# 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=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 AinRn*n, with lambda_i(A)>1, solve the following

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=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 $A\in {\mathbb{R}}^{n×n}$, with ${\lambda }_{i}\left(A\right)>1$, solve the following

From my understanding, since all eigenvalues of A are outside of unit circle (discrete-time system), we can change condition (A,B) is stabilizable with (A,B) is controllable, which is equivalent to rank([$\mathrm{r}\mathrm{a}\mathrm{n}\mathrm{k}\left(\left[B\phantom{\rule{1em}{0ex}}AB\phantom{\rule{1em}{0ex}}{A}^{2}B\phantom{\rule{1em}{0ex}}\dots \phantom{\rule{1em}{0ex}}{A}^{n-1}B\right]\right)=n$
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.
You can still ask an expert for help

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

Jaylen Mcmahon
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.
$\mathrm{trace}\left({B}^{\prime }XB\right)=\mathrm{trace}\left(B{B}^{\prime }X\right)=\mathrm{trace}\left(A{X}^{-1}{A}^{\prime }X\right)-\mathrm{trace}\left(I\right)$, so we can minimize trace $\mathrm{trace}\left(A{X}^{-1}{A}^{\prime }X\right)$ instead of trace $\mathrm{trace}\left({B}^{\prime }XB\right)$.
Rewrite $X={A}^{\prime }X\left(I+B{B}^{\prime }X{\right)}^{-1}A$ to $B{B}^{\prime }-A{X}^{-1}{A}^{\prime }+{X}^{-1}=0$, then Lagrangian function:
$\begin{array}{rl}& \mathrm{\Lambda }\left(B,X,V\right)=\mathrm{trace}\left(A{X}^{-1}{A}^{\prime }X\right)+\mathrm{trace}\left({V}^{\prime }B{B}^{\prime }\right)-\mathrm{trace}\left({V}^{\prime }A{X}^{-1}{A}^{\prime }\right)+\mathrm{trace}\left({V}^{\prime }{X}^{-1}\right),\\ & \frac{\mathrm{\partial }\mathrm{\Lambda }\left(B,X,V\right)}{\mathrm{\partial }B}=\left({V}^{\prime }+V\right)B=0,\\ & \frac{\mathrm{\partial }\mathrm{\Lambda }\left(B,X,V\right)}{\mathrm{\partial }B}=\left(A{X}^{-1}{A}^{\prime }-{X}^{-1}{A}^{\prime }XA{X}^{-1}\right)+\left({X}^{-1}{A}^{\prime }VA{X}^{-1}\right)-\left({X}^{-1}V{X}^{-1}\right)=0.\end{array}$
then we define $g\left(B,X,V\right)=\underset{B,X}{inf}\mathrm{\Lambda }\left(B,X,V\right)$ and dual problem becomes: maxVg(B,X,V).
First we need to find what B and X minimize g(B,X,V). Assuming that ${V}^{\prime }+V$ and B are both nonzero, I found that either ${V}^{\prime }+V$ or B must be singular to satisfy $\left({V}^{\prime }+V\right)B=0$