TITLE: Converse linear quadratic optimal control
QUESTION [5 upvotes]: It is well known that for a linear time invariant system
$$
\dot{x} = A x + B u \tag{1}
$$
with $(A, B)$ controllable, there exists a static state feedback $u = -K x$ such that the cost function
$$
J = \int_0^{\infty} x^T Q x + u^T R u \, dt \tag{2}
$$
is minimized, assuming $Q \geq 0$ (positive semi-definite) and $R > 0$ (positive definite). The gain $K$ is the solution of the algebraic Riccati equation:
$$
\begin{align}
0 &= A^T P + P A - P B R^{-1} B^T P + Q \\
K &= R^{-1} B^T P \\
P &= P^T \geq 0
\end{align}
$$
known as linear quadratic regulator (LQR). However, I wonder whether the converse also holds?
That is, given a stabilizing $K_s$ (such that $A - B K_s$ is Hurwitz), do there exist matrices $Q \geq 0$ and $R > 0$ such that $u = -K_s x$ minimizes $(2)$ given $(1)$? Or put differently:
Question: Is every stabilizing linear state feedback optimal in some sense?

REPLY [3 votes]: The problem is called the Inverse Problem of Optimal Control (see page 147 - 148). 
Given a system
$$\dot{x}=Ax+Bu,\qquad x(t_0)=x_0$$
$$
z = \begin{bmatrix}
Q^{1/2}& 0 \\
0 & R^{1/2}\\
\end{bmatrix}\begin{bmatrix}x\\u \end{bmatrix},$$
with $(A,B)$ is stabilizable, $(Q,A)$ is detectable and $R>0$ (positive definite). The linear quadratic regulator problem is given by minimizing 
$$\int_{0}^{\infty}z^Tz dt.$$

From Boyd et al 1994: The inverse problem of optimal control is the following. Given a
  matrix $K$, determine if there exists $Q ≥ 0$ and $R > 0$, such that
  $(Q, A)$ is detectable and $u = Kx$ is the optimal control for the
  corresponding LQR problem. Equivalently, we seek $R > 0$ and $Q ≥ 0$
  such that there exists P nonnegative and P1 positive-definite
  satisfying
$$(A + BK)^T P + P(A + BK) + K^TRK + Q = 0, \quad B^T P + RK = 0$$
and $A^T P_1 + P_1A < Q$. This is an LMIP in $P$, $P_1$, $R$ and $Q$.
  (The condition involving $P_1$ is equivalent to $(Q, A)$ being
  detectable.)