Controllability matrix — rank test: Question bank

Definition Which system are we talking about — continuous or discrete?

Unless a question explicitly says "discrete", everything on this page is about the continuous-time system $\overset{x}{˙} = A x + B u$ (state changes smoothly in time, $\overset{x}{˙}$ = rate of change). The same matrix $C$ and the same rank- $n$ test decide controllability in both worlds — but the two domains differ in how fast you can reach a target:

Continuous-time $\overset{x}{˙} = A x + B u$ : if rank $= n$ , you can reach the target for any finite time $T > 0$ , however small (at the cost of large control effort).
Discrete-time $x_{k + 1} = A x_{k} + B u_{k}$ (state jumps step by step): if rank $= n$ , you reach any target in at most $n$ steps — and you generally cannot do it in fewer, because after $k$ steps you can only span $B, A B, \dots, A^{k - 1} B$ . The Edge cases section returns to this discrete-time difference explicitly.

Picture the ideas first

True or false — justify

Every reachable direction is some combination of the columns of

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

True — the matrix exponential's power series collapses onto those blocks by the Cayley–Hamilton theorem, so nothing outside their span is ever reachable.

A controllable system must be stable.

False — controllability says you can move every mode; an unstable mode can still be controllable, which is exactly the case where feedback can rescue it.

For continuous-time

\overset{x}{˙} = A x + B u

, if

rank (C) = n

then you can reach the target in any time

T > 0

, however small.

True — continuous-time controllability gives reachability for every positive

T

(the price of a tiny

T

is huge control effort, but it's still possible).

Adding more inputs (widening

B

) can never reduce controllability.

True — extra columns can only enlarge or preserve the column span of

C

, so rank cannot drop; more actuators never lock a system.

A single-input system (

m = 1

) can never be controllable when

n > 1

False — Example 1's double integrator has

m = 1

n = 2

and is controllable because

A

mixes the one push into both directions.

B = 0

the system is uncontrollable for every

A

True — with no actuator every block

B, A B, \dots

is the zero column, so

rank (C) = 0 < n

; you have nothing to push with.

Controllability of

(A, B)

is unchanged if we replace

A

A^{T}

False in general —

A^{T}

describes a different dynamics; swapping to the transpose is the recipe for observability (the dual), not controllability.

Two systems with the same

C

up to invertible column operations have the same controllability verdict.

True — controllability is a rank statement, and rank is invariant under multiplying by invertible matrices, so the yes/no answer is preserved.

An uncontrollable system can still be stabilizable.

True — stabilizability only demands that the unstable modes be controllable; harmless (stable) uncontrollable modes are allowed, which is the whole point of Stabilizability.

The rank-

n

test decides controllability for the discrete-time system

x_{k + 1} = A x_{k} + B u_{k}

too.

True — the same

C

and the same rank condition apply; only the timing differs, since discrete-time reaches the target in at most

n

steps rather than any

T > 0

Spot the error

A

3 \times 3

, so I'll build

C = [B A B A^{2} B A^{3} B]

to be safe."

The

A^{3} B

block is wasted — Cayley–Hamilton makes

A^{3}

a combination of

I, A, A^{2}

, so it adds no new direction; stop at

A^{n - 1} B = A^{2} B

det C \neq = 0

, therefore controllable — I checked one square block."

For

m > 1

C

n \times (nm)

and not square, so "the determinant" is undefined; you must check

rank (C) = n

, e.g. via row reduction, not a single determinant.

B

has no zero entries, so the system is controllable."

A full

B

guarantees nothing — controllability is about how

A

rotates

B

over powers; you must still compute the rank of

C

(a full

B

can still yield rank

< n

if the mixing is degenerate).

"Row 2 of

C

is all zeros, so state

x_{2}

is uncontrollable."

A zero row of $C$ in the raw coordinates does signal a lost direction, but you can't blindly blame the labelled state

x_{2}

— after a change of coordinates the untouched direction may be a mixture of states; the rank tells you a mode is lost, not which physical variable.

"The system is unstable, so it can't be controllable."

Stability and controllability are independent; instability describes free evolution, controllability describes what inputs can do — an unstable system is often the one you most want to be controllable so you can place its poles.

"I found the rank is

n

, so all poles will settle by themselves."

Rank

n

means you can design feedback to move the poles anywhere; it says nothing about the open-loop poles, which may still sit in the unstable region until you actually close the loop.

