3
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Let's say that I have a system matrix A and to find out the eigenvalues $\lambda$ ,I do this:

$$ \hbox{det}(\lambda I - A) = 0 $$

Then to find out if the system are controllable, I uses the Hautus Lemma test . This thest is mutch better that the regular $\hbox{rank}(\hbox{ctrb}(A, B)) = n\ $ test. Anyway! Here it is:

$$ \hbox{rank}([\lambda_i I - A, B]) = n$$

Let's say that I got 3 eigenvalues of A. They are $\lambda_1 = -2$ , $\lambda_2 = -10$ and $\lambda_3 = -0.5$. Now I test if the system is controllable: \begin{align} \hbox{rank}([\lambda_1 I - A, B]) &= 3\, ,\\ \hbox{rank}([\lambda_2 I - A, B]) &= 1\, ,\\ \hbox{rank}([\lambda_3 I - A, B]) &= 3 \end{align}

So something went wrong here! I got 3 eigenvalues, which mean that my state vector is the length 3. That means that my rank of the system should be number 3. But this: $$ \hbox{rank}([\lambda_2 I - A, B]) = 1$$ gives number 1 buy using $\lambda_2 = 10$.

Question: Does this mean that something is wrong with my state vector at row number 2 beacuse the eigenvalue $\lambda_2 = 10$ must reprecent the state vector $x_2$ ?

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  • $\begingroup$ It sure looks that way, but there’s no way for us to tell since you’ve provided nothing but the eigenvalues you computed. $\endgroup$ – amd Jun 10 '17 at 23:27
  • $\begingroup$ Don't you know state space models? $\endgroup$ – Daniel Mårtensson Jun 11 '17 at 9:48
  • $\begingroup$ When you ask what you did wrong, no one but a mind reader will be able to tell you unless you show your work. $\endgroup$ – amd Jun 12 '17 at 23:34
  • $\begingroup$ @DanielMårtensson you need to provide the $A$ matrix. Before anything usefull can be said. $\endgroup$ – WG- Sep 18 '17 at 14:33
  • $\begingroup$ @DanielMårtensson This result only shows that your system is not controllable. According to Hautus, you should have gotten $rank=3$ for ALL eigenvalues iff your system is controllable. Doing Kalman's test should hold the same result ($rank \neq 3$) $\endgroup$ – bertozzijr Jul 7 '18 at 16:14

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