Worked examples — Observability matrix — rank test

We only ever touch the parent's core objects: the observability matrix $\mathcal{O}$, its rank and null space, and — for the last example — the PBH test. Nothing new is assumed.

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The scenario matrix

Every observability problem falls into one of these case classes. The goal of this page is to hit every row at least once.

#	Case class	What's tricky about it	Hit by
A	Full-rank, distinct eigenvalues, single output	The "textbook yes"	Ex 1
B	Hidden decoupled state	A whole column of $\mathcal{O}$ dies	Ex 2
C	Repeated eigenvalue, 2 eigenvectors	Scalar output cannot separate them	Ex 3
D	Repeated eigenvalue, 1 eigenvector (Jordan block)	Coupling rescues observability	Ex 4
E	Zero / degenerate input: $C=0$	The blind sensor — limiting case	Ex 5
F	$3\times 3$ system, does adding a 2nd sensor help?	Rank climbs with $p$	Ex 6
G	Real-world: spacecraft attitude + gyro bias	Word problem, physical meaning of $\ker \mathcal{O}$	Ex 7
H	Exam twist: rank looks full but PBH disagrees?	Cross-check two tests agree	Ex 8

Recall The recipe (memorise once, reuse below)

1. Read off $n$ (size of $A$), $p$ (rows of $C$).
2. Compute $C,CA,C{A}^2,\dots ,C{A}^{n-1}$.
3. Stack them into $\mathcal{O}$ (size $pn\times n$).
4. $\mathrm{rank}(\mathcal{O})=n\Rightarrow$ observable; less $\Rightarrow$ find the unobservable direction in $\ker \mathcal{O}$.

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Example 1 — Case A: the textbook "yes"

Step 1 — build the second row $CA$. $CA=\left[\begin{array}{cc}1& 1\end{array}\right]\left[\begin{array}{cc}-1& 0\\ 0& -3\end{array}\right]=\left[\begin{array}{cc}-1& -3\end{array}\right].$ Why this step? With $n=2$ we need exactly $C$ and $CA$ — one power up, no more (recipe step 2).

Step 2 — stack. $\mathcal{O}=\left[\begin{array}{cc}1& 1\\ -1& -3\end{array}\right].$ Why this step? Rows of $\mathcal{O}$ are the directions the sensor can "see" through $x(0)$ and its derivative.

Step 3 — rank. $\det \mathcal{O}=(1)(-3)-(1)(-1)=-3+1=-2\ne 0$, so $\mathrm{rank}=2=n$ ⇒ observable. ✅ Why this step? A nonzero determinant means no direction of $x(0)$ maps to the zero output.

Verify: the two modes have distinct eigenvalues $-1,-3$, and $C$ has a nonzero entry on each eigenvector ($C=[11]$). Distinct eigenvalues + $C$ touching every eigenvector is exactly the condition the PBH test wants — the two answers agree.

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Example 2 — Case B: a hidden decoupled state

Step 1 — $CA$. $CA=\left[\begin{array}{cc}1& 0\end{array}\right]\left[\begin{array}{cc}-1& 0\\ 0& -3\end{array}\right]=\left[\begin{array}{cc}-1& 0\end{array}\right].$ Why this step? Recipe — grab derivative information.

Step 2 — stack & look at columns. $\mathcal{O}=\left[\begin{array}{cc}1& 0\\ -1& 0\end{array}\right].$ The entire second column is zero. Why this matters: a zero column means the direction $x_0=[0,1{]}^{\top }$ multiplies into nothing.

Step 3 — rank. $\mathrm{rank}=1<2$ ⇒ not observable. ❌

Step 4 — name the ghost. $\ker \mathcal{O}=\mathrm{span}\{[0,1{]}^{\top }\}$. Check: $C{e}^{At}[0,1{]}^{\top }=\left[\begin{array}{cc}1& 0\end{array}\right]\left[\begin{array}{c}0\\ e^{-3t}\end{array}\right]=0$ for all $t$.

