\documentclass[12pt]{article}
\usepackage{latexsym}
\usepackage{epsfig}
\renewcommand{\baselinestretch}{1.27}
\setlength{\topmargin}{-0.25in}
\setlength{\textheight}{8.5in}
\setlength{\oddsidemargin}{0in}
\setlength{\textwidth}{6.5in}
\setlength{\fboxrule}{0.01in}
\newcommand{\markex}{\nopagebreak\begin{center}\rule{5in}{.01in}\end{center}}
\newcommand{\markdefn}{\nopagebreak\hfill$\Diamond$}
\newcommand{\mb}{\mbox}
\newcommand{\bm}{\boldmath}
\def\b#1{\mb{\bm$#1$}}
\newfont{\Bb}{msbm10 scaled\magstep1}
\begin{document}






We use the commands of the LMI Control Toolbox discussed above to
write a program that finds $\b P> 0.5 \b I_2$ such that
\begin{eqnarray*}
\b A_1^{\top}\b P+\b P\b A_1 &<& 0\\
\b A_2^{\top}\b P+\b P\b A_2 &<& 0
\end{eqnarray*}
where
\[
\b A_1=\left[\begin{array}{cc}
-1 & 0\\
0 & -1
\end{array}\right]\quad\mb{and}\quad \b A_2=\left[\begin{array}{cc}
-2 & 0\\
1 & -1
\end{array}\right].
\]



{\bf Solution:} The program has the form,

\begin{verbatim}
A_1=[-1 0;0 -1];
A_2=[-2 0;1 -1];
setlmis([])
P=lmivar(1,[2,1])
lmiterm([1 1 1 P],A_1',1,'s')
lmiterm([2 1 1 P],A_2',1,'s')
lmiterm([3 1 1 0],.5)
lmiterm([-3 1 1 P],1,1)
lmis=getlmis;
[tmin,xfeas]=feasp(lmis);
P=dec2mat(lmis,xfeas,P)
\end{verbatim}

-------------------------

Write a MATLAB program that finds satisfying, $0.1\b I_3<\b P< \b
I_3$, such that
\[
\b A^{\top}\b P+\b P\b A < 0,
\]
where
\[
\b A=\left[\begin{array}{ccc}
-0.9501 & -0.4860 & -0.4565\\
-0.2311 & -0.8913 & -0.0185\\
-0.6068 & -0.7621 & -0.8214
\end{array}\right].
\]

{\bf Solution:}
\begin{verbatim}
A =[-0.9501   -0.4860   -0.4565
   -0.2311   -0.8913   -0.0185
   -0.6068   -0.7621   -0.8214];
setlmis([]);
P=lmivar(1,[3 1]);
lmiterm([1 1 1 P],1,A,'s')
lmiterm([2 1 1 0],0.1)
lmiterm([-2 1 1 P],1,1)
lmiterm([3 1 1 P],1,1)
lmiterm([-3 1 1 0],1)
lmis=getlmis;
[tmin,xfeas]=feasp(lmis);
P=dec2mat(lmis,xfeas,P)
\end{verbatim}
---------------------------------------

Consider a square matrix $\b A$ partitioned as follows:
\[
\b A=\left[\begin{array}{cc}
\b A_{11} & \b A_{12}\\
\b A_{21} & \b A_{22}
\end{array}\right],
\]
where $\b A_{11}$ and $\b A_{22}$ are square submatrices. Assume
that $\b A_{22}$ is nonsingular. Find an expression for the Schur
complement of $\b A_{22}$.



{\bf Solution:} We have
\[
\left[\begin{array}{cc}
\b A_{11} & \b A_{12}\\
\b A_{21} & \b A_{22}
\end{array}\right]=\left[\begin{array}{cc}
\b I & \b A_{12}\b A_{22}^{-1}\\
\b O & \b I
\end{array}\right]\left[\begin{array}{cc}
\hat{\b{\Delta}} & \b O\\
\b O & \b A_{22}
\end{array}\right]\left[\begin{array}{cc}
\b I & \b O\\
\b A^{-1}_{22}\b A_{21} & \b I
\end{array}\right],
\]
where $\hat{\b{\Delta}}=\b A_{11}-\b A_{12}\b A_{22}^{-1}\b
A_{21}$ is the Schur complement of $\b A_{22}$.




--------------------------------------------


Represent the following quadratic inequality:
\[
\b A^{\top}\b P+\b P\b A+\b P\b B\b R^{-1}\b B^{\top}\b P < 0,
\]
in  the form of LMIs, where $\b R=\b R^{\top}>0$.


{\bf Solution:} The quadratic inequality,
\[
\b A^{\top}\b P+\b P\b A+\b P\b B\b R^{-1}\b B^{\top}\b P < 0,
\]
can be equivalently represented by the following LMI,
\[
\left[\begin{array}{cc} -\b R & \b B^{\top}\b P\\
\b P\b B & \b A^{\top}\b P + \b P\b A \end{array}\right] < 0,
\label{eq:first_lmi}
\]
or as the following LMI,
\[
\left[\begin{array}{cc}
\b A^{\top}\b P + \b P\b A & \b P\b B\\
\b B^{\top}\b P & -\b R\\
\end{array}\right] < 0.
\]
It is easy to verify, using the Schur complements, that the above
two LMIs are equivalent to the following quadratic inequality:
\[
\left[\begin{array}{cc}
\b A^{\top}\b P + \b P\b A + \b P\b B\b R^{-1}\b B^{\top}\b P &\b O\\
\b O & -\b R\\
\end{array}\right] < 0.
\]
\end{document}
