Assume that $(A_2,B_2)$ is stabilizable and $(C,A)$ is detectable. Under these conditions, there exist a regulator if the equations
\begin{aligned} TA_1-A_2T-B_2V&=A_3\\ D_1+D_2T+EV&=0 \end{aligned}
have a solution $(T,V)$.
If $A_1$ is antistable, the solvability of these equations is a necessary condition for a regulator to exist.
Specifically, if for $G:Y\to X$ we have $\sigma(A+GC)\subset\mathbb{C}^{-}$, and for $F_2:X\to U$ we have $\sigma(A_2+B_2F_2)\subset\mathbb{C}^{-}$, and if $F_1 :=-F_2T+V$, $F:=(F_1,F_2)$ where $(T,V)$ is a solution eqution of above then a regulator is given by
\begin{aligned} \dot{w}&=(A+GC+BF)w(t)-Gy(t)\\ u(t)&=Fw(t). \end{aligned}
Now I have a question: Show that if $A_1=a_1$ is scalar with $a_1>0$, then the $a_1$ is a eigenvalue of the matrix $A+GC+BF$?