# Can a state matrix have a row of zeros?

While formulating a state matrix of a system, say a system of a typical cruise controller, $$\begin{bmatrix} \dot{v} \end{bmatrix} = \begin{bmatrix} -\frac{b}{m} \end{bmatrix} \begin{bmatrix} v \end{bmatrix} + \begin{bmatrix} -\frac{1}{m} \end{bmatrix} \begin{bmatrix} u \end{bmatrix}$$ $$y = \begin{bmatrix} 1 \end{bmatrix} \begin{bmatrix} v \end{bmatrix}$$ If we consider $b = 0$ (negligible), then does this state space matrix make sense any more and is it viable? A state matrix without state variable influence in equation?

Yes, a state matrix with zero rows and/or columns makes sense and is viable. It typically signify pure integrators in the system. In the example you give, $$\dot{v} = -\frac{b}{m} v +\frac{1}{m} u$$ where $v$ is the speed, $u$ is the externally applied force, and $bv$ is some viscous damping force. Now if the viscous damping coefficient is zero (no retarding force acting on the object of mass $m$, the differential equation becomes $$\dot{v} = \frac{1}{m} u$$ The $\dot{v}$ is integrated kinematically to obtain $v$, and your state-space model becomes \begin{eqnarray*} [\dot{v}] &=& [0][v] + [1/m][u] \\ y &=& [1][v] \end{eqnarray*}
Another way to look at it is to obtain the transfer function from $u$ to $y$ (the transfer function of the plant which you want to control), which becomes $$\frac{Y}{U} = \frac{1}{m}\cdot\frac{1}{s+b/m}$$ which is a constant term ($1/m$) times a first-order lag. For $b=0$, the transfer function becomes $$\frac{Y}{U} = \frac{1}{m}\cdot\frac{1}{s}$$ which is simply a constant term ($1/m$) times an integrator ($1/s$).
If the externally applied force is zero (i.e., $u=0$) the output of the integrator ($v$) remains constant; if $u>0$, the resulting acceleration ($\dot{v}=u/m$) is integrated "up" and the speed increases; if $u<0$, $\dot{v}$ is integrated "down" and the speed decreases. Speed, of course, is the state variable in your system. Thus, it is possible to transfer the system state from any initial state, $v(t_0)$, to any other state, $v(t_f)$, in a finite time by means of the input, irrespective of the value of $b$. So, by purely physical reasoning we can say thet the system is controllable.
Since this is a linear time-invariant system, the controllability matrix $$\mathbf{Q} = \left[ \begin{array}{cccc} \mathbf{B} & \mathbf{A}\mathbf{B} & \cdots & \mathbf{A}^{n-1}\mathbf{B} \end{array} \right]$$ where $\mathbf{A}=[-b/m]$ is the state matrix, $\mathbf{B}=[1/m]$ is the input matrix, and $n=1$ is the number of state variables. For your system, $$\mathbf{Q} = \mathbf{B} = \left[ \begin{array}{c} 1/m \end{array} \right]$$ i.e., a non-zero scalar which can be inverted (it is non-singular). Thus, the rank of $\mathbf{Q}$ is one, and the system is controllable.
• I am pretty clear now. Can you comment on the controllability of the system if $b = 0$? I found the rank of controllability matrix is 1. Jun 7 '17 at 16:37