Question bank — Observability — when KF can estimate state

False — observability is a rank property. Extra sensors that measure already-visible linear combinations add rows to $\mathcal{O}$ but no rank, so a tall $C$ guarantees nothing.

False — observable only guarantees $x(0)$ is eventually recoverable over an interval; a poorly conditioned (weakly observable) $\mathcal{O}$ lets sensor noise swamp the estimate for a long time.

True — we set $u=0$ because known inputs can be subtracted off; only the unforced response $y=C{e}^{At}x(0)$ carries the information about $x(0)$.

False — if the unobservable modes are already stable (they decay on their own) the system is detectable, which is enough for a bounded steady-state covariance.

True in form, false in meaning — they are duals: $(A,C)$ observable $⟺(A^{\top },C^{\top })$ controllable. One is about reading the state out, the other about driving it in.

True — their difference $x_1(0)-x_2(0)$ lies in $\ker (\mathcal{O})$ and produces $y(t)\equiv 0$, so no estimator can distinguish them.

False — observability is a property of $(A,C)$ alone. Noise $Q$ only makes the covariance grow in that blind direction; it never opens a measurement pathway.

False — stability is about $A$'s eigenvalues, observability is about how $C$ couples to them. A decaying but unmeasured decoupled mode is stable yet unobservable (it's detectable, not observable).

The extra block $C{A}^n$ is redundant by Cayley–Hamilton ($A^n$ is a linear combo of lower powers), so it can never raise the rank. Stopping at $C{A}^{n-1}$ is not a shortcut, it's complete.

Observability needs full column rank $=n$, not full row rank. The matrix is tall; we only ever require its columns to be independent so that $\mathcal{O}x(0)=\text{data}$ has a unique solution.

If $A_{12}=0$ and $C$ doesn't touch state 2, nothing about it leaks into $y$; the measurement update leaves that variance untouched, so a small reported variance there is a filter lying to you.

Invertibility (full rank) only decides the yes/no question. A near-singular $\mathcal{O}$ (tiny smallest singular value) is technically observable but weakly so — check the Gramian eigenvalues for conditioning.

With $A=\left[\begin{array}{cc}0& -1\\ 0& 0\end{array}\right]$, $C=[10]$, we get $CA=[0-1]$, so $\mathcal{O}=\left[\begin{array}{cc}1& 0\\ 0& -1\end{array}\right]$ has rank 2 — the bias drifts the angle, so watching the angle backs it out. See Gyro bias estimation / star-tracker calibration.

The rank test is independent of $T$; if $x(0)\in \ker (\mathcal{O})$ it produces $y\equiv 0$ for all time. A longer window cannot conjure information that was never coupled to the output.

Each derivative $y^{(k)}(0)=C{A}^{k}x(0)$ is one more independent linear equation constraining the unknown $x(0)$; stacking them is how we accumulate enough equations to pin it down.

An $x(0)$ in the kernel makes every block $C{A}^{k}x(0)=0$, so all derivatives of $y$ vanish and $y(t)\equiv 0$ — it is exactly the set of "silent" initial states invisible to any output.

The measurement update $P\leftarrow (I-KC)P$ does nothing there (that direction carries no output), while the time update keeps adding process noise $Q$ and the unstable dynamics inflate it — so it never converges.

Detectability only demands the unobservable modes be stable, so they shrink by themselves; the filter can afford to be blind to something that fades on its own, so $P$ still settles to a bounded steady state.

Because information reaches $y$ through the interaction $C{A}^k$ — how the dynamics $A$ shuffle hidden states into the measured ones — not through how many raw channels $C$ has.

Observability (plus stabilizability of the noise input) is precisely the condition guaranteeing the Riccati equation has a unique stabilizing steady-state solution, so $P$ neither diverges nor stalls at a useless value.

Yes — if its coupling into the dynamics is nonzero (e.g. velocity driving position, $A_{12}\ne 0$), it leaks into the output through $CA$ and can be reconstructed even though no sensor sees it directly.

Unobservable — $\mathcal{O}=[c]=[0]$ has rank 0. With no measurement of the only state, the sensor is disconnected and nothing can be recovered.

No — the second block sits in a sealed box with $A_{12}=0$ and no $C$ coupling; its states form the unobservable subspace and never appear in $y$.

Yes — then $\mathcal{O}$ already contains the identity in its first block, giving rank $n$ regardless of $A$; direct measurement of all states is the trivially observable case.

Then $C{A}^k=0$ for $k\ge 1$, so $\mathcal{O}$ collapses to just $C$. Only the states $C$ touches are observable; with no dynamics, nothing leaks, so unmeasured states are permanently blind.

Not detectable — detectability requires unobservable modes to be asymptotically stable (strictly decaying); a marginal (undamped, non-decaying) unobservable mode fails that, so the covariance cannot be guaranteed bounded.

For an LTI system no — $(A,C)$ are fixed constants, so observability is global. Point-dependence only arises when linearizing a nonlinear system, where $A,C$ change with the trim state.