# LMIs in Control/Matrix and LMI Properties and Tools/Dilation

## Dilation

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

${\begin{bmatrix}{\mathbf {PA}}+{\mathbf {A^{T}P}}-{\mathbf {P}}&{\mathbf {P}}\\*&{\mathbf {-P}}\end{bmatrix}}<0,$ where ${\mathbf {P}}\in \S ^{n\times m}$ , ${\mathbf {A}}\in \mathbb {R} ^{n\times n}$ , with ${\mathbf {P}}>0.$ This can be rewritten as

${\begin{bmatrix}{\mathbf {A^{T}}}&{\mathbf {1}}&{\mathbf {0}}\\{\mathbf {1}}&{\mathbf {0}}&{\mathbf {1}}\end{bmatrix}}{\begin{bmatrix}{\mathbf {0}}&{\mathbf {P}}&{\mathbf {0}}\\*&{\mathbf {-P}}&{\mathbf {0}}\\*&*&{\mathbf {-P}}\end{bmatrix}}{\begin{bmatrix}{\mathbf {A}}&{\mathbf {1}}\\{\mathbf {1}}&{\mathbf {0}}\\{\mathbf {0}}&{\mathbf {1}}\end{bmatrix}}<0.$ (1)

Then since ${\mathbf {P}}>0,$ ${\begin{bmatrix}{\mathbf {-P}}&{\mathbf {0}}\\*&{\mathbf {-P}}\end{bmatrix}}<0,$ which is equivalent to

${\begin{bmatrix}{\mathbf {0}}&{\mathbf {1}}&{\mathbf {0}}\\{\mathbf {0}}&{\mathbf {0}}&{\mathbf {1}}\end{bmatrix}}{\begin{bmatrix}{\mathbf {0}}&{\mathbf {P}}&{\mathbf {0}}\\*&{\mathbf {-P}}&{\mathbf {0}}\\*&*&{\mathbf {-P}}\end{bmatrix}}{\begin{bmatrix}{\mathbf {0}}&{\mathbf {0}}\\{\mathbf {1}}&{\mathbf {0}}\\{\mathbf {0}}&{\mathbf {1}}\end{bmatrix}}<0.$ (2)

These expanded inequalities (1) and (2) are now in the form of the strict projection lemma, meaning they are equivalent to

${\mathbf {\Phi }}({\mathbf {P}})+{\mathbf {G}}({\mathbf {A}}){\mathbf {VH^{T}}}+{\mathbf {HV^{T}G^{T}}}({\mathbf {A}}),$ (3)

where $N({\mathbf {G^{T}}}({\mathbf {A}}))=R({\mathbf {N}}_{G}({\mathbf {A}})),N({\mathbf {H^{T}}})=R({\mathbf {N}}_{H}),$ and $V\in \mathbb {R} ^{n\times n}.$ By choosing

${\mathbf {G}}({\mathbf {A}})={\begin{bmatrix}{\mathbf {-1}}\\{\mathbf {A^{T}}}\\{\mathbf {1}}\end{bmatrix}},{\mathbf {H}}={\begin{bmatrix}{\mathbf {1}}\\{\mathbf {0}}\\{\mathbf {0}}\end{bmatrix}},$ we can now rewrite the inequality (3) as

${\begin{bmatrix}-({\mathbf {V}}+{\mathbf {V^{T}}})&{\mathbf {V^{T}A}}+{\mathbf {P}}&{\mathbf {V^{T}}}\\*&{\mathbf {-P}}&{\mathbf {0}}\\*&*&{\mathbf {-P}}\end{bmatrix}}<0,$ which is the new dilated inequality.

## Examples

Some useful examples of dilated matrix inequalities are presented here.

