Question bank — Linear Quadratic Regulator (LQR) — Riccati equation, optimal gains

False. Only the ratio $Q/R$ sets aggressiveness; scaling both by the same factor leaves the minimizer of $J$ unchanged, so $K$ is identical.

True. $P$ is the matrix of the quadratic value function $V=x^{\top }Px$; only its symmetric part contributes to a quadratic form, so we take $P=P^{\top }$ by construction.

False. Stability is not optimality — an already-stable plant can still be sluggish or fuel-wasteful, and LQR finds the control that minimizes the combined error-plus-effort cost.

False. Larger $R$ penalizes control effort more, so the optimizer uses less actuation, giving a slower, gentler response. Larger $Q$ speeds it up.

False. $R$ must be strictly positive definite so $R^{-1}$ exists in $K=R^{-1}{B}^{\top }P$ and the cost is strictly convex in $u$; a zero direction in $R$ would allow infinite free control.

True. Under $(A,B)$ controllable and $(A,\sqrt{Q})$ observable, the CARE has a unique symmetric $P⪰0$ that makes $A-BK$ Hurwitz; the other algebraic roots are non-stabilizing.

True (under the standard conditions). The stabilizing Riccati solution is defined precisely as the one rendering $A-BK$ Hurwitz, so every closed-loop eigenvalue has negative real part.

False. In the infinite-horizon steady state $\dot{P}=0$, so $P$ and hence $K$ are constant — a single fixed gain applied for all time.

True. The law $u=-Kx$ needs every state; when only outputs are measured you must estimate $x$ (e.g. a Kalman Filter), producing LQG Control.

False. Cost-to-go from the origin is zero and the problem is symmetric under $x\to -x$, so $V$ is a pure quadratic $x^{\top }Px$ with no linear or constant term.

Missing the $R^{-1}$: the correct gain is $K=R^{-1}{B}^{\top }P$. This factor is exactly why $R$ must be invertible.

The quadratic term must be negative: $A^{\top }P+PA-PB{R}^{-1}{B}^{\top }P+Q=0$. That minus sign is what the optimizer "saves" by controlling; a plus sign has no stabilizing solution.

You must pick the stabilizing root — the unique $P⪰0$ making $A-BK$ Hurwitz. A root with $P/⪰0$ yields a gain that destabilizes the plant.

It minimizes the integral ${\int }_0^{\infty }(x^{\top }Qx+u^{\top }Ru)dt$ over the whole trajectory, not a terminal value; the running cost trades error against effort at every instant.

The quadratic cost is what makes the value function quadratic and the optimal control linear ($u=-Kx$). A linear (L1) cost breaks that clean closed form and gives bang-bang-type control.

Only $Q⪰0$ (positive semidefinite) is required; you may penalize just some states. The strict condition is on $R\succ 0$, plus $(A,\sqrt{Q})$ observability for uniqueness.

It gives a PD controller: $K=[k_p,k_d]$ acting on position and velocity. There is no integral term because the standard LQR cost contains no integral-of-error state.

The quadratic cost forces a quadratic value function $V=x^{\top }Px$; minimizing the HJB bracket over $u$ is minimizing a quadratic in $u$, whose stationary point is linear in $x$.

In steady state the cost-to-go stops changing, so $\dot{P}=0$ and the differential Riccati ODE collapses to a purely algebraic matrix equation with no time derivative.

The term $-PB{R}^{-1}{B}^{\top }P$ is quadratic in the unknown $P$, so despite linear dynamics the equation for $P$ is nonlinear.

The closed-loop pole $a-bK=a-\frac{b^2}{r}p$ is always negative for the positive Riccati root, so feedback overrides the unstable dynamics regardless of the sign of $a$.

In a quadratic form $x^{\top }Qx$ only the symmetric part of $Q$ contributes, so any antisymmetric part is wasted; we take them symmetric without loss of generality.

Controllability guarantees you can move every state; observability of $\sqrt{Q}$ ensures every unstable mode is actually penalized by the cost, so the optimizer has a reason to damp it — together they give a unique stabilizing $P$.

The HJB bracket is convex in $u$ because $R\succ 0$, so its single stationary point is the global minimum.

The infinite-horizon steady-state gain is constant, so the entire optimization reduces to solving one CARE ahead of time; runtime is just the cheap product $u=-Kx$.

$u=-K\cdot 0=0$. No error means no push needed; the controller idles exactly at the setpoint.

The gain $K\to 0$, so the controller does almost nothing and the closed loop approaches the open-loop dynamics — fine if the plant is already stable, disastrous if it is not.

The controller becomes extremely aggressive with very large gains and fast poles; in the limit $R^{-1}$ blows up, which is why $R$ must be strictly positive to keep $K$ well defined.

With nothing penalizing error, the cheapest way to minimize $\int u^{\top }Rudt$ is $u=0$; but if $(A,\sqrt{Q})$ is unobservable the solution may leave unstable modes uncontrolled, so $Q=0$ is degenerate and not generally used.

No. The gain $K$ is the same for every starting point; the value $J$ achieved equals $x(0{)}^{\top }Px(0)$, which scales with $x(0)$, but the control rule does not.

No. If an unstable mode cannot be influenced by $u$, no feedback can move it into the left half-plane, so no stabilizing $P⪰0$ exists — controllability of the unstable part is essential.

Test the closed loop: the optimal one is the $P⪰0$ for which $A-BK$ has all eigenvalues in the left half-plane (Hurwitz); the other yields an anti-stable closed loop.