# Minimize $\mathrm{tr}(R^{-1}B^TXB)$ over $R$ subject to $X=A^TXA-A^TXB(R+B^TXB)^{-1}B^TXA$

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)


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.

• What exactly is $X$ ? Jul 17, 2022 at 8:13
• @P.Quinton X is the unique solution to DARE (discrete-time algebraic Riccati equation), thus it is symmetric and positive definite
– Lee
Jul 17, 2022 at 9:38
• 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/… Jul 17, 2022 at 13:21
• @P.Quinton assume that $B^TXB$ is full rank. we can also assume that both $A$ and $B$ are full rank
– Lee
Jul 17, 2022 at 13:57
• @user1551 done.
– Lee
Jul 20, 2022 at 4:01

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{ }