How can one determine the stability of a given dynamical system?

Question

How can one determine the stability of a given dynamical system? For example, for what values of $a$ is the following matrix stable?

\begin{bmatrix} 0&a \\ a&0 \end{bmatrix}

I'm just confused about how to deduce a matrix's stability from its eigenvalues. I know this is a dumb question, but help would be greatly appreciated.

Answer 1

If the real part of the eigenvalues are negative, then the system is attracted to some sort sink, and if you give your particle a small kick in its initial conditions, it is headed to the same sink.

If the real part of the eigenvalues are positive, then everything is expanding away from some source, and distances between initial conditions will increase over time.

If the real part of the eigenvalues are 0, then your system will orbit, and a small perturbation will remain a small perturbation, neither growing nor shrinking.

If some of your eigenvalues are positive and some are negative, then two particles with nearby initial conditions, may stay nearby for some time, but will eventually start to diverge.

Answer 2

For a continuous-time dynamical system: to find the stability of the system we first find the equilibrium solution of the system (i.e. for a continuous system $\dot x=f(x)$, by solving $f(x)=0$ ), then finding the Jacobian matrix of the system at the equilibrium solution. Now if the eigenvalues of the Jacobian matrix(at the equilibrium points) are negative or negative real part (if complex) then the equilibrium points are locally stable.

For a discrete-time dynamical system: Here equilibrium solution are solution of $f(x)=x$ and equilibrium points are locally stable if eigenvalues of the Jacobian matrix(at the equilibrium points) are of modulus less than $1$.

Here note that Jacobian matrix is nothing but a tool of local linearization in a dynamical system. Fore more details see this problems in stack exchange: 1 2

Answer 3

Compute the eigenvalues as follows:

$$ \det(A-\lambda I) = 0 \implies \lambda^2- a^2=0 \implies (\lambda-a)(\lambda+a)=0 $$

The eigenvalues are $\lambda_{1,2} = \pm a$. Whether you choose $a$ as a positive or negative, you will end up with a positive eigenvalue, therefore, the system is unstable.