advanced31 min studyLesson 352 in Mathematics

Controllability and reachability Gramians

Connect algebraic rank with finite-time state transfer and quantify the input energy required in each direction.

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What you’ll understand

  • Use the Kalman rank condition for finite-dimensional continuous-time LTI systems.
  • Derive the reachability Gramian and minimum-energy input.
  • Distinguish reachability from a practical energy budget or a stabilization requirement.

An actuator may directly push only one state variable yet influence several others through the dynamics. Reachability asks whether these indirect paths can produce every desired final state. A rank test decides the exact algebraic question, while a Gramian measures how difficult the transfer is. A system can be mathematically controllable and still require prohibitive energy in a poorly actuated direction. We allow unrestricted square-integrable inputs on a fixed positive time interval; actuator bounds or forbidden intermediate states define a different reachability problem and can invalidate the unconstrained conclusions.

Dynamics spread an input across state directions

For ẋ=Ax+Bu on Rn\mathbb R^n, start with the columns of B, then the columns of AB, A²B and so on. They identify directions available directly and through repeated action of the dynamics. The Kalman controllability matrix stops at A^{n−1}B because Cayley–Hamilton expresses all higher powers using those already listed. The pair is controllable exactly when this matrix has rank n. For continuous-time LTI systems with unrestricted inputs, this is equivalent to reaching every final state from zero over any prescribed positive horizon.

Reaching any final state from any initial state is equivalent here, because the required displacement is x_f−e^{AT}x₀ and e^{AT} is invertible. This equivalence requires care in discrete time: if A is singular, driving an initial state to zero and reaching every state from zero need not be the same property. Throughout this lesson, controllability refers to the continuous-time finite-dimensional convention. A rank-deficient pair may still be stabilizable if every uncontrollable mode is already strictly stable; full state transfer is a stronger demand than feedback stabilization.

C=[B AB ⋯ An−1B],(A,B) controllable  ⟺  rank⁡C=n\mathcal C=[B\ AB\ \cdots\ A^{n-1}B],\qquad (A,B)\text{ controllable}\iff\operatorname{rank}\mathcal C=n
This test concerns finite-dimensional continuous-time LTI dynamics with unrestricted inputs. Constraints and numerical tolerances require additional analysis.↔ Scroll wide equations horizontally.

The Gramian is the geometry of the transfer map

On horizon T>0, the endpoint map sends an input u to the integral of e^{A(T−t)}Bu(t). Its adjoint sends a state-space vector v to Bᵀe^{Aᵀ(T−t)}v. Composing endpoint map with adjoint gives the reachability Gramian W_T. The quadratic form vᵀW_Tv is the integral of the squared norm of the adjoint signal, hence is nonnegative. It is zero precisely when that direction is orthogonal to every endpoint reachable through the input. The Gramian is positive definite exactly for a controllable pair.

When W_T is invertible, the minimum-energy input is the adjoint signal associated with W_T^{-1} times the required endpoint displacement. Substituting it into the endpoint map verifies that it reaches the target. Any other reaching input differs by an element of the endpoint map’s kernel, which is orthogonal to this minimum-energy signal. Pythagoras in the input function space then proves minimality, not merely stationarity. The resulting energy is a quadratic form in the target displacement. Small Gramian eigenvalues correspond to expensive state directions.

WT=∫0TeAτBB⊤eA⊤τ dτ,u∗(t)=B⊤eA⊤(T−t)WT−1d,Emin⁡=d⊤WT−1d,d=xf−eATx0W_T=\int_0^T e^{A\tau}BB^\top e^{A^\top\tau}\,d\tau,\quad u_*(t)=B^\top e^{A^\top(T-t)}W_T^{-1}d,\quad E_{\min}=d^\top W_T^{-1}d,\quad d=x_f-e^{AT}x_0
The formulas require W_T positive definite. Energy means the integral of uᵀu over the chosen horizon, not peak actuator magnitude.↔ Scroll wide equations horizontally.

Rank is qualitative; energy is quantitative

A nearly singular Gramian creates a distinction between exact algebra and numerical practice. A theoretical rank calculation may say all directions are reachable, while finite precision makes an inverse unreliable and the corresponding input extremely large. State and input units also affect raw Gramian magnitudes, so compare energies only after choosing physically meaningful scaling. If the target lies in a proper reachable subspace, a pseudoinverse can compute minimum energy within that subspace; it cannot make an unreachable component disappear while still satisfying the original endpoint requirement.

