Question

In Exercises 9–14, classify the origin as an attractor, repeller, or saddle point of the dynamical system \({{\bf{x}}_{k + 1}} = A{{\bf{x}}_k}\). Find the directions of greatest attraction and/or repulsion.

\(A = \left( {\begin{aligned}{}{.4}&{}&{.5}\\{ - .4}&{}&{1.3}\end{aligned}} \right)\)

Short answer

The origin is an attractor as two eigenvalues obtained, that is, \(0.8{\rm{ and }}0.9\) are less than 1. The direction of the greatest attraction passes through the origin, which is \({v_1} = \left( {\begin{aligned}{}5\\4\end{aligned}} \right)\).

Step-by-step solution

Discrete Dynamical System 

For any Dynamical System with matrix \(A\), the eigenvalues of \(A\) will determine the system of:

\(\begin{aligned}{}{\rm{Greatest attraction }} \to {\rm{if }}\lambda < 1\\{\rm{Greatest repulsion }} \to {\rm{if }}\lambda > 1\end{aligned}\)

Evaluating Eigenvalues of the given matrix 

The given matrix for the dynamic system \({{\bf{x}}_{k + 1}} = A{{\bf{x}}_k}\) is:

\(A = \left( {\begin{aligned}{}{.4}&{}{.5}\\{ - .4}&{}&{1.3}\end{aligned}} \right)\)

For eigenvalues, we have:

\(\begin{aligned}{}\det \left( {A - \lambda I} \right) = 0\\\left| {\begin{aligned}{}{0.4 - \lambda }&{}&{0.5}\\{ - 0.4}&{}&{1.3 - \lambda }\end{aligned}} \right| = 0\\\left( {0.4 - \lambda } \right)\left( {1.3 - \lambda } \right) + 0.2 = 0\\{\lambda ^2} - 1.7\lambda + 0.72 = 0\\\lambda = 0.8,0.9\end{aligned}\)

Here, both eigenvalues are less than 1.

Hence, the origin is an attractor for the given system

Eigenvector for eigenvalue 0.8

Now, for eigenvectors, we have:

At \(\lambda = 0.8\).

\(\begin{aligned}{}\left( {A - \left( {0.8} \right)I} \right){v_1} = 0\\\left( {\begin{aligned}{}{0.4 - 0.8}&{}&{0.5}\\{ - 0.4}&{}&{1.3 - 0.8}\end{aligned}} \right)\left( {\begin{aligned}{}x\\y\end{aligned}} \right) = 0\\\left( {\begin{aligned}{}{ - 0.4}&{}&{0.5}\\{ - 0.4}&{}&{0.5}\end{aligned}} \right)\left( {\begin{aligned}{}x\\y\end{aligned}} \right) = 0\end{aligned}\)

Also, the system of equation will be:

\(\begin{aligned}{} - 0.4x + 0.5y = 0\\ - 0.4x + 0.5y = 0\end{aligned}\)

For this, the general solution, we have:

\(\begin{aligned}{}{v_1} &= \left( {\begin{aligned}{}x\\y\end{aligned}} \right)\\ &= \left( {\begin{aligned}{}{\frac{5}{4}y}\\y\end{aligned}} \right)\\ &= \left( {\begin{aligned}{}5\\4\end{aligned}} \right)\end{aligned}\)

Here, the eigenvector is: \({v_1} = \left( {\begin{aligned}{}5\\4\end{aligned}} \right)\).