Exercises — Observability matrix — rank test

$\mathcal{O}$ stacks $n=4$ blocks ($C,CA,C{A}^2,C{A}^3$), each of size $p\times n=2\times 4$. So $\mathcal{O}$ is $(pn)\times n=(2\cdot 4)\times 4===8\times 4==$. Full column rank means $\mathrm{rank}(\mathcal{O})=n===4==$. What it looks like: a tall skinny slab, 8 rows tall, 4 columns wide — we only ever need 4 independent rows to "see" all 4 states.

$CA=\left[\begin{array}{cc}1& 0\end{array}\right]\left[\begin{array}{cc}3& 0\\ 0& 5\end{array}\right]=\left[\begin{array}{cc}3& 0\end{array}\right]$. $\mathcal{O}=\left[\begin{array}{cc}1& 0\\ 3& 0\end{array}\right].$ The second column is all zeros ⇒ $\mathrm{rank}=1<2=n$ ⇒ not observable. Why: $C$ only touches state 1; state 2 lives on its own eigenvector, never influencing $y$. $x_0=\left[\begin{array}{c}0\\ 1\end{array}\right]$ is unobservable.

$CA=\left[\begin{array}{cc}0& 5\end{array}\right]$, so $\mathcal{O}=\left[\begin{array}{cc}0& 1\\ 0& 5\end{array}\right]$. First column zero ⇒ rank $1$ ⇒ not observable. Symmetric to L1.2: a single scalar output on one diagonal channel can only ever see that one mode.

$CA=\left[\begin{array}{cc}0& 1\end{array}\right]\left[\begin{array}{cc}0& 1\\ 0& 0\end{array}\right]=\left[\begin{array}{cc}0& 0\end{array}\right]$. $\mathcal{O}=\left[\begin{array}{cc}0& 1\\ 0& 0\end{array}\right],\mathrm{rank}=1<2\Rightarrow \text{not observable}.$ Interpret: state 1 feeds state 2 (via the "1" in $A$), but the output watches state 2 while state 2's dynamics (${\dot{x}}_2=0$) never carry state 1's info back to $y$. Contrast with Example 1 in the parent, where $C=[10]$ was observable — measuring the "upstream" state lets derivatives reveal the "downstream" one, but not vice-versa. Direction of the chain matters.

$CA=\left[\begin{array}{cc}1& 0\end{array}\right]\left[\begin{array}{cc}0& 1\\ -2& -3\end{array}\right]=\left[\begin{array}{cc}0& 1\end{array}\right]$. $\mathcal{O}=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]=I\Rightarrow \mathrm{rank}=2=n\Rightarrow \text{observable}.$ Why it works: this is a position/velocity pair. $y=x_1$ is position; $\dot{y}=CAx=x_2$ is velocity. Two measurements, two states, done.

$CA=\left[\begin{array}{cc}0& 1\end{array}\right]\left[\begin{array}{cc}1& 1\\ 0& 1\end{array}\right]=\left[\begin{array}{cc}0& 1\end{array}\right]$. $\mathcal{O}=\left[\begin{array}{cc}0& 1\\ 0& 1\end{array}\right],\mathrm{rank}=1\Rightarrow \text{not observable}.$ Kernel: solve $\mathcal{O}x=0\Rightarrow x_2=0$, $x_1$ free. So $x_0=\left[\begin{array}{c}1\\ 0\end{array}\right]\in \ker \mathcal{O}$. Verify invisibility: with $x_2(0)=0$, the coupling in $A$ pushes into $x_2$ via the "1", but let us check the output directly. $y=x_2$. Starting from $x_0=(1,0)$: ${\dot{x}}_2=x_2$, and $x_2(0)=0\Rightarrow x_2(t)=0$ for all $t$. Since ${\dot{x}}_2$ depends only on $x_2$ (the second row of $A$ is $[01]$, no $x_1$ term), $x_1$'s value never reaches the output. Hence $y(t)\equiv 0$. ✅

