Consider the state space system $\dot{z} = Az + Bu$ with $A = \begin{bmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 1\\0 & 0 & 0 & 0\end{bmatrix}$ and $B = \begin{bmatrix} 0 & 0\\ 1 & 0\\ 0 & 0 \\ 0 & 1\end{bmatrix}$. We need to design a control law $u=−Kz$ so that all eigenvalues of $A−BK$ have a real part that is less than $−10$, where $K=\begin{bmatrix} k_{11} & k_{12} & k_{13}& k_{14} \\ k_{21} & k_{22} & k_{23}& k_{24}\end{bmatrix}$.
I am trying to look for a mathematical manual approach as opposed to one involving coding. I am not sure that such a method even exists without any assumptions. For me, I got stuck after computing $\det(\lambda I-(A-BK))=0$, which gives $\lambda(\lambda^2+k_{24}\lambda+k_{23})(k_{11}+k_{12}+\lambda)-(k_{14}\lambda+k_{13})(\lambda k_{22}+k_{21})=0$. This seems quite unapproachable to me from here and I think there should be a nicer approach to solving the problem, since the $A$ matrix has a rank of $2$ which I haven't used anywhere. Any leads and hints would be appreciated.
Thanks
Since $$A-BK=\begin{pmatrix}0&1&0&0\\-k_{11}&-k_{12}&-k_{13}&-k_{14}\\0&0&0&1\\-k_{21}&-k_{22}&-k_{23}&-k_{24}\end{pmatrix}$$ it is enough to take $k_{21}=k_{22}=0$; then the eigenvalues of $A-BK$ are those of the two sub-matrices $$\begin{pmatrix}0&1\\-k_{11}&-k_{12}\end{pmatrix},\qquad\qquad\begin{pmatrix}0&1\\-k_{23}&-k_{24}\end{pmatrix}$$ on the diagonal. The requirement is then that the eigenvalues of each sub-matrix $\begin{pmatrix}0&1\\-a&-b\end{pmatrix}$ (where $a$ represents $k_{11}$, $k_{23}$, and $b$ represents $k_{12}$, $k_{24}$) have real part less than $-10$. Looking at the trace and determinant, this is the case when $b>20$ and $a>100$.
Conclusion: A sufficient condition for all eigenvalues of $A-BK$ having real part less than $-10$ is $$k_{21}=k_{22}=0,\qquad k_{11}, k_{23}>100,\qquad k_{12},k_{24}>20$$