What is the finite eigenvalue of a controlled system as $\epsilon \rightarrow 0$? Consider a state space system (A, B) as follows:
$$A = \begin{bmatrix}
    a_1 & a_2 \\
    -a_2 & 0 \\
    \end{bmatrix},
B= \begin{bmatrix}
    0 \\
    1 \\
    \end{bmatrix}$$
where $a_1$ and $a_2$ are real constants and $a_1 \ne 0$. Furthermore, let
$$Q=\begin{bmatrix}
    0 & 0 \\
    0 & 1 \\
    \end{bmatrix},
R = \epsilon^2$$
where $\epsilon > 0$. Let $P(\epsilon)$ be the positive definite solution, assuming it exists, of the algebraic Riccati equation:
$$A^TP + PA - PBR^{-1}B^TP + Q = 0.$$
What is the finite eigenvlue(s) of $A- \epsilon^{-2}BB^TP(\epsilon)$ as $\epsilon \rightarrow 0$?
I have figured out that if a positive definite solution $P(\epsilon)$ exists, then $a_2 \ne 0$. However, I am unable to calculate $P(\epsilon)$. Is there a trick to find the eigenvalues of $A- \epsilon^{-2}BB^TP(\epsilon)$ as $\epsilon \rightarrow 0$?
Thanks in advance!
 A: Since you are dealing with LQR and they ask about the closedloop eigenvalues, the first thing that comes to mind is to use the associated Hamiltonian matrix
$$
H = 
\begin{bmatrix}
A & -\epsilon^{-2} B\,B^\top \\ -Q & -A^\top
\end{bmatrix},
$$
which should have negated eigenvalue pairs and the two negative eigenvalues should match the closedloop eigenvalues. Solving directly for the eigenvalues of $H$ as a function of $a_1$, $a_2$ and $\epsilon$ would require solving a quartic equation. However, it is easier to make use of the stated property of $H$, namely from that it follows that its characteristic polynomial would be of the form
$$
\det(\lambda\,I - H) = (\lambda - p_1)\,(\lambda + p_1)\,(\lambda - p_2)\,(\lambda + p_2) = p_1^2\,p_2^2 - (p_1^2 + p_2^2)\,\lambda^2 + \lambda^4.
$$
Matching the actual characteristic polynomial with this form allows one to find expressions for $p_1^2$ and $p_2^2$ relatively easily. Now, when one applies the limit of $\epsilon \to 0$ to these solutions one will remain bounded while the other does not.
