Observability matrix — rank test: Question bank

Reminder of the objects in play (built fully in the parent note): the system is $\overset{x}{˙} = A x + B u, y = C x + D u$ with state $x \in R^{n}$ and output $y \in R^{p}$ .

Figure — Observability matrix — rank test

The figure shows what "an unobservable direction" looks like: the sensor row $C$ (cyan) spans one axis; the amber state vector along the untouched axis casts zero shadow onto it — no matter how the dynamics $A$ push it, the meter reads nothing.

True or false — justify

TF: A system with more outputs (larger

p

) is always more observable than one with fewer.

False.

p

only makes

O

taller (

p n

rows), but observability needs full column rank

n

; a badly-aligned

C

with many rows can still miss a state direction, while a single well-chosen row can capture all of them.

TF: If

rank (O) = n

then the initial state

x (0)

is uniquely recoverable from the output.

True. Full column rank means

ker (O) = {0}

, so the map

x (0) \mapsto (y (0), \overset{y}{˙} (0), \dots)

is injective and can be inverted for a unique

x (0)

TF: Observability guarantees the system is stable.

False. Observability is purely a geometric relationship between

C

and

A

; it says nothing about eigenvalue locations. An unstable system can be perfectly observable, and a stable one can hide an entire state.

TF: A stable system is automatically observable because its states settle down predictably.

False. Settling behaviour (eigenvalues) and visibility to the sensor (kernel of

O

) are independent. The build above is structurally fine yet has a whole invisible state.

TF: If

C = 0

(no sensors read anything), the system can never be observable for

n \geq 1

True. Every block

C A^{k} = 0

, so

O = 0

has rank

0 < n

; with no output the free response is identically zero and

x (0)

is completely hidden.

TF: The observability matrix must be square.

False. It is

p n \times n

; it is square only when

p = 1

. For

p > 1

it is tall, and rank is still capped at

n

TF:

(A, C)

observable is equivalent to

(A^{⊤}, C^{⊤})

being controllable.

True — this is the duality statement. Transposing swaps "stack

C

vertically with

C A^{k}

" into "stack

C^{⊤}

horizontally with

A^{⊤ k} C^{⊤}

TF: Adding the row block

C A^{n}

O

can increase its rank.

False. By the Cayley–Hamilton theorem — which states that a matrix satisfies its own characteristic equation, so

A^{n}

equals a combination of

I, \dots, A^{n - 1}

— the block

C A^{n}

is a linear combination of existing rows and contributes zero new rank.

TF: If any single scalar output

y_{i}

is unobservable on its own, the whole system is unobservable.

False. Observability is a property of the full

O

; different output channels can cover different state directions, so a set of individually-weak sensors can jointly be observable.

Spot the error

Error: "I computed

O

up to

C A^{n - 1}

, got rank

n - 1

, then added

C A^{n}

and

C A^{n + 1}

to reach rank

n

— so the system is observable."

The added rows are Cayley–Hamilton-redundant and cannot raise the rank; the true rank was

n - 1

, so the system is genuinely not observable. The reasoning fabricated rank out of dependent rows.

Error: "

O

has

p n = 6

rows for

n = 2

, so its rank is 6 and the system is observable."

Rank counts independent columns, which is at most

n = 2

; the number of rows is irrelevant. Observable requires rank

= n = 2

, not

= 6

Error: "Two identical decoupled modes (repeated eigenvalue

λ

with two eigenvectors) can be separated by a single scalar output if I take enough derivatives."

No number of derivatives helps: every

C A^{k}

acts as

C λ^{k}

on that eigenspace, so all rows are proportional there — the two modes are indistinguishable. You need

p \geq 2

independent output rows.

Error: "The kernel of

O

contains a nonzero

x_{0}

, but the system is still observable because

y (t)

eventually becomes nonzero."

x_{0} \in ker O

then

C e^{A t} x_{0} \equiv 0

for all

t

— the output is identically zero, never "eventually" nonzero. A nonzero kernel is exactly the definition of unobservability.

Error: "Observability depends on

B

and

D

, so I included them when forming

O

The contribution of

u

B

D

y

is known and subtractable; only the free response

C e^{A t} x (0)

carries hidden information, so

O

is built from

A

and

C

alone.

