Worked examples — Controllability matrix — rank test

Everything here uses only ideas already built in the parent: the State-space representation $\dot{x}=Ax+Bu$, matrix multiplication, determinant/rank, and Cayley–Hamilton theorem to know we stop at $A^{n-1}B$.

The scenario matrix

Before any numbers, let's map out the space of possible cases. Every controllability problem lives in one of these cells. Our 8 examples below each carry a tag like (C2) telling you which cell it fills.

C1 — Full rank, single input

1. Compute $AB$ (we need up to $A^{n-1}B=A^{1}B$ because $n=2$). $AB=\left[\begin{array}{cc}0& 1\\ -2& -3\end{array}\right]\left[\begin{array}{c}0\\ 1\end{array}\right]=\left[\begin{array}{c}1\\ -3\end{array}\right].$ Why this step? Cayley–Hamilton theorem guarantees $A^{2}B$ adds nothing new, so $\{B,AB\}$ is the complete list.

2. Stack into $\mathcal{C}$. $\mathcal{C}=[BAB]=\left[\begin{array}{cc}0& 1\\ 1& -3\end{array}\right].$ Why this step? The reachable subspace is the column span of $\mathcal{C}$; stacking side by side is how we collect all reachable directions.

3. Take the determinant (fastest rank test for a square matrix). $\det \mathcal{C}=0\cdot (-3)-1\cdot 1=-1\ne 0.$ Why this step? Non-zero determinant $\Rightarrow$ full rank $=2=n$.

✅ Controllable.

C2 — Decoupled modes, actuator misses one

1. Compute $AB$. $AB=\left[\begin{array}{cc}-1& 0\\ 0& -4\end{array}\right]\left[\begin{array}{c}1\\ 0\end{array}\right]=\left[\begin{array}{c}-1\\ 0\end{array}\right].$ Why this step? Diagonal $A$ scales each coordinate; the zero in row 2 of $B$ stays zero.

2. Stack. $\mathcal{C}=\left[\begin{array}{cc}1& -1\\ 0& 0\end{array}\right].$ Why this step? Row 2 is entirely zero — a giant hint.

3. Rank. Second row all zeros $\Rightarrow \mathrm{rank}=1<2$. Why this step? One independent row means one reachable dimension. $x_2$ is the dead direction.

❌ Not controllable. The mode ${\dot{x}}_2=-4x_2$ runs with no input.

C3 — Same $A$, coupling in $B$ saves it

1. Compute $AB$. $AB=\left[\begin{array}{c}-1\\ -4\end{array}\right].$ Why this step? Each mode scales by its own eigenvalue, so $B$ and $AB$ now differ in proportion, not just magnitude.

2. Stack. $\mathcal{C}=\left[\begin{array}{cc}1& -1\\ 1& -4\end{array}\right].$

3. Determinant. $\det =1\cdot (-4)-(-1)\cdot 1=-4+1=-3\ne 0.$ Why this step? Non-zero $\Rightarrow$ rank $2$.

✅ Controllable.

C4 — Degenerate input $B=0$

1. Compute $AB$. $AB=A\cdot 0=\left[\begin{array}{c}0\\ 0\end{array}\right].$ Why this step? $A$ times the zero vector is zero — every power of $A$ applied to $0$ stays $0$.

2. Stack. $\mathcal{C}=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right].$

3. Rank. All zeros $\Rightarrow \mathrm{rank}=0<2$.

❌ Uncontrollable — reachable subspace is just the origin. This is the extreme limiting case: zero rank, zero reachable dimensions.

C5 — Multi-input: two weak inputs, one strong system

1. Compute $AB$. Since $A=0$, $AB=0$ and $A^{2}B=0$. Why this step? With $A=0$, higher power blocks vanish — only $B$ itself carries reachable directions.

2. Stack $\mathcal{C}=[BAB{A}^{2}B]$, size $3\times 6$: $\mathcal{C}=\left[\begin{array}{cccccc}1& 0& 0& 0& 0& 0\\ 0& 1& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0\end{array}\right].$

3. Rank. Only two non-zero rows $\Rightarrow \mathrm{rank}=2<3$.

❌ Uncontrollable. Direction 3 is unreachable — no thruster, no coupling.

C6 — $n=3$ where $A^{2}B$ genuinely matters

1. Compute $AB$. $AB=\left[\begin{array}{ccc}0& 1& 0\\ 0& 0& 1\\ 0& 0& 0\end{array}\right]\left[\begin{array}{c}0\\ 0\\ 1\end{array}\right]=\left[\begin{array}{c}0\\ 1\\ 0\end{array}\right].$ Why this step? $A$ shifts the "1" one row up — this is the cascade in action.

