Winigefx

2022-07-01

The question reads: For which real numbers k is the zero state a stable equilibrium of the dynamic system ${x}_{t+1}=A{x}_{t}$?
$A=\left[\begin{array}{cc}0.1& k\\ 0.3& 0.3\end{array}\right]$
So, my thought is I need to find the eigenvalues. In order to do this I calculated the characteristic polynomial as
${x}^{2}-0.4x+0.03-0.3k=0$, with $x$ representing eigenvalues.
Using the quadratic formula I found that the (real) eigenvalues are
$x=\frac{2±\sqrt{1+30k}}{10}$
and for the zero state to be in stable equilibrium $\sqrt{1+30k}<8$. Hence, $k<21/10$ (for stable equilibrium).
My question is how do I figure out the values for $k$ if the eigenvalues are complex?
Do I solve the inequality $\sqrt{-1-30k}<8$?

frethi38

In general, the zero state of a dynamical system of the form ${x}_{t+1}=A{x}_{t}$ is a stable equilibrium when all of the eigenvalues of A are inside the unit circle in the complex plane. So if you have an eigenvalue of the form $a+bi$, you get stability when ${a}^{2}+{b}^{2}<1$. Since this is a homework question, I'll let you work out the rest of the problem from here. (This isn't going to be equivalent to solving $\sqrt{-1-30k}<8$, though.)

George Bray

Good answer. Just to be clear, though, one has asymptotic stability if ${a}^{2}+{b}^{2}<1$. The system is generally considered stable if ${a}^{2}+{b}^{2}=1$.