf {P}}+{\mathbf {X}}&{\mathbf {0}}&{\mathbf {C^{T}}}&{\mathbf {0}}\\*&*&-{\mathbf {R}}&{\mathbf {D^{T}}}&{\mathbf {0}}\\*&*&*&-{\mathbf {S}}&{\mathbf {0}}\\*&*&*&*&-{\mathbf {X}}\\\end{bmatrix}}<0}
implies the inequality
[
P
A
+
A
T
P
P
B
C
T
∗
−
R
D
T
∗
∗
−
S
]
{\displaystyle {\begin{bmatrix}{\mathbf {PA}}+{\mathbf {A^{T}P}}&{\mathbf {PB}}&{\mathbf {C^{T}}}\\*&-{\mathbf {R}}&{\mathbf {D^{T}}}\\*&*&-{\mathbf {S}}\end{bmatrix}}}
Example 3
Consider matrices
A
,
V
∈
R
n
×
n
,
Q
,
X
∈
§
n
,
B
∈
R
n
×
m
,
C
∈
R
p
×
n
,
D
∈
R
p
×
m
,
R
∈
§
m
,
{\displaystyle {\mathbf {A,V}}\in \mathbb {R} ^{n\times n},{\mathbf {Q,X}}\in \S ^{n},{\mathbf {B}}\in \mathbb {R} ^{n\times m},C\in \mathbb {R} ^{p\times n},{\mathbf {D}}\in \mathbb {R} ^{p\times m},{\mathbf {R}}\in \S ^{m},}
and
S
∈
§
p
,
{\displaystyle {\mathbf {S}}\in \S ^{p},}
where
Q
,
R
,
S
,
X
>
0.
{\displaystyle {\mathbf {Q,R,S,X}}>0.}
The matrix inequality
[
−
V
−
V
T
V
T
A
T
+
Q
0
V
T
C
V
T
∗
−
2
Q
+
X
B
0
0
∗
∗
−
R
D
T
0
∗
∗
∗
−
S
0
∗
∗
∗
∗
−
X
]
<
0
{\displaystyle {\begin{bmatrix}-{\mathbf {V}}-{\mathbf {V^{T}}}&{\mathbf {V^{T}A^{T}}}+{\mathbf {Q}}&{\mathbf {0}}&{\mathbf {V^{T}C}}&{\mathbf {V^{T}}}\\*&-2{\mathbf {Q}}+{\mathbf {X}}&{\mathbf {B}}&{\mathbf {0}}&{\mathbf {0}}\\*&*&-{\mathbf {R}}&{\mathbf {D^{T}}}&{\mathbf {0}}\\*&*&*&-{\mathbf {S}}&{\mathbf {0}}\\*&*&*&*&-{\mathbf {X}}\\\end{bmatrix}}<0}
implies the inequality
[
A
Q
+
Q
A
T
B
Q
C
T
∗
−
R
D
T
∗
∗
−
S
]
{\displaystyle {\begin{bmatrix}{\mathbf {AQ}}+{\mathbf {QA^{T}}}&{\mathbf {B}}&{\mathbf {QC^{T}}}\\*&-{\mathbf {R}}&{\mathbf {D^{T}}}\\*&*&-{\mathbf {S}}\end{bmatrix}}}