# Zero State, Stable Equilibrium, Dynamic System

The question reads: For which real numbers $k$ is the zero state a stable equilibrium of the dynamic system $x_{t+1} = Ax_t$?

$A = \begin{bmatrix} 0.1 &k \\ 0.3 & 0.3 \end{bmatrix}$

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 $$x = \frac{2 \pm \sqrt{1+30k}}{10}$$ 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 $\sqrt{-1-30k} < 8$?

Thanks!

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Just so you know, I reformatted your question using LaTeX for easier viewing. –  Mike Spivey Jan 6 '11 at 0:09
For the real case, you also need $k\geq \cfrac{-1}{30}$ to keep them real. –  Ross Millikan Jan 6 '11 at 0:24
Thanks Dr. Spivey! –  user5430 Jan 6 '11 at 1:39
You're quite welcome. Glad it was helpful. –  Mike Spivey Jan 6 '11 at 3:43

In general, the zero state of a dynamical system of the form $x_{t+1} = Ax_t$ 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.)
Good answer. Just to be clear, though, one has asymptotic stability if $a^2 + b^2 <1$. The system is generally considered stable if $a^2 + b^2 = 1$. –  cch Jan 6 '11 at 0:29