2. Compute $A^{2}B=A(AB)$. $A^{2}B=\left[\begin{array}{ccc}0& 1& 0\\ 0& 0& 1\\ 0& 0& 0\end{array}\right]\left[\begin{array}{c}0\\ 1\\ 0\end{array}\right]=\left[\begin{array}{c}1\\ 0\\ 0\end{array}\right].$ Why this step? Now the "1" reaches the top state — this final power is what fills the last dimension. Skipping it would falsely look rank-2.

3. Stack and check. $\mathcal{C}=\left[\begin{array}{ccc}0& 0& 1\\ 0& 1& 0\\ 1& 0& 0\end{array}\right].$ This is the anti-diagonal identity: $\det =+1\cdot (0\cdot 0-1\cdot 1)\cdots$ — cleaner to note it's a permutation matrix, so $\det =\pm 1\ne 0$, rank $3$. Why this step? Full rank $=3=n$.

✅ Controllable. This is why the parent says "stop at $A^{n-1}B$" and not earlier — here $A^{2}B$ was essential.

C7 — The repeated-eigenvalue trap

1. Case (a): compute $A_{a}B$. $A_{a}B=\left[\begin{array}{cc}2& 1\\ 0& 2\end{array}\right]\left[\begin{array}{c}0\\ 1\end{array}\right]=\left[\begin{array}{c}1\\ 2\end{array}\right].$ Stack: ${\mathcal{C}}_a=\left[\begin{array}{cc}0& 1\\ 1& 2\end{array}\right]$, $\det =0\cdot 2-1\cdot 1=-1\ne 0$. Controllable ✅ Why this step? The off-diagonal $1$ mixes the two states, so a single input reaches both.

2. Case (b): compute $A_{b}B$. $A_{b}B=\left[\begin{array}{cc}2& 0\\ 0& 2\end{array}\right]\left[\begin{array}{c}1\\ 1\end{array}\right]=\left[\begin{array}{c}2\\ 2\end{array}\right].$ Stack: ${\mathcal{C}}_b=\left[\begin{array}{cc}1& 2\\ 1& 2\end{array}\right]$, $\det =1\cdot 2-2\cdot 1=0$. Uncontrollable ❌ Why this step? With a diagonalizable repeated eigenvalue, the two modes are indistinguishable to any single input — $B$ and $A_{b}B$ are parallel.

C8 — Word problem + exam twist

1. Write $(A,B)$ from the physics. $A=\left[\begin{array}{cc}0& 1\\ -k& 0\end{array}\right],B=\left[\begin{array}{c}0\\ b\end{array}\right].$ Why this step? Reading $\dot{x}=Ax+Bu$ off the two scalar equations — the State-space representation step.

2. Compute $AB$. $AB=\left[\begin{array}{cc}0& 1\\ -k& 0\end{array}\right]\left[\begin{array}{c}0\\ b\end{array}\right]=\left[\begin{array}{c}b\\ 0\end{array}\right].$ Why this step? Only need $A^{1}B$ since $n=2$.

3. Stack and take determinant symbolically. $\mathcal{C}=\left[\begin{array}{cc}0& b\\ b& 0\end{array}\right],\det \mathcal{C}=0\cdot 0-b\cdot b=-b^2.$ Why this step? Symbolic determinant lets us solve for the failure condition directly.

4. Solve $\det =0$. $-b^2=0⟺b=0$. Why this step? Uncontrollable exactly when the determinant vanishes.

Answer: controllable for every $b\ne 0$ (any value of $k$!), and uncontrollable only when $b=0$ — the physically obvious "dead thruster" case. The stiffness $k$ never affects controllability.

Recap: which cell taught what

Practice reflexes:

Connections

• Controllability matrix — rank test (parent — the machine these examples exercise)
• Observability matrix — rank test (dual: run the same drill on $A^T,C^T$)
• State-space representation (the C8 translation step)
• Cayley–Hamilton theorem (why we stop at $A^{n-1}B$ in every example)
• Pole placement & state feedback (C1/C3/C6 controllability lets you place all poles)
• Kalman decomposition (C2/C4/C5 uncontrollable modes get split off here)
• Matrix exponential $e^{At}$ (the propagator behind reachability)
• Stabilizability (C7b's uncontrollable mode is unstable → not even stabilizable)

Concept Map