For a Hurwitz A, the infinite-horizon Gramian converges and solves AW+WAᵀ+BBᵀ=0. Hurwitz stability is a sufficient convergence hypothesis, not a necessary one for arbitrary B: an unreachable unstable mode can contribute zero to the integrand. Finite-horizon reachability itself does not require stable dynamics. A longer horizon often reduces minimum energy because a feasible earlier transfer can be embedded in a longer zero-start experiment, but constraints and a nonzero prescribed initial state change the comparison one must make. In design, inspect both structural rank and energy geometry before concluding that an actuator placement is adequate.

Move position while ending at rest

For the double integrator ẋ₁=x₂, ẋ₂=u, transfer zero to (1,0) in time T=1 with minimum ∫u²dt.

  1. Here eAτB=(τ,1)⊤e^{A\tau}B=(\tau,1)^\top, so W1=(1/31/21/21)W_1=\begin{pmatrix}1/3&1/2\\1/2&1\end{pmatrix} and W1−1=(12−6−64)W_1^{-1}=\begin{pmatrix}12&-6\\-6&4\end{pmatrix}. Its positive determinant is 1/12.

  2. Multiply W_1^{-1}(1,0)ᵀ=(12,−6)ᵀ. Then u(t)=(1−t,1)(12,−6)ᵀ=6−12t.

  3. The final velocity is ∫₀¹(6−12t)dt=0. The final position is ∫₀¹(1−t)(6−12t)dt=1, and the energy is ∫₀¹(6−12t)²dt=12.

The result

The optimal command accelerates and then brakes. Constant positive input could move the position but could not meet the zero-final-velocity requirement.

Explore the model

Interactive lab

Interactive lab · matrix

Directional distortion and difficult inverses

Pause and Reset, then set a=1,b=0; predict both singular values equal one. For a fresh comparison Reset and set a=0.25,b=0; predict singular values 1 and 0.25 and condition number 4. The transformed square compresses one direction. This triangular matrix is not the reachability Gramian or endpoint operator; it illustrates why small singular values make inverse problems directionally expensive.

Directional distortion and difficult inversesThe selected matrix is A = [[a,b],[0,1]]. Playback interpolates geometry between the identity and A; readouts always describe A. Its determinant is a and its singular values come from AᵀA. Positive a makes every supported matrix invertible. This two-dimensional family is not a general eigensolver or a learned network. Current readings: Area scale · det A: 1.4; Singular values · σmax, σmin: 1.597, 0.876; Condition number · κ₂(A): 1.823. All parameters are adjustable with the labeled controls below.Matrix A with first row 1.4, 0.6 and second row zero, one; display interpolates between identity and A while final determinant remains 1.4.−4−2024−2−1012x · equal geometric scaleyA = [ 1.4 0.6 ; 0 1 ]Display blend: 0% · final values below
Interpolated geometryOriginal unit squareSelected final transform
Area scale · det A
1.4
Singular values · σmax, σmin
1.597, 0.876
Condition number · κ₂(A)
1.823
t = 0 s

The selected matrix is A = [[a,b],[0,1]]. Playback interpolates geometry between the identity and A; readouts always describe A. Its determinant is a and its singular values come from AᵀA. Positive a makes every supported matrix invertible. This two-dimensional family is not a general eigensolver or a learned network. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forFull rank does not imply that all directions are equally easy to produce.

Check your understanding

A continuous-time LTI pair has a full-rank controllability matrix but a very small Gramian eigenvalue. What follows?

Try it yourself

For ẋ=−x+u, x(0)=0, T=log 2 and target x(T)=1, compute the Gramian, minimum energy and optimal input.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Integrate exp(−2τ) from zero to T, then use the scalar endpoint-adjoint formula.

Compare with the solution

The Gramian is (1−exp(−2T))/2=(1−1/4)/2=3/8. Hence minimum energy is 1/W=8/3 and u*(t)=(8/3)exp(t−T). Substitution into the endpoint response gives (8/3)∫₀ᵀexp(−2(T−t))dt=(8/3)(3/8)=1. Its squared-input integral is (64/9)(3/8)=8/3, verifying both target and energy directly.

Take it with you

  • Continuous-time controllability rank decides unrestricted finite-time state transfer.
  • The reachability Gramian supplies a minimum-energy metric.
  • A small Gramian eigenvalue can make an exactly reachable direction practically difficult.

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