Error: "I built

O = [C A C A^{2} C]

by putting

A

on the left."

That is the controllability pattern (

A^{k} B

A

on the left, stacked horizontally). Observability stacks vertically with

A

on the right:

C, C A, C A^{2}

Why questions

Why do we differentiate

y = C e^{A t} x (0)

t = 0

instead of just using

y (0)

A single instant gives only

y (0) = C x (0)

, the projection of

x (0)

onto the rows of

C

. Derivatives

C A^{k} x (0)

reveal new directions, and stacking them spans all directions the sensor can possibly reach.

Why is

n - 1

the exact stopping power — not

n

2 n

The Cayley–Hamilton theorem (a matrix satisfies its own characteristic polynomial) makes

A^{n}

and everything above a combination of

I, \dots, A^{n - 1}

, so rows

C A^{k}

for

k \geq n

are dependent. Powers

0

n - 1

already span the entire reachable row space.

Why does an unobservable state produce exactly zero output rather than just a small one?

Because

x_{0} \in ker O

means

C A^{k} x_{0} = 0

for every

k

, and

e^{A t}

is a power series in

A

; so

C e^{A t} x_{0}

is a sum of zero terms — identically zero, not merely small.

Why can the PBH test sometimes be preferred over the rank test?

The PBH test (Popov–Belevitch–Hautus) checks one eigenvalue

λ

at a time by asking whether

[A - λ I C]

has full column rank; a rank drop pinpoints which specific mode is unobservable — information the single rank number of

O

hides.

Why do only

A

and

C

matter, not

B

and

D

Because

u (t)

is a signal we choose and know, its entire effect on

y

(through

B

and the feed-through

D

) can be computed and removed; what is left to reconstruct

x (0)

from is the free response, governed solely by

C

and

A

Why does observability matter for a Kalman Filter?

The Kalman Filter reconstructs the state from noisy outputs; if a state lives in

ker O

it emits no signal, so its estimate cannot converge. Observability is a prerequisite for the estimation error to be driven to zero.

Why does Kalman decomposition split the system into observable and unobservable parts?

Because

ker O

is an

A

-invariant subspace: states inside it stay hidden forever, so the Kalman decomposition separates the system into a visible block (reconstructable) and an invisible block that no sensor record can ever expose.

Edge cases

Edge: What is

O

and its rank when

n = 1

(scalar state)?

O = C

, a single

p \times 1

block. Observable iff

C \neq = 0

; any nonzero sensor reading of a one-dimensional state already sees everything.

Edge: If

A = 0

(all states frozen), when is

(0, C)

observable?

O = [C; 0; \dots; 0]

(semicolons stack the zero blocks below

C

), so rank

= rank (C)

. Observable iff

C

alone has rank

n

, because with frozen dynamics derivatives add nothing.

Edge: If

A = λ I

(a single repeated eigenvalue filling all of

R^{n}

), how observable can a scalar output be?

Every

C A^{k} = λ^{k} C

is a scalar multiple of

C

, so

O

has rank

1

. A scalar sensor can resolve at most one direction; you need

p \geq n

to observe such a fully-degenerate mode.

Edge: Can a system be simultaneously observable and completely uncontrollable?

Yes — observability (property of

A, C

) and controllability (property of

A, B

) are independent. You may perfectly reconstruct a state you cannot steer at all.

Edge: If

O

has rank exactly

n - 1

, how big is the hidden subspace?

The unobservable subspace

ker O

has dimension

n - rank (O) = 1

; exactly one independent state direction is invisible, while the remaining

n - 1

can be reconstructed. See Rank and null space of a matrix.

Edge: Does scaling

C

by a nonzero constant change observability?

No. Multiplying

C

c \neq = 0

scales every row of

O

c

, which leaves the column rank unchanged — observability is a property of the row space, not row magnitudes.

Recall One-line self-test

If someone hands you a tall $O$ and says "look how many rows — surely observable," what do you check first? Whether its column rank equals $n$ — rows and their count are a distraction; only independent columns decide observability.

Connections

Observability matrix — rank test (parent)
Controllability matrix — rank test · Cayley–Hamilton theorem · PBH test
Kalman Filter · Kalman decomposition · State-space representation · Rank and null space of a matrix