Controllability matrix — rank test: Visual walkthrough

Everything below is built from zero. If a symbol appears, it was drawn first.

Step 1 — What "state" and "input" even look like

WHAT. Before any matrices, picture the state as a single dot living in a room. If the system has $n = 2$ numbers describing it (say position $x_{1}$ and velocity $x_{2}$ ), the dot lives on a flat 2D floor. Its coordinates ARE the state.

WHY. Controllability is a geographic question — "can this dot reach every point on the floor?" — so we must first agree that the state is a point, and a change of state is an arrow (a direction the dot could move).

PICTURE. The floor is the whole plane $R^{2}$ . The dot sits at the origin (system "at rest"). Three example target points are marked — the question is whether we can march the dot to each.

Step 2 — A push only moves you along the columns of $B$

WHAT. Freeze the drift for a second (pretend $A = 0$ ). Then $\overset{x}{˙} = B u$ . Whatever numbers you dial into $u$ , the resulting motion arrow is $B u$ — a blend of the columns of $B$ .

WHY. We need to know the very first set of directions available. With a single input ( $m = 1$ ), $B$ is one column, so your only instantaneous move is "forward or backward along that one arrow." That is the seed of everything reachable.

PICTURE. From the origin, the arrow $B$ points one way; dialing $u$ positive or negative slides the dot up and down that single line. Everything off that line is, for this instant, unreachable.

Step 3 — The drift bends the push into a NEW direction: enter $A B$

WHAT. Now switch the drift back on. Once the dot has moved a hair along $B$ , the room's field $A$ grabs that displacement and pushes it somewhere else. The rate at which the push-direction gets bent is $A B$ .

WHY. This is the whole secret of controllability. A single steering arrow $B$ seems to give a single line — but if $A$ rotates that arrow into a fresh direction $A B$ , then by pushing, waiting, and pushing again, you can reach a second dimension. We need $A B$ because it is the first new direction the drift manufactures for free.

PICTURE. Start along $B$ (blue). Apply the field: each point on the blue arrow gets nudged by $A$ , tipping the arrow over toward $A B$ (pink). Blue + pink now open up a whole patch of the plane, not just a line.

Step 4 — Keep rippling: $A^{2} B, A^{3} B, \dots$

WHAT. Repeat. Feed $A B$ through the field once more to get $A^{2} B$ ; again for $A^{3} B$ . Each is the next ripple of your single original push.

WHY. For $n > 2$ states, two directions aren't enough to fill the room. We keep asking the field for one more fresh direction. The list of all directions a push can eventually reach is the growing family $B, A B, A^{2} B, A^{3} B, \dots$

PICTURE. A chain of arrows fanning outward: blue $B$ , pink $A B$ , yellow $A^{2} B$ , each tipped further by the field. In a 3D room these three can point along three independent axes — filling all of space.

Step 5 — The list STOPS at $A^{n - 1} B$ (Cayley–Hamilton)

WHAT. The ripples do not keep giving new directions forever. Once you've computed up to $A^{n - 1} B$ , the very next one $A^{n} B$ is just a recycled blend of the earlier arrows — no new direction.

WHY. The Cayley–Hamilton theorem says every $n \times n$ matrix $A$ obeys its own characteristic equation, which lets you rewrite $A^{n}$ as a mix of $I, A, A^{2}, \dots, A^{n - 1}$ . So $A^{n} B$ folds back into arrows you already have. This is why the parent's formula stops at $A^{n - 1} B$ — going further is wasted chalk.

PICTURE. The fan of arrows from Step 4, but the would-be $A^{n} B$ (dashed grey) lands inside the region already spanned — it adds no new reach.

Step 6 — Stack them: the controllability matrix

WHAT. Lay all the surviving direction-arrows side by side as columns of one big matrix. That matrix is $C$ .

WHY. "Do these arrows fill the room?" is exactly the question "how many independent columns does this matrix have?" — and that number is the rank. So the geometry of Step 1–5 becomes one clean algebra check.

PICTURE. The individual arrows slide together into labelled columns of a grid; a caption reads "rank = how many point in genuinely different directions."

Step 7 — The degenerate case: rank $< n$ (a locked room)

WHAT. Sometimes all the ripples stay squashed into a smaller flat — a line inside a plane, a plane inside 3D. Then $rank (C) < n$ and part of the room is sealed off.

WHY. We must see the failure, not just the success. If $A$ never bends $B$ off its own line (e.g. $B$ is already an eigen-direction of a diagonal $A$ ), every ripple lies on that same line. The dot is trapped on it forever.

PICTURE. Two examples on one board.

Left (uncontrollable): $A = [1002], B = [10]$ . Both $B$ and $A B = [10]$ lie on the horizontal axis — the vertical mode $x_{2}$ is a locked door. Rank $= 1 < 2$ .
Right (controllable): same $A$ but $B = [11]$ . Now $A B = [12]$ tips off-axis; blue + pink span the whole plane. Rank $= 2$ .

The one-picture summary

Every idea on one board: your single push $B$ ; the field $A$ tipping it into ripples $A B, A^{2} B, \dots$ ; Cayley–Hamilton snipping the list at $A^{n - 1} B$ ; the arrows stacked into $C$ ; and the verdict — full rank fills the room (controllable), squashed rank leaves a locked door (uncontrollable).

Recall Feynman: the whole walkthrough in plain words

You've got one dot in a room and one lever. Push the lever and the dot slides along one arrow — that arrow is $B$ . But the room has a current: the moment the dot moves, the current sweeps it into a new direction, and that new direction is $A B$ . Wait a bit more and the current gives you yet another, $A^{2} B$ , and so on. So even with one lever you slowly collect a bouquet of directions. The magic (Cayley–Hamilton) is that after exactly $n$ of them the current stops inventing anything new — everything after is just a rerun. So you line up those first $n$ direction-arrows into a grid called $C$ and ask one question: do they point in enough independent ways to reach every corner? Count the independent ones — that count is the rank. If it equals the number of coordinates $n$ , the whole room is yours: controllable. If it's smaller, some direction is a locked door the current never opens no matter how you jiggle the lever: uncontrollable. That single count is the entire test.

Connections

Controllability matrix — rank test — the parent result these pictures derive.
Cayley–Hamilton theorem — the reason the ripple list stops at $A^{n - 1} B$ (Step 5).
Matrix exponential $e^{At}$ — the propagator whose expansion produces the very same powers of $A$ .
State-space representation — where $\overset{x}{˙} = A x + B u$ comes from.
Pole placement & state feedback — needs full rank to move every mode.
Kalman decomposition — formalises the "locked room" of Step 7.
Stabilizability — the softer cousin: only the unstable locked doors must be openable.
Observability matrix — rank test — the mirror question, seeing instead of steering.

Visual walkthrough — Controllability matrix — rank test