# 0-controllability of three simple systems.

Consider the following three discrete-time state-space realizations $$(A_1,B_1,C_1), (A_2,B_2,C_2) \ \text{and} \ (A_3,B_3,C_3)$$ with

$$A_1=\begin{bmatrix}0&1\\0&1 \end{bmatrix}, \ \ \ \quad B_1=\begin{bmatrix}0\\2 \end{bmatrix}, \quad C_1=\begin{bmatrix}0&1 \end{bmatrix}$$

$$A_2=\begin{bmatrix}0&1\\-1&1 \end{bmatrix}, \quad B_2=\begin{bmatrix}0\\2 \end{bmatrix}, \quad C_2=\begin{bmatrix}0&1 \end{bmatrix}$$

$$A_3=\begin{bmatrix}0&1\\0&1 \end{bmatrix}, \ \ \ \quad B_3=\begin{bmatrix}1\\1 \end{bmatrix}, \quad C_3=\begin{bmatrix}1&1 \end{bmatrix}$$

The question is: Which discrete-time state space realizations are $$0$$-controllable? The answers state that all the systems are $$0$$-controllable.

The ranks of the controllability matrices are: $$\text{rank (}\mathcal{C}_1) = 2$$, $$\text{rank (}\mathcal{C}_2) = 2$$ and $$\text{rank (}\mathcal{C}_3) = 1$$.

I know that for discrete-time systems controllability and $$0$$-controllability are equivalent if the $$A$$-matrix is invertible. $$A_2$$ is the only invertible matrix so $$A_2$$ is $$0$$-controllable.

But I don't know how to test the other matrices. I've read several definitions but these don't give the correct answer.

In order to check whether a discrete LTI system is 0-controllable you either want a mode to be controllable or have an eigenvalue of zero, such any uncontrollable modes would die out in finite time. The easiest tool to check this in my opinion would be the Hautus lemma, namely for all $$\lambda\neq0$$ the following matrix should have full rank $$[A-\lambda\,I \quad B]$$.