Controllability matrix — rank test: Exercises — Physics deep dive

Intuition How to use this page

These problems climb a ladder. L1 just asks you to recognize the pieces; L5 asks you to invent and combine ideas. Every problem has a full worked solution hidden in a collapsible callout — try it yourself first, then reveal. This page builds directly on the parent note.

First, so nobody trips on a symbol, here is every letter we use — with a picture:

$x$ is the state vector: the full list of numbers that describes the system right now (positions, velocities, angles, …). Geometrically, $x$ is a point in an $n$ -dimensional space; as time passes it traces a path.
$u$ is the input (control): the numbers you choose to push the system — thruster levels, voltages. Geometrically, $u$ chooses which push directions get switched on.
$\overset{x}{˙} = A x + B u$ is the rule for how the point $x$ moves: $A$ tells you how $x$ drifts on its own, $B$ tells you how your push $u$ nudges it.
$A$ is the $n \times n$ system matrix (how the state evolves on its own).
$B$ is the $n \times m$ input matrix (how your controls push the state).
$C = [B A B A^{2} B \dots A^{n - 1} B]$ is the controllability matrix.
$rank (C)$ = the number of independent directions its columns reach. Controllable $⟺ rank = n$ .

The whole game is geometric: your inputs can only ever produce motion inside the span of the columns of $C$ . Controllable means that span fills the entire state space — nothing is locked out. The picture below is the mental image to carry through every problem.

Figure — Controllability matrix — rank test

Level 1 — Recognition

Recall Solution L1.1

(a) $x$ has 2 rows, so $n = 2$ . (b) $u$ is a single scalar (one column in $B$ ), so $m = 1$ . (c) $C$ is $n \times (n \cdot m) = 2 \times (2 \cdot 1) = 2 \times 2$ . (d) You compute $B, A B, \dots, A^{n - 1} B$ , i.e. up to $A^{1} B$ . That is 2 blocks: $B$ and $A B$ . (Cayley–Hamilton says $A^{2} B$ adds nothing — see Cayley–Hamilton theorem.)

Recall Solution L1.2

The second row is entirely zero, so the columns only ever reach vectors of the form $[a 0]$ . That spans one direction, so $rank (C) = 1 < 2 = n$ . Not controllable. The dead direction is the $x_{2}$ -axis $[01]$ : no input combination ever produces motion there.

Level 2 — Application

Recall Solution L2.1

Step 1 — $A B$ (only block we need since $n = 2$ ): $A B = [0 - 1 2 - 3] [01] = [2 - 3] .$ Step 2 — stack: $C = [B A B] = [01 2 - 3] .$ Step 3 — determinant (fastest full-rank test for a square $C$ ): $det C = 0 \cdot (- 3) - 2 \cdot 1 = - 2 \neq = 0.$ Nonzero determinant $\Rightarrow rank = 2 = n$ . ✅ Controllable.

Recall Solution L2.2

Step 1: $A B = [40 0 - 1] [30] = [120] .$ Step 2: $C = [30120] .$ Step 3: $det = 3 \cdot 0 - 12 \cdot 0 = 0$ . Second row all zero $\Rightarrow rank = 1 < 2$ . ❌ Not controllable. Physically: $A$ is diagonal, so the two modes never talk to each other, and $B$ 's second entry is $0$ , so the input never touches mode 2. Mode 2 evolves as $\overset{x}{˙}_{2} = - x_{2}$ with no input — untouchable. (See Kalman decomposition for splitting such systems.)

Recall Solution L2.3

Step 1: $A B = [4 - 1] .$ Step 2: $C = [11 4 - 1] .$ Step 3: $det = 1 \cdot (- 1) - 4 \cdot 1 = - 5 \neq = 0 \Rightarrow rank = 2$ . ✅ Controllable. The input now pushes both modes, and because the two modes have different eigenvalues ( $4$ and $- 1$ ), $B$ and $A B$ point in different directions and span the plane.

Level 3 — Analysis

Recall Solution L3.1

Step 1: $A B = [1 λ] .$ Step 2: $C = [11 1 λ] .$ Step 3: $det C = 1 \cdot λ - 1 \cdot 1 = λ - 1.$ Uncontrollable exactly when $det = 0$ , i.e. $λ = 1$ . Why: at $λ = 1$ the two diagonal modes share the same eigenvalue. With repeated eigenvalues and a single input, the input cannot distinguish the two modes — $B$ and $A B = λ B$ become parallel. This is the intuition behind the PBH test (Popov–Belevitch–Hautus): a mode is uncontrollable when $B$ lies "flat" against a repeated/shared eigen-direction.

Look at the figure below: as $λ$ slides from $3$ down toward $1$ , the vector $A B$ swings closer and closer to $B$ . At $λ = 1$ they coincide — $C$ 's two columns lie on one line, so their span is a line, not the plane, and rank drops to $1$ .

