first of all, excuse my poor English :( Anyway, I need to factorize the equation below in terms of Schur Complement.
$\begin{equation} { x }^{ T }\left( PA+{ A }^{ T }P \right) x<-\gamma { x }^{ T }Px \end{equation}$
Where $\lambda$ is an scalar and P a symmetric matrix $P>0$.
I was wondering if I could just factorize as follows:
${ x }^{ T }\left( PA+{ A }^{ T }P+\gamma P \right) x<0$
And forget about x terms ($x^T(something)x$) because it doesn't gives any further information because:
${ x }^{ T }\left( PA+{ A }^{ T }P+\gamma P \right) x<0$ is truth if and just if $PA+{ A }^{ T }P+\gamma P<0$ (if I'm not wrong).
then, my main problem is... How to get that expression in terms of shchur complement if there isn't any, lets say $Q^{-1}$ term.
I've tried this:
$M=\begin{pmatrix} -{ \gamma P }^{ -1 } & I \\ I & PA+{ A }^{ T }P \end{pmatrix}$
with lead us the problem that isn't linear due to the inverted term. Working around, making some variable changes, I got a pretty simple solution, doing:
$PA+{ A }^{ T }P+\gamma P{ P }^{ -1 }P$
with it's extraordinary easy to write in S-Complement.
Anyway, I got this matrix:
$$\begin{pmatrix} -\gamma P & P \\ P & PA+{ A }^{ T }P \end{pmatrix}<0$$
Having this, I must solve with:
$M<0$, $P>0$, and solve for $\lambda$ (I think)
Oh, I forgot: A is given.
$$ A=\begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ -3000 & -650 & -45 \end{pmatrix}$$
My professor gave us just something like:
$$\lambda>0 (?)$$
So... ¿? I'm really concerned about.
I've tried solve this using YALMIP with SeDuMi, doing something like:
A = [0 1 0; 0 0 1; -3000 -650 -45];
P = sdpvar(3,3); #defining sdp variables
gamma = sdpvar(1,1);
M = [-P*gamma P; P M*A+A'P];
constraints = [M<0, P>0]
solvesdp(constraints)
which lead me to the error "No suitable solver".
And I'm mad about it. I've seen A LOT of problems like this, with multiple sdpvars expressed in terms of Schur Complement, which lead to a linear matrix variables. By just adding the gamma sdpvar of ONE dimension, my problem gets up to a "Bilinear Matrix Variable".
Please, Help! :(
Your initial problem is equivalent to
$$ x^\top \left(P\,A + A^\top P\right) x = -\mu\,x^\top P\,x \tag{1} $$
and $\mu > \gamma$. Where $(1)$ is essentially a "homogeneous" continuous Lyapunov equation and can also be written as
$$ P\,\mathcal{A} + \mathcal{A}^\top P = 0, \tag{2} $$
with $\mathcal{A} = A + \frac{\mu}{2}I$. Homogeneous is referring to that the right hand side is zero. A Lyapunov equation can only have positive definite solution for $P$ if $\mathcal{A}$ is Hurwitz, so the real part of all eigenvalues of $\mathcal{A}$ have negative real part. Using the definition of $\mathcal{A}$ it can be shown that if $\lambda$ is an eigenvalue of $A$, $\lambda + \frac{\mu}{2}$ is an eigenvalue of $\mathcal{A}$. So all values for $\mu$ such that $\mathcal{A}$ is Hurwitz would be
$$ \mu < -2\max_{\lambda\in\Omega} Re(\lambda), \tag{3} $$
with $\Omega$ the set containing all eigenvalues of $A$. Combining the inequality between $\mu$ and $\gamma$ with $(3)$ would be the same as replacing $\mu$ with $\gamma$ in $(3)$. So this would solve finding feasible values for $\gamma$.
Finding $P$ could now be done with a LMI. It can also be done by solving the Lyapunov equation
$$ P\,\mathcal{A} + \mathcal{A}^\top P = -Q, \tag{4} $$
with $Q$ any positive definite matrix. Namely this would mean that
$$ x^\top \left(P\,A + A^\top P\right) x = -\mu\,x^\top P\,x - x^\top Q\,x < -\mu\,x^\top P\,x, \tag{5} $$
which does satisfy the initial inequality from the question.
So choose any $\mu$ according to $(3)$ and positive definite $Q$. After this $\mu$ can be used to evaluate $\mathcal{A}$ which combined with $Q$ can be used to solve for a positive definite $P$ using $(4)$.