Verify: the unobservable state gives $y(t)\equiv 0$ — invisible exactly as the parent note's boxed definition of an unobservable state promised.

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Example 3 — Case C: repeated eigenvalue, two eigenvectors

Step 1 — $CA$. $CA=\left[\begin{array}{cc}1& 1\end{array}\right]\left[\begin{array}{cc}2& 0\\ 0& 2\end{array}\right]=\left[\begin{array}{cc}2& 2\end{array}\right].$

Step 2 — stack. $\mathcal{O}=\left[\begin{array}{cc}1& 1\\ 2& 2\end{array}\right].$ Row 2 $=2\times$ Row 1. Why this matters: rows are parallel, so they span only a 1-D space.

Step 3 — rank. $\det =1\cdot 2-1\cdot 2=0$ ⇒ $\mathrm{rank}=1<2$ ⇒ not observable. ❌

Step 4 — the ghost. $\ker \mathcal{O}=\mathrm{span}\{[1,-1{]}^{\top }\}$: the difference of the two states is invisible.

Verify: eigenvalue $2$ has geometric multiplicity $2$ (two independent eigenvectors). A scalar output can never resolve that — you'd need $p\ge 2$ rows in $C$. Matches the parent's "modal degeneracy" warning.

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Example 4 — Case D: repeated eigenvalue, ONE eigenvector (coupling rescues it)

Step 1 — $CA$. $CA=\left[\begin{array}{cc}1& 0\end{array}\right]\left[\begin{array}{cc}2& 1\\ 0& 2\end{array}\right]=\left[\begin{array}{cc}2& 1\end{array}\right].$ Why this step? The coupling term $1$ leaks state 2 into the derivative of the measured state 1.

Step 2 — stack. $\mathcal{O}=\left[\begin{array}{cc}1& 0\\ 2& 1\end{array}\right].$

Step 3 — rank. $\det =1\cdot 1-0\cdot 2=1\ne 0$ ⇒ $\mathrm{rank}=2=n$ ⇒ observable! ✅

Verify: the difference from Ex 3 is geometric multiplicity. A repeated eigenvalue with a single eigenvector (a genuine Jordan block) has geometric multiplicity $1$; the PBH test then only needs $C$ to touch that one eigenvector — which $C=[10]$ does. So repeated eigenvalues are not automatically fatal: it's the eigenvector count, not the eigenvalue count, that decides.

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Example 5 — Case E: the blind sensor ($C=0$)

Step 1 — $CA$. $CA=\left[\begin{array}{cc}0& 0\end{array}\right]A=\left[\begin{array}{cc}0& 0\end{array}\right].$ Why this step? Any power of $A$ left-multiplied by the zero row stays zero.

Step 2 — stack. $\mathcal{O}=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right].$

Step 3 — rank. $\mathrm{rank}=0<2$ ⇒ not observable — maximally so. ❌

Step 4 — the ghost is everything. $\ker \mathcal{O}={\mathbb{R}}^2$: every initial state gives $y\equiv 0$.

Verify: rank $0$ is the floor of the rank test; the "number of observable directions" equals $\mathrm{rank}(\mathcal{O})=0$, so zero of the $n=2$ states are recoverable — the correct limiting answer for a dead sensor.

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Example 6 — Case F: does a second sensor help? ($3\times 3$)

Step 1 — powers of $A$ times $C_1$. $C_1=\left[\begin{array}{ccc}1& 0& 0\end{array}\right],C_{1}A=\left[\begin{array}{ccc}0& 1& 0\end{array}\right],C_1{A}^2=\left[\begin{array}{ccc}0& 0& 1\end{array}\right].$ Why this step? $n=3$ so we go up to $A^2$.

Step 2 — stack for the single sensor. ${\mathcal{O}}_1=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]=I_3,\mathrm{rank}=3=n.$ Observable already! Why: position → velocity → acceleration; each derivative peels off the next state.