Recall Solution L3.2

Step 1: $A B = [24] = 2 B .$ ( $A = 2 I$ , so $A B = 2 B$ — parallel to $B$ !) Step 2: $C = [1224] .$ Step 3: $det = 1 \cdot 4 - 2 \cdot 2 = 0 \Rightarrow rank = 1$ . ❌ Not controllable. Reachable subspace = span of $B = [12]$ , i.e. the line $x_{2} = 2 x_{1}$ . Unreachable target: any point off that line, e.g. $x_{f} = [10]$ . Check: is $[10]$ a multiple of $[12]$ ? No. So it can never be reached from the origin. Deep reason: when $A = 2 I$ (a scalar times identity), every power $A^{k} B$ is just a scalar times $B$ — all blocks stay on one line. A single input can never leave that line.

In the figure below, the magenta line is the reachable subspace $x_{2} = 2 x_{1}$ ; both $B$ (navy) and $A B = 2 B$ (orange) lie on it, while the violet target $(1, 0)$ floats off the line — provably out of reach.

Level 4 — Synthesis

Recall Solution L4.1

Here $n = 3$ , so we need $B, A B, A^{2} B$ . $A B$ : $A B = 000100010001 = 010 .$ $A^{2} B = A (A B)$ : $A^{2} B = 000100010010 = 100 .$ Stack: $C = 001010100 .$ Determinant: this is a permutation (anti-diagonal) matrix; $det = - 1 \neq = 0 \Rightarrow rank = 3 = n$ . ✅ Controllable. Why it works: the single input hits the bottom mode, and each power of $A$ shifts it up the chain — the "ripple" $B \to A B \to A^{2} B$ fills all three directions. This is why chains of integrators are the poster child of controllability, and why Pole placement & state feedback can freely assign every pole here.

Recall Solution L4.2

Why single input fails: $A = I$ , so $A B = I B = B$ . Then $C = [B B]$ — the two columns are identical, rank $\leq 1 < 2$ . No single-column $B$ can win. Smallest working $B$ : we need $m = 2$ columns whose combined span is $R^{2}$ . Take $B = [1001] .$ Then $A B = B$ , and $C = [B A B] = [10011001]$ . Rank of a rectangular $C$ — use the minors rule from the definition box above. This $2 \times 4$ matrix contains the $2 \times 2$ sub-block (columns 1 and 2) $[1001]$ with $det = 1 \neq = 0$ , so $rank = 2 = n$ . ✅ Controllable. (Row reduction gives the same: the two non-zero rows confirm rank 2.) Lesson: when $A = I$ every mode has the same eigenvalue and $A$ does no mixing at all; you must supply the directions directly through $B$ . You need at least as many independent input columns as the eigenvalue multiplicity.

Level 5 — Mastery

Recall Solution L5.1

$A B$ : $A 01 a = 1 a 0 .$ $A^{2} B = A (A B)$ : $A 1 a 0 = a 00 .$ Stack: $C = 01 a 1 a 0 a 00 .$ Determinant (expand along the bottom row — only one nonzero entry, $a$ in position $(3, 1)$ ): $det C = a \cdot det [1 a a 0] = a (1 \cdot 0 - a \cdot a) = a (- a^{2}) = - a^{3} .$ Controllable $⟺ det \neq = 0 ⟺ - a^{3} \neq = 0 ⟺ a \neq = 0$ . Interpretation: here $B = 01 a$ injects the input into the middle state (and, if $a \neq = 0$ , also the bottom state). At $a = 0$ the input enters only the middle rung, so it can never reach the third state — uncontrollable. This is a different injection point from L4.1 (whose $B = 001$ enters at the bottom and stays controllable). Any $a \neq = 0$ here re-connects the bottom rung and restores full rank. This is exactly the kind of edge case Stabilizability cares about.

Recall Solution L5.2

Key idea (duality): $(A, C)$ is observable $⟺ (A^{T}, C^{T})$ is controllable. So build the controllability matrix of the transposed pair. (This is the bridge to the Observability matrix — rank test.) Let $A^{T} = [- 1 2 0 - 4]$ and $C^{T} = [10]$ . $A^{T} C^{T}$ : $[- 1 2 0 - 4] [10] = [- 1 2] .$ Stack: $C_{dual} = [10 - 1 2] .$ Determinant: $1 \cdot 2 - (- 1) \cdot 0 = 2 \neq = 0 \Rightarrow rank = 2 = n$ . Controllable dual $\Rightarrow$ observable original. ✅ You reused the same rank test with $A \to A^{T}, B \to C^{T}$ .

Recall Solution L5.3

$A B$ : $[0 - a_{0} 1 - a_{1}] [01] = [1 - a_{1}] .$ Stack: $C = [01 1 - a_{1}] .$ Determinant: $0 \cdot (- a_{1}) - 1 \cdot 1 = - 1 \neq = 0$ for all $a_{0}, a_{1}$ . So $rank = 2 = n$ always. ✅ Always controllable. Why this matters: companion form is by construction controllable, which is exactly why controllable systems can be transformed into it — and why Pole placement & state feedback uses companion form to read off feedback gains directly. Note $a_{0}$ never even appeared in $det C$ : it changes the poles, not the controllability.

Recall One-line summary

Every problem reduced to the same three moves: compute the $A^{k} B$ blocks up to $A^{n - 1} B$ , stack them into $C$ , and test the rank (determinant when $C$ is square; minors or row-reduction when it is rectangular). Full rank $= n$ ⇒ controllable; anything less ⇒ a mode is forever locked.