Let
$$A = \begin{bmatrix} -1 & -N\\ 6 & 0\end{bmatrix}$$
from the state space realization of an LTI system. For this system to be stable, all eigenvalues must be within the unit circle, i.e., for all eigenvalues $|\lambda_i|<1$ must be satisfied. Matrix $A$ has eigenvalues
$$\lambda_{1,2} = \frac{-1 \pm \sqrt{1-24N} }{2}$$
How can I derive $N$ in such a way that both eigenvalues lie within the unit circle? The solution should probably be $$0<N<1/6$$ This seems obvious, but since the eigenvalues can be complex I don't know how to interpret this.
Best Answer
The two roots are $$ \lambda_1(N) = -\frac{-1+ \sqrt{1-24N}}{2}, \;\;\; \lambda_2(N) = -\frac{-1- \sqrt{1-24N}}{2}. $$ It is easily seen that the condition for real roots are $N \le \frac{1}{24}$. We consider real eigenvalue case first. The real eigenvalues are within the unit disc if $$ -1 \le \lambda_i(N) \le 1, \;\; i=1,2 $$ The root $\lambda_1(N)$ decreases monotonically with respect to $N$ and $\lambda_2(N)$ increases monotonically with $N$. We study the two roots when $N$ decreases from $N=\frac{1}{24}$.
For the boundary condition $\lambda_1(N) \le 1$, we get $$ N \ge -\frac{1}{3} $$ and it satisfies the condition for real roots. For the condition $\lambda_2(N) \ge -1$, we get $$ N \ge 0 $$ which also satisfies the condition for real root. However, when $N$ is decreasing, $\lambda_2(N)$ hits the boundary condition $-1$ first (before $\lambda_1(N)$ hits $1$). Thus, we discard the solution for $\lambda_1(N)$.
Thus, the required condition in the real eigenvalue case is $$ 0 \le N \le \frac{1}{24} . $$
For complex eigenvalues, the moduli of the two eigenvalues are the same. $$ |\lambda_1(N)| = |\lambda_2(N)| . $$ The moduli have to be less than or equal to unity for the eigenvalues to be in the unit disc. The condition is $$ \frac{1+ (\sqrt{|1-24N|})^2}{4} \le 1 $$ Equivalently, $$ \frac{1+|1-24N|}{4} \le 1 . $$ Since, $1-24N$ is negative in the range we are interested, we obtain the condition for complex roots to be in the unit disc as, $$ \frac{1+24N-1}{4} \le 1 $$ which gives us $$ N \le \frac{1}{6} . $$ By incorporating the condition for complex roots, we have the complete solution for complex roots as, $$ \frac{1}{24} < N \le \frac{1}{6} . $$ Thus, the complete solution for the eigenvalues (real or complex) to be in the unit disc is $$ 0 \le N \le \frac{1}{6} . $$