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Given is a single input single output, time invariant state space system. \begin{equation} x(t) = \left(\begin{array}{r} 5 \\ -1 \\ 4\end{array}\right)e^{-2t} \end{equation} \begin{equation} x(t) = \left(\begin{array}{c} 3 \\ 3 \\ 1 \end{array}\right)e^{-2t} \end{equation} \begin{equation} x(t) = \left(\begin{array}{r} 6 \\ 2 \\ -7\end{array}\right) e^{-t} \end{equation}

The above state trajectories arise from different initial states at time $0$. In each case the input to the system $u(t)=0$.

How do you determine whether the system is controllable or not?

There is not enough information to try to calculate matrix $A$ and $B$ is not given, so the controllability matrix can't be constructed. I am thinking the answer has to do with the fact that the state trajectories do not have the same form (i.e $e^{-2t}$ and $e^{-t}$).

Any ideas?

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  • $\begingroup$ This question seems ill-posed without a $B$ matrix. For example, regardless of what $A$ is, a system with $B = 0$ is trivially uncontrollable. $\endgroup$ – yoknapatawpha Nov 6 '14 at 1:59
  • $\begingroup$ I agree @yoknapatawpha but this is the entirety of the question and I have arrived at the same conclusion as you. $\endgroup$ – ebejko Nov 6 '14 at 5:18
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    $\begingroup$ So it turns out that it is possible to prove that the system is not controllable for any value of B. Any ideas? $\endgroup$ – ebejko Nov 7 '14 at 7:20
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You can construct the $A$ matrix because eigenvectors and eigenvalues are given. Like

$$T = \begin{bmatrix} 5 & 3 & 6 \\ -1 & 3 & 2 \\ 4 & 1 & -7 \end{bmatrix}$$ $$\Lambda = \begin{bmatrix} -2 & 0 & 0 \\ 0 & -2 & 0 \\ 0 & 0 & -1 \end{bmatrix}$$

Then, $A = T \Lambda T^{-1}$. Now you can show that the system is not controllable for any $b^T = \begin{bmatrix} b_1 & b_2 & b_3 \end{bmatrix}$.

Note that any $n$th order system is always controllable if $B$ is a rank $n$ matrix so I believe this system has 1 input.

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