$CA=\left[\begin{array}{cc}4a& 4b\end{array}\right]=4C$. $\mathcal{O}=\left[\begin{array}{cc}a& b\\ 4a& 4b\end{array}\right].$ Row 2 $=4\times$ row 1, so the rows are parallel ⇒ $\mathrm{rank}=1<2$ for every $(a,b)$. Why fundamental: $A=4I$ has eigenvalue $4$ with geometric multiplicity $2$ — a whole 2-D eigenspace. A single output cannot distinguish two identical, decoupled modes (see parent Example 3). By the PBH test, at eigenvalue $\lambda =4$ we'd need $\mathrm{rank}\left[\begin{array}{c}A-4I\\ C\end{array}\right]=2$, but $A-4I=0$, leaving only the one row $C$ ⇒ rank $1$. You need $p\ge 2$ outputs.

Duality: $(A,C)$ observable $⟺(A^{\top },C^{\top })$ controllable. Here $A^{\top }=\left[\begin{array}{cc}0& 0\\ 1& 0\end{array}\right],B:=C^{\top }=\left[\begin{array}{c}1\\ 0\end{array}\right].$ Controllability matrix $\mathcal{C}=[B{A}^{\top }B]$. $A^{\top }B=\left[\begin{array}{cc}0& 0\\ 1& 0\end{array}\right]\left[\begin{array}{c}1\\ 0\end{array}\right]=\left[\begin{array}{c}0\\ 1\end{array}\right]$. $\mathcal{C}=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right],\mathrm{rank}=2\Rightarrow \text{controllable}.✅$ Consistent: observable original $\Rightarrow$ controllable dual, as promised.

Rank test: $CA=\left[\begin{array}{cc}2& 3\end{array}\right]$, $\mathcal{O}=\left[\begin{array}{cc}1& 1\\ 2& 3\end{array}\right]$. Determinant $=1\cdot 3-1\cdot 2=1\ne 0$ ⇒ rank $2$ ⇒ observable. PBH: check each eigenvalue $\lambda \in \{2,3\}$. Need $\mathrm{rank}\left[\begin{array}{c}A-\lambda I\\ C\end{array}\right]=2$.

• $\lambda =2$: $\left[\begin{array}{cc}0& 0\\ 0& 1\\ 1& 1\end{array}\right]$ — rows $(0,1)$ and $(1,1)$ independent ⇒ rank $2$. ✔
• $\lambda =3$: $\left[\begin{array}{cc}-1& 0\\ 0& 0\\ 1& 1\end{array}\right]$ — rows $(-1,0)$ and $(1,1)$ independent ⇒ rank $2$. ✔ Both eigenvalues pass ⇒ observable. Both methods agree. Distinct eigenvalues + a $C$ that "touches" every eigenvector ⇒ observable.

Eigenvalues $1,2,3$ are distinct, each with a 1-D eigenspace (the coordinate axes). With no modal degeneracy, a single output ($p=1$) can work provided it touches all three eigenvectors. Choose $C=\left[\begin{array}{ccc}1& 1& 1\end{array}\right]$. $\mathcal{O}=\left[\begin{array}{ccc}1& 1& 1\\ 1& 2& 3\\ 1& 4& 9\end{array}\right].$ This is a Vandermonde matrix in the nodes $1,2,3$; its determinant $=(2-1)(3-1)(3-2)=2\ne 0$ ⇒ rank $3$ ⇒ observable. Smallest is $p=1$. ✅ Why Vandermonde appears: row $k$ is $C{A}^{k-1}=(1\cdot {\lambda }_1^{k-1},1\cdot {\lambda }_2^{k-1},1\cdot {\lambda }_3^{k-1})$ — powers of the eigenvalues. Distinct nodes ⇒ nonzero determinant.