Example 1

Consider matrices ${\mathbf {A,G}}\in \mathbb {R} ^{n\times n},{\mathbf {\Delta }}\in \mathbb {R} ^{m\times m},{\mathbf {P}}\in \S ^{n},\delta _{1},\delta _{2},a,b\in \mathbb {R} _{>0},$ where ${\mathbf {P}}>0$ and $b=a^{-1}.$ The following matrix inequalities are equivalent:

${\mathbf {AP}}+{\mathbf {PA^{T}}}+\delta _{1}{\mathbf {P}}+\delta _{2}{\mathbf {APA^{T}}}+{\mathbf {P\Delta ^{T}\Delta P}}<0;$ ${\begin{bmatrix}{\mathbf {0}}&{\mathbf {-P}}&{\mathbf {P}}&{\mathbf {0}}&{\mathbf {P\Delta ^{T}}}\\*&{\mathbf {0}}&{\mathbf {0}}&{\mathbf {-P}}&{\mathbf {0}}\\*&*&-\delta _{1}^{-1}{\mathbf {P}}&{\mathbf {0}}&{\mathbf {0}}\\*&*&*&-\delta _{2}^{-1}{\mathbf {P}}&{\mathbf {0}}\\*&*&*&*&{\mathbf {-1}}\\\end{bmatrix}}+He({\begin{bmatrix}{\mathbf {A}}\\{\mathbf {1}}\\{\mathbf {0}}\\{\mathbf {0}}\\{\mathbf {0}}\\\end{bmatrix}}{\mathbf {G}}{\begin{bmatrix}{\mathbf {1}}&-b{\mathbf {1}}&b{\mathbf {1}}&{\mathbf {1}}&b{\mathbf {\Delta }}^{T}\end{bmatrix}})<0.$ Example 2

Consider matrices ${\mathbf {A,V}}\in \mathbb {R} ^{n\times n},{\mathbf {P,X}}\in \S ^{n},{\mathbf {B}}\in \mathbb {R} ^{n\times m},C\in \mathbb {R} ^{p\times n},{\mathbf {D}}\in \mathbb {R} ^{p\times m},{\mathbf {R}}\in \S ^{m},$ and ${\mathbf {S}}\in \S ^{p},$ where ${\mathbf {P,R,S,X}}>0.$ The matrix inequality

${\begin{bmatrix}-{\mathbf {V}}-{\mathbf {V^{T}}}&{\mathbf {VA}}+{\mathbf {P}}&{\mathbf {VB}}&{\mathbf {0}}&{\mathbf {V}}\\*&-2{\mathbf {P}}+{\mathbf {X}}&{\mathbf {0}}&{\mathbf {C^{T}}}&{\mathbf {0}}\\*&*&-{\mathbf {R}}&{\mathbf {D^{T}}}&{\mathbf {0}}\\*&*&*&-{\mathbf {S}}&{\mathbf {0}}\\*&*&*&*&-{\mathbf {X}}\\\end{bmatrix}}<0$ implies the inequality

${\begin{bmatrix}{\mathbf {PA}}+{\mathbf {A^{T}P}}&{\mathbf {PB}}&{\mathbf {C^{T}}}\\*&-{\mathbf {R}}&{\mathbf {D^{T}}}\\*&*&-{\mathbf {S}}\end{bmatrix}}$ Example 3

Consider matrices ${\mathbf {A,V}}\in \mathbb {R} ^{n\times n},{\mathbf {Q,X}}\in \S ^{n},{\mathbf {B}}\in \mathbb {R} ^{n\times m},C\in \mathbb {R} ^{p\times n},{\mathbf {D}}\in \mathbb {R} ^{p\times m},{\mathbf {R}}\in \S ^{m},$ and ${\mathbf {S}}\in \S ^{p},$ where ${\mathbf {Q,R,S,X}}>0.$ The matrix inequality

${\begin{bmatrix}-{\mathbf {V}}-{\mathbf {V^{T}}}&{\mathbf {V^{T}A^{T}}}+{\mathbf {Q}}&{\mathbf {0}}&{\mathbf {V^{T}C}}&{\mathbf {V^{T}}}\\*&-2{\mathbf {Q}}+{\mathbf {X}}&{\mathbf {B}}&{\mathbf {0}}&{\mathbf {0}}\\*&*&-{\mathbf {R}}&{\mathbf {D^{T}}}&{\mathbf {0}}\\*&*&*&-{\mathbf {S}}&{\mathbf {0}}\\*&*&*&*&-{\mathbf {X}}\\\end{bmatrix}}<0$ implies the inequality

${\begin{bmatrix}{\mathbf {AQ}}+{\mathbf {QA^{T}}}&{\mathbf {B}}&{\mathbf {QC^{T}}}\\*&-{\mathbf {R}}&{\mathbf {D^{T}}}\\*&*&-{\mathbf {S}}\end{bmatrix}}$ 