I'm trying to understand the CARE/DARE solution and have a few questions. For example, for the CARE the equation is $$A^tP + PA - PBR^{-1}B^tP+Q=0$$ where $P,Q,R,A,B$ are $n\times n$ matrices and $P,Q,R$ are symmetric; P is the unknown for which a solution is sought. This equation can be conveniently written as a "quadratic form" $$X^t JZ X=0$$ where $$X=\begin{pmatrix} 1 \\ P \end{pmatrix} \qquad J=\begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} \qquad Z=\begin{pmatrix} A & -BR^{-1}B^t \\ -Q & -A^t \end{pmatrix}$$ and $$JZ=\begin{pmatrix} -Q & -A^t \\ -A & BR^{-1}B^t\end{pmatrix}=(JZ)^t \, .$$ Since $JZ$ is symmetric, its $2n$ eigenvalues are all real. However, this doesn't imply that the eigenvalues of $Z$ are all real, but since the characteristic polynomial $$\det\left(Z-t1 \right)=\det\left(JZ-tJ\right)=\det\left( (JZ-tJ)^t \right)=\det\left( JZ+tJ \right)$$ is an even function (with real coefficients) in $t$, every eigenvalue $\lambda$ actually occurs as $\lambda,-\lambda,\lambda^*,-\lambda^*$. Thus, if no eigenvalue is on the imaginary axis, half of all the eigenvalues have negative real part. Then we encode the corresponding subspace in the first column (block) of the eigenmatrix $$U=\begin{pmatrix} U_{11} & U_{12} \\ U_{21} & U_{22} \end{pmatrix}$$ and see that $$X=\begin{pmatrix} 1 \\ U_{21} U^{-1}_{11}\end{pmatrix}$$ actually solves the equation $$X^t JZ X=\begin{pmatrix} 1 & U_{21} U^{-1}_{11} \end{pmatrix} JZ \begin{pmatrix} 1 \\ U_{21} U^{-1}_{11} \end{pmatrix}=\begin{pmatrix} - U_{21} U^{-1}_{11} & 1 \end{pmatrix} Z \begin{pmatrix} U_{11} \\ U_{21} \end{pmatrix} U^{-1}_{11} = \begin{pmatrix} - U_{21} U^{-1}_{11} & 1 \end{pmatrix} \begin{pmatrix} U_{11} \\ U_{21} \end{pmatrix} \Lambda_- U^{-1}_{11} = 0$$ where $\Lambda_-$ is the $n\times n$ matrix of negative eigenvalues on the diagonal. Now I have two questions:
Why are only the negative eigenvalues of interest? I mean, obviously the positive eigenvalues also give rise to solutions. What does it have to do with the loop transfer matrix $$A-BR^{-1}B^t P$$ for the system to be stable?
How do you see that this loop transfer matrix has only negative eigenvalues? I mean, by putting the corresponding subspace into the first column of U, is just a convention and not a conclusion, isn't it?