We want $\ker \mathcal{O}=\mathrm{span}\{(1,1{)}^{\top }\}$, i.e. rank $1$, with $(1,1{)}^{\top }$ giving $y\equiv 0$. Pick $C=\left[\begin{array}{cc}1& -1\end{array}\right]$ so that $C(1,1{)}^{\top }=0$. Now we must ensure $CA(1,1{)}^{\top }=0$ too (so the whole line stays in the kernel — an $A$-invariant subspace). Choose $A=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]=I$. Then $CA=C=\left[\begin{array}{cc}1& -1\end{array}\right]$. $\mathcal{O}=\left[\begin{array}{cc}1& -1\\ 1& -1\end{array}\right],\mathrm{rank}=1,\ker =\mathrm{span}\{(1,1{)}^{\top }\}.✅$ Check: $\mathcal{O}(1,1{)}^{\top }=(0,0{)}^{\top }$ and $(1,-1{)}^{\top }$ spans the row space, so only the $(1,1)$ direction is invisible. This is a mini Kalman decomposition: the observable part is the $(1,-1)$ direction, the unobservable part is $(1,1)$.

The unobservable state is $(0,1{)}^{\top }$. We need a new row that has a nonzero projection onto it. The cheapest: add $C_2=\left[\begin{array}{cc}0& 1\end{array}\right]$. $C=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]=I.$ Then $\mathcal{O}$ already contains $I$ in its first two rows ⇒ rank $2$ ⇒ observable. ✅ Why two outputs are forced here: $A=2I$ has a repeated eigenvalue with a 2-D eigenspace (L3.1). No scalar output can ever fix it — you must have $p\ge 2$. This is not laziness of the designer; it's a hard structural requirement.

Compute a block row of the transformed matrix: $\tilde{C}{\tilde{A}}^k=(C{T}^{-1})(TA{T}^{-1}{)}^k=C{T}^{-1}T{A}^k{T}^{-1}=C{A}^k{T}^{-1},$ using $(TA{T}^{-1}{)}^k=T{A}^k{T}^{-1}$ (the $T^{-1}T$ pairs cancel in the middle). Stacking all $k=0,\dots ,n-1$: $\tilde{\mathcal{O}}=\mathcal{O}{T}^{-1}.$ Since $T^{-1}$ is invertible, right-multiplying by it does not change rank. Hence $\mathrm{rank}(\tilde{\mathcal{O}})=\mathrm{rank}(\mathcal{O})$. ∎ Meaning: observability is a property of the system, not of the basis you happen to write it in — exactly what the Kalman decomposition relies on.

$C=\left[\begin{array}{ccc}1& 0& 0\end{array}\right]$. $CA=\left[\begin{array}{ccc}0& 1& 0\end{array}\right]$ (first row of $A$). $C{A}^2=(CA)A=\left[\begin{array}{ccc}0& 1& 0\end{array}\right]A=\left[\begin{array}{ccc}0& 0& 1\end{array}\right]$ (second row of $A$). $\mathcal{O}=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]=I\Rightarrow \det =1\ne 0\Rightarrow \mathrm{rank}=3=n.$ Observable. ✅ This is the "observer companion" pattern: measuring the first state of a chain of integrators reveals successive derivatives $x_2,x_3$, so the whole chain unrolls.

$CA=\left[\begin{array}{cc}1& 0\end{array}\right]\left[\begin{array}{cc}0& 1\\ -1& 0\end{array}\right]=\left[\begin{array}{cc}0& 1\end{array}\right]$. $\mathcal{O}=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]=I\Rightarrow \mathrm{rank}=2\Rightarrow \text{observable}.✅$ Interpretation: for a harmonic oscillator, $x_1$ is position and ${\dot{x}}_1=x_2$ is velocity. Watching position and its slope gives both states at any instant, so an estimator can lock onto the true trajectory. Eigenvalues here are $\pm i$ (undamped), yet observability holds — reinforcing the parent's mistake note: observability is independent of stability.

• L1: size $=pn\times n$; a zero column kills rank.
• L2: coupling has direction — always compute $C{A}^k$.
• L3: repeated eigenvalues need $p\ge 2$; distinct eigenvalues still need $C$ to touch every eigenvector; PBH cross-checks the rank test.
• L4: design $C$ via Vandermonde; unobservable subspace = $A$-invariant kernel of $\mathcal{O}$; repeated modes force more sensors.
• L5: $\tilde{\mathcal{O}}=\mathcal{O}{T}^{-1}$ ⇒ coordinate-free; observability ⟂ stability.