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For $\alpha$, $\beta$, $\gamma$ $\in\mathbb{R}$, consider the system $$\frac{d}{dt}x=\underbrace{\begin{bmatrix}0 & \alpha & \gamma & 0 \\ -\alpha & 0 & 0 & 0 \\ 0 & 0 & 0 & \beta \\ 0 & 0 & -\beta & 0 \end{bmatrix}}_{A}x$$

I want to determine for which values of $\alpha,\beta,\gamma\in\mathbb{R}$ the system represents a stable, asymptotically stable, or unstable system.

Then $$\begin{aligned}\det(I\xi-A)&=\begin{vmatrix}\xi & -\alpha & -\gamma & 0 \\ \alpha & \xi & 0 & 0 \\ 0 & 0 & \xi & -\beta \\ 0 & 0 & \beta & \xi\end{vmatrix}\\&=\xi\begin{vmatrix}\xi & 0 & 0 \\ 0 & \xi & -\beta \\ 0 & \beta & \xi \end{vmatrix}-\alpha\begin{vmatrix}-\alpha & -\gamma & 0 \\ 0 & \xi & -\beta \\ 0 & \beta & \xi \end{vmatrix} \\&=\xi^{2}\begin{vmatrix}\xi & -\beta \\ \beta & \xi \end{vmatrix}+\alpha^{2}\begin{vmatrix}\xi & -\beta \\ \beta & \xi\end{vmatrix} \\ &=\xi^{2}(\xi^{2}+\beta^{2})+\alpha^{2}(\xi^{2}+\beta^{2}) \\ &=\xi^{4}+\beta^{2}\xi^{2}+\alpha^{2}\xi^{2}+\alpha^{2}\beta^{2} \\&=(\xi^{2}+\alpha)(\xi^{2}+\beta)\end{aligned}$$

Hence the eigenvalues are given by $$\lambda_{1,2}=\pm\sqrt{-\alpha}=\pm i\sqrt{\alpha}$$

$$\lambda_{3,4}=\pm\sqrt{-\beta}=\pm i\sqrt{\beta}$$

Thus $\Re(\lambda_{i})=0$ for $i\in[1,4]$ and are nonsemisimple, thus the system is unstable for all $\alpha,\beta,\gamma\in\mathbb{R}$. Is that correct?

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  • $\begingroup$ You have a mistake in the computation. The eigenvalues should be $\mp i \alpha, \mp i \beta$. $\endgroup$ – obareey Oct 30 '15 at 18:52
  • $\begingroup$ @obareey In which line of computation is the error? $\endgroup$ – Jason Born Oct 30 '15 at 19:52
  • $\begingroup$ The polynomial should be $(\xi^2 + \alpha^2)(\xi^2 + \beta^2)$ $\endgroup$ – obareey Oct 30 '15 at 20:06
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Let's say $\alpha \neq \beta$. Then there is no repetitive eigenvalues exist, so the system is marginally stable, but not asymptotically stable.

Now assume $\alpha = \beta$ and $\gamma = 0$. In this case even if the eigenvalues have algebraic multiplicity 2, they also have geometric multiplicity 2, so the system is again marginally stable.

İf $\alpha = \beta$ and $\gamma \neq 0$, then the eigenvalues have geometric multiplicity 1 while having algebraic multiplicity 2, hence the system is unstable.

Note: Characteristic polynomial isn't enough to conclude the stability in case of repeated eigenvalues. For example let's look at the following systems: $$\dot{x} = \begin{bmatrix}0 & 0 \\ 0 & 0\end{bmatrix} x$$ $$\dot{x} = \begin{bmatrix}0 & 1 \\ 0 & 0\end{bmatrix} x$$ Both matrices have the same characteristic equation, but the first system is stable and the second system is not. Because in the first system there are two linearly independent eigenvectors for the eigenvalue $0$, i.e. geometric multiplicity 2, but in the second system there is only one linearly independent eigenvector, i.e. geometric multiplicity 1.

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  • $\begingroup$ Can I ask how you derived those answers, given that the characteristic polynomial is expressed only in terms of $\alpha$ and $\beta$? $\endgroup$ – Jason Born Oct 30 '15 at 23:07
  • $\begingroup$ @user3482534 See my updated answer. $\endgroup$ – obareey Nov 1 '15 at 9:12

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