Relation between Riccati Algebraic Equation and optimization problem

Question

Reading this page: http://www.mathworks.com/help/robust/ug/minimizing-linear-objectives-under-lmi-constraints.html I got stuck in the result that says it can be show that minimizing Trace of X (a matrix variable) over the constraint A'X + XA + XBB'X + Q < 0 is equivalent to solve the algebraic Ricatti equation A'X + XA + XBB'X + Q = 0 How can I approach the argument to prove this?

Answer 1

Suppose that $(A,B)$ is stabilizable and $Q$ is Hermitian. Define the operator $\mathcal{R}(X)$ as $$ \mathcal{R}(X) = A^TX + XA - XBB^TX+ Q. $$

If there exists a Hermitian solution $X$ of the inequality $\mathcal{R}(X) \ge 0$, then there exists a Hermitian solution $X_+$ of the continuous-time algebraic Riccati equation $\mathcal{R}(X) = 0$, such that $X_+ \ge X$ for all $X$ such that $\mathcal{R}(X) \ge 0$.

Furthermore, $X_+$ is the stabilizing solution of algebraic Riccati equation. Thus, one can retrieve this solution by optimizing the trace functional on the set $\mathcal{R}(X) \ge 0$.

For a proof, see Theorem 2.1 of the paper Existence and comparison theorems for algebraic Riccati equations for continuous- and discrete-time systems or Theorem 9.1.1 of the book Algebraic Riccati Equations by Peter Lancaster and Leiba Rodman, which preview is available in google books.

In the case where $$ \mathcal{R}(X) = A^TX + XA + XBB^TX+ Q, $$ which is the one in the MATLAB documentation link, change $A$ to $-A$, $Q$ to $-Q$ and adapt the argument mutatis mutandis.