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Given $A\in\mathbb{R}^{n\times n}$ and $B\in\mathbb{R}^{n\times 2}$, I am interested in solving the following problem:

\begin{array}{ll} \underset{R \in \mathbb{R}^{2\times 2}}{\text{minimize}} & \mathrm{tr} \left( R^{-1}B^T X B \right)\\ \text{subject to} & X=A^TXA-A^TXB(R+B^TXB)^{-1}B^TXA.\end{array}

Here $X$ is unique stabilizing solution to DARE, thus X is positive definite. It is required that $R>0$ (i.e. positive definite) and $B^TXB$ to be full rank.

EDIT: For a fixed $A,B,R$, we can get unique $X$ by solving DARE, for example by using matlab "idare" or "dare" command. However, here $R$ is not fixed, it is a variable, thus for each $R$, there is corresponding $X$.


My attempt: I wanted to start with simpler case when we put additional constraints on $R$. Assume that $R$ is diagonal and positive definite. WLOG we can assume that $R=\mathrm{diag}\{r_1,r_2\}$, such that $r_1+r_2=1$ and $1>r_i>0$ for $i=1,2.$

A=[3 0 0 0; 0 2 1 0; 0 0 2 0; 0 0 0 2];
B=rand(4,2);
Q=zeros(4,4);    
r1=linspace(0.01, 0.99);
for i=1:100  
    R=[r1(i) 0; 0 1-r1(i)];
    [X,~,~] = idare(A,B,Q,R);
    T(i)=trace(inv(R)*B'*X*B);
end 
plot(T)

enter image description here

It looks like that as we increase $r_1$ from $0$ to $1$, then $\mathrm{tr} \left( R^{-1}B^T X B \right)$ is continuous, moreover, it is convex. However, I am unable to prove it.

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  • $\begingroup$ What exactly is $X$ ? $\endgroup$
    – P. Quinton
    Jul 17, 2022 at 8:13
  • $\begingroup$ @P.Quinton X is the unique solution to DARE (discrete-time algebraic Riccati equation), thus it is symmetric and positive definite $\endgroup$
    – Lee
    Jul 17, 2022 at 9:38
  • $\begingroup$ Do we have a rank constraint about $B^T X B$ ? I was thinking aobut trying something like this on $(R+B^T X B)^{-1}$ : math.stackexchange.com/questions/17776/… $\endgroup$
    – P. Quinton
    Jul 17, 2022 at 13:21
  • $\begingroup$ @P.Quinton assume that $B^TXB$ is full rank. we can also assume that both $A$ and $B$ are full rank $\endgroup$
    – Lee
    Jul 17, 2022 at 13:57
  • $\begingroup$ @user1551 done. $\endgroup$
    – Lee
    Jul 20, 2022 at 4:01

1 Answer 1

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Given the matrix inner product $(:)$ $$\eqalign{ \def\c#1{\color{red}{#1}} \def\CLR#1{\c{\LR{#1}}} \def\LR#1{\left(#1\right)} \def\trace#1{\operatorname{Tr}\LR{#1}} \def\p{\partial} \def\grad#1#2{\frac{\p #1}{\p #2}} \def\qiq{\quad\implies\quad} A:B &= \trace{A^TB} \\ }$$ The objective function can be written as $$\eqalign{ \phi &= \trace{R^{-1}B^T X B} \\ &= R^{-1}:B^TXB \\ }$$ For typing convenience, define the matrix variable $$\eqalign{ &M = (R+B^TXB)^{-1} \;=\; M^T \\ &dM = -M\LR{dR+B^TdX\,B}M \\ &\c{dm} = -\LR{M\otimes M}\c{dr} - \LR{MB^T\otimes MB^T}\c{dx} \\ }$$ Use the DARE constraint to relate the differentials of $M,R,$ and $X,\,$ and vectorize them. $$\eqalign{ &X = A^TXA-A^TXB\c{M}B^TXA \\ &dX = A^TXB\,\c{dM}\,B^TXA + A^T\c{dX}\,A - A^T\c{dX}\,BMB^TXA - A^TXBMB^T\c{dX}\,A \\ &\c{dx} = \LR{A^TXB\otimes A^TXB}\c{dm} + \LR{A^T\otimes A^T}\c{dx} \\ &\qquad\qquad\qquad - \LR{A^TXBMB^T\otimes A^T}\c{dx} - \LR{A^T\otimes A^TXBMB^T}\c{dx} \\ &\LR{A^TXB\otimes A^TXB}\LR{M\otimes M}\c{dr} \\&\qquad\qquad\qquad = \LR{A^T\otimes A^T-I\otimes I}\c{dx} \\ &\qquad\qquad\qquad - \LR{A^TXB\otimes A^TXB}\LR{MB^T\otimes MB^T}\c{dx} \\&\qquad\qquad\qquad - \LR{A^TXBMB^T\otimes A^T}\c{dx} \\&\qquad\qquad\qquad - \LR{A^T\otimes A^TXBMB^T}\c{dx} \\ &P^T\c{dr} = Q^T\c{dx} \\ }$$ Calculate the gradient of the objective function $$\eqalign{ \phi &= R^{-1}:B^TXB \\ \c{d\phi} &= R^{-1}:B^T\c{dX}\,B - B^TXB:R^{-1}\c{dR}\,R^{-1} \\ &= BR^{-1}B^T:\c{dX} - R^{-1}B^TXBR^{-1}:\c{dR} \\ &= E:\c{dX} - F:\c{dR} \\ &= e:\c{dx} - f:\c{dr} \\ &= \LR{PQ^{+}e - f}:\c{dr} \\ \grad{\phi}{r} &= {PQ^{+}e - f} \\ }$$ Solving the zero gradient condition for an optimal $R$ value seems hopeless, but a numerical solution via gradient descent is feasible given this closed-form expression for the gradient.

$$\eqalign{ }$$

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