"Discrete-time: rank is

n

, so I can reach the target in a single step."

A single step only spans the columns of

B

; you generally need up to

n

steps to build up

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

, so claiming one step is the error — one step suffices only if

B

alone already has rank

n

Why questions

Why does the reachable set collapse onto only the powers

B

through

A^{n - 1} B

, not all infinitely many powers of

A

Because Cayley–Hamilton writes

A^{n}

as a combination of

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

; every higher power recycles those same directions (see the second figure), so the span stops growing at

A^{n - 1} B

Why is the target rank exactly

n

and never more?

The state lives in

R^{n}

, so its reachable span can be at most

n

-dimensional; rank

= n

means the columns already fill the whole state space and there is nothing left to reach.

Why does a diagonal

A

with a zero in the matching row of

B

give an uncontrollable mode?

A diagonal

A

never mixes modes, so a mode whose

B

-entry is zero is pushed by no input and is never fed by neighbours — it drifts on its own forever, untouchable.

Why can the same

A

be uncontrollable with one

B

and controllable with another?

Controllability is a property of the pair

(A, B)

; changing where the actuator pushes (Example 3 vs Example 2) can make the push spread into a mode that

A

's mixing then propagates across the whole space.

Why do we care about controllability before doing pole placement?

Pole placement can only relocate poles that live in the controllable subspace; uncontrollable modes keep their original poles no matter the feedback gain.

Why is observability called the "dual" of controllability rather than a separate idea?

Because the observability test is the controllability test on the transposed system (

A \to A^{T}

B \to C^{T}

); every theorem about one mirrors into the other by transposition.

Why does discrete-time controllability come with a "number of steps" while continuous-time comes with "any

T > 0

A discrete system moves in finite jumps, so it can only span one more power of

A

per step and needs up to

n

steps; a continuous system flows through infinitely many intermediate instants, so it fills the reachable subspace in any positive time.

Edge cases

n = 1

A = [a]

B = [0]

: controllable?

No —

C = [0]

has rank

0 < 1

; with no actuator the single state cannot be moved regardless of

a

n = 1

A = [a]

B = [b]

with

b \neq = 0

: controllable?

Yes —

C = [b]

has rank

1 = n

; a scalar system is controllable exactly when its lone input entry is nonzero.

A = 0

(the zero matrix),

B = I_{n}

: controllable?

Yes —

C = [I 0 \dots 0]

already has rank

n

from the

B = I

block; every state has its own dedicated actuator, so no mixing is even needed.

A = 0

B

a single column: controllable when

n > 1

No —

A B, A^{2} B, \dots

are all zero, so

C = [B 0 \dots]

has rank

\leq 1 < n

; with no dynamics to spread one push, a single input can only reach one direction.

Two identical decoupled modes (same eigenvalue) each fed by the same scalar input through equal weights: controllable?

No — the input pushes both copies identically, so their difference direction is never touched; repeated eigenvalues with a single input generically leave a mode unreachable.

Discrete-time controllable system,

n = 3

: what is the fewest steps guaranteed to reach any target from rest?

At most

3

steps — after

k

steps you span

B, A B, \dots, A^{k - 1} B

, so

n = 3

steps guarantee the full span; fewer steps suffice only if the earlier blocks already reach rank

3

A stable but uncontrollable system — is it useless?

Not necessarily — if the untouchable modes are stable they settle on their own, so the system is stabilizable and perfectly usable; you just cannot arbitrarily reshape those modes.

A purely marginal mode (eigenvalue exactly

0

) that is uncontrollable — problem?

Yes for stabilizability — a mode at zero neither decays nor can be fixed, so an uncontrollable marginal or unstable mode is exactly what breaks stabilizability.

Recall One-line self-test before you close the page

Say aloud: "Controllable means == $rank (C) = n$ ==, it is about moving states not settling them, it depends on the pair $(A, B)$ , higher powers past $A^{n - 1} B$ add nothing by Cayley–Hamilton, and the same test works for both continuous- and discrete-time (discrete reaches in $\leq n$ steps)." If any clause felt shaky, revisit that group above.

Connections

Controllability matrix — rank test
Observability matrix — rank test
State-space representation
Cayley–Hamilton theorem
Pole placement & state feedback
Kalman decomposition
Matrix exponential $e^{At}$
Stabilizability