Find $K$ equal to the eigenvalues of $A – B K$

control theoryeigenvalues-eigenvectorslinear algebralinear-controlmatrices

I've a little trouble with this old exam question. It is a multiple choice exam question!


A discrete state-space is represented as $$x(t-1) = Ax(t) + Bu(t)$$ where $$A = \begin{bmatrix}-3&1&-1\\0&-2&-1\\1&0&-2\end{bmatrix} \qquad B = \begin{bmatrix}0.5\\0.5\\-0.5\end{bmatrix}$$ Find a $K \in \mathbb{R}^{1×3}$ equal to the eigenvalues of $A-BK$.


This problem looks pretty hard because it is a 3th order system. It isn't possible to use Matlab or a calculator. The only way I can think of when solving this problem would be to take for K:

$K = \begin{bmatrix}k1\\k2\\k3\end{bmatrix}$ and work out the $det((A-BK)-I\lambda)=0$ equation where the solution got a form of: $(\lambda -k_1)(\lambda -k_2)(\lambda -k_3)$ and thus the eigenvalues would be equal to $\lambda_1 = k_1,\lambda_2 = k_2,\lambda_3 = k_3$. The equation $(A-BK)-I\lambda$ is:
$(A-BK)-I\lambda =\begin{bmatrix}-3-\lambda&1&-1\\0&-2-\lambda&-1\\1&0&-2-\lambda\end{bmatrix} + \begin{bmatrix}-0.5k_1&-0.5k_2&-0.5k_3\\-0.5k_1&-0.5k_2&-0.5k_3\\0.5k_1&0.5k_2&0.5k_3\end{bmatrix}$

I calculated the controllability matrix, which is $M_c = \begin{bmatrix}0.5&-0.5&-0.5\\0.5&-0.5&-0.5\\-0.5&1.5&-3.5\end{bmatrix}$ which is rank 2 (< rank n = 3) and thus uncontrollable.

To get to a solution for $det((A-BK)-I\lambda)=0$ it will take ages. There is probably a trick which needs to be used or something I missed. Thanks in advance!


This question is solved. The system (A,B) has one uncontrollable eigenvalue, which is at -3. This eigenvalue can't be relocated. There was only one answer of the multiple choice options which had this eigenvalue. A bit silly, because the equation did not have to be worked out completely. The possible answers were needed to get to the final answer.

Best Answer

This question is solved. The system (A,B) has one uncontrollable eigenvalue, which is at -3. This eigenvalue can't be relocated. There was only one answer of the multiple choice options which had this eigenvalue. A bit silly, because the equation did not have to be worked out completely. The possible multiple choice answers were needed to get to the final answer.

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