Control system problem I am studying control systems, and I want to solve following problem.
Given full rank state matrix $A$ (with all unstable eigenvalues), design input matrix $B$, such that cost function $J = \operatorname{trace}(B'XB)$ is minimized, where $X$ is the solution to discrete-time Ricatti equation (DARE). I have contraint that $(A,B)$ is stabilizable, i.e.
For a given full rank $A \in \mathbb{R}^{n \times n}$, with $\lambda_i(A)>1$, solve the following
\begin{array}{ll} \underset{X\in \mathbb{R}^{n\times n},B \in \mathbb{R}^{n\times m}}{\text{minimize}} & \mathrm{tr} \left( B' X B \right)\\ \text{subject to} & X=A'X(I+BB'X)^{-1}A\\&(A,B)\text{ is stabilizable}\end{array}
From my understanding, since all eigenvalues of $A$ are outside of unit circle (discrete-time system), we can change condition $(A,B)\text{ is stabilizable}$ with $(A,B)\text{ is controllable}$, which is equivalent to $\mathrm{rank}([B\quad AB\quad A^2B\quad \ldots\quad A^{n-1}B])=n$.
The problem is for sure feasible, since for any full rank $A$, there is $B$ such that rank condition is satisfied and we can solve DARE.
 A: I tried to use dual problem, maybe someone can help me to finish it. By creating random $B$, usually we get stabilizable pair $(A,B)$, so lets ignore the second constraint for now.
$\operatorname{trace}(B'XB)=\operatorname{trace}(BB'X)=\operatorname{trace}(AX^{-1}A'X)-\operatorname{trace}(I)$, so we can minimize $\operatorname{trace}(AX^{-1}A'X)$ instead of $\operatorname{trace}(B'XB)$.
Rewrite $X=A'X(I+BB'X)^{-1}A$ to $BB'-AX^{-1}A'+X^{-1}=0$, then Lagrangian function:
\begin{align}
&\Lambda(B,X,V)=\operatorname{trace}(AX^{-1}A'X)+\operatorname{trace}(V'BB')-\operatorname{trace}(V'AX^{-1}A')+\operatorname{trace}(V'X^{-1}),\\
&\frac{\partial \Lambda(B,X,V)}{\partial B}=(V'+V)B=0,\\
&\frac{\partial \Lambda(B,X,V)}{\partial B}=(AX^{-1}A'-X^{-1}A'XAX^{-1})+(X^{-1}A'VAX^{-1})-(X^{-1}VX^{-1})=0.
\end{align}
then we define $g(B,X,V)=\inf_{B,X} \Lambda(B,X,V)$ and dual problem becomes: $\max_V g(B,X,V)$.
First we need to find what $B$ and $X$ minimize $g(B,X,V)$. Assuming that $V'+V$ and $B$ are both nonzero, I found that either $V'+V$ or $B$ must be singular to satisfy $(V'+V)B=0$.