Step 3 — the two-sensor case (does it hurt?). $C_{2}A=\left[\begin{array}{ccc}0& 1& 0\\ 0& 0& 0\end{array}\right],C_2{A}^2=\left[\begin{array}{ccc}0& 0& 1\\ 0& 0& 0\end{array}\right].$ ${\mathcal{O}}_2=\left[\begin{array}{ccc}1& 0& 0\\ 0& 0& 1\\ 0& 1& 0\\ 0& 0& 0\\ 0& 0& 1\\ 0& 0& 0\end{array}\right],\mathrm{rank}=3=n.$ Why this step? Extra sensor rows can only raise or keep the rank — never lower it — but rank is capped at $n=3$.

Verify: both give rank $3$. The lesson: a chain of integrators is observable from its first state alone; the extra sensor is redundant here (rank already saturated). Adding rows never reduces observability — it just can't exceed $n$.

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Example 7 — Case G: real-world spacecraft attitude + gyro bias

Step 1 — $CA$. $CA=\left[\begin{array}{cc}1& 0\end{array}\right]\left[\begin{array}{cc}0& -1\\ 0& 0\end{array}\right]=\left[\begin{array}{cc}0& -1\end{array}\right].$ Why this step? The bias couples into $\dot{\theta }$, so the derivative of the measured angle carries information about $b$.

Step 2 — stack. $\mathcal{O}=\left[\begin{array}{cc}1& 0\\ 0& -1\end{array}\right],\det =-1\ne 0,\mathrm{rank}=2=n.$

Step 3 — verdict. Observable ✅ — the star tracker can separate attitude from bias.

Verify (physical + units): $\dot{\theta }=-b$ (with $u=0$), so a nonzero constant bias makes the measured angle drift linearly: $\theta (t)={\theta }_0-bt$. The slope of the drift reveals $b$ (units: rad/s) and the intercept reveals ${\theta }_0$ (rad) — two observations, two unknowns, exactly what rank $2$ guarantees. This is why a Kalman filter (Kalman Filter) can estimate and remove gyro bias from a single angle measurement.

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Example 8 — Case H: exam twist, do rank test and PBH agree?

Step 1 — rank test. $CA=\left[\begin{array}{cc}-1& 0\end{array}\right],\mathcal{O}=\left[\begin{array}{cc}1& 0\\ -1& 0\end{array}\right],\mathrm{rank}=1<2.$ ⇒ not observable. ❌

Step 2 — PBH cross-check. Why this step? The PBH test checks, at each eigenvalue $\lambda$, whether $\left[\begin{array}{c}A-\lambda I\\ C\end{array}\right]$ has full column rank $n$. At $\lambda =-1$: $\left[\begin{array}{c}A-(-1)I\\ C\end{array}\right]=\left[\begin{array}{cc}0& 0\\ 0& 0\\ 1& 0\end{array}\right],\mathrm{rank}=1<2.$

Step 3 — reconcile. Both tests say not observable. The student's error: $-1$ is a repeated eigenvalue with two eigenvectors (it's $-I$, geometric multiplicity $2$), and a scalar $C$ cannot resolve two identical modes — exactly the Case C failure in disguise.

Verify: $\ker \mathcal{O}=\mathrm{span}\{[0,1{]}^{\top }\}$, and $C{e}^{At}[0,1{]}^{\top }=\left[\begin{array}{cc}1& 0\end{array}\right]\left[\begin{array}{c}0\\ e^{-t}\end{array}\right]=0$. The rank test and PBH must agree — they are two windows on the same geometry, as noted in the parent's connections.

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Connections

• Observability matrix — rank test (parent)
• PBH test (used in Ex 8 as a cross-check)
• Kalman Filter (Ex 7 bias estimation)
• Kalman decomposition (splits out the ghost states from Ex 2, 3, 5)
• Rank and null space of a matrix (every step lives here)
• Cayley–Hamilton theorem (why we stop at $A^{n-1}$ throughout)
• Controllability matrix — rank test (dual)
• State-space representation