Question

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

9. \(A = \left( {\begin{aligned}{}{1.7}&{}&{ - .3}\\{ - 1.2}&{}&{.8}\end{aligned}} \right)\)

\(\)

Short answer

The direction of greatest attraction and or repulsion is eigenvector \({{\rm{v}}_1}\) and \({{\rm{v}}_2}\).

Here, \({{\rm{v}}_1} = \left( {\begin{aligned}{}{ - 1}\\1\end{aligned}} \right)\) and \({{\rm{v}}_2} = \left( {\begin{aligned}{}1\\4\end{aligned}} \right)\).

Step-by-step solution

Find eigenvalue

\(A = \left( {\begin{aligned}{}{1.7}&{}&{ - .3}\\{ - 1.2}&{}&{.8}\end{aligned}} \right)\)

For finding eigenvalue,

\(\begin{aligned}{}det\left( {{\rm{A - }}\lambda {\rm{I }}} \right){\rm{ = }}\left( {\left( {{\rm{1}}{\rm{.7 - }}\lambda } \right)\left( {.8 - \lambda } \right)} \right) - \left( {\left( { - .3} \right)\left( { - 1.2} \right)} \right)\\0 = {\lambda ^2}{\rm{ - 2}}.5\lambda {\rm{ + }}1{\rm{ }}\end{aligned}\)

From this characteristic equation

\(\begin{aligned}{}\lambda &= \frac{{2.5 \pm \sqrt {{{2.5}^2} - 4\left( 1 \right)} }}{2}\\ &= \frac{{2.5 \pm \sqrt {2.25} }}{2}\\ &= \frac{{2.5 \pm 1.5}}{2}\\ &= 2,.5\end{aligned}\)

The origin is a saddle point because one eigenvalue is greater than 1 and the other eigenvalue is less than 1 in magnitude.

Find the eigenvector

The direction of the greatest repulsion is through the origin and the eigenvector \({{\rm{v}}_1}\).

\(\begin{aligned}{}\left( {A - 2I} \right)x = 0\\\left( {\begin{aligned}{}{ - .3}&{}&{ - .3}&{}&0\\{ - 1.2}&{}&{ - 1.2}&{}&0\end{aligned}} \right) \sim \left( {\begin{aligned}{}1&{}&1&{}&0\\0&{}&0&{}&0\end{aligned}} \right)\end{aligned}\)

\({x_1} = - {x_2}\)

So, \({x_2}\) is free.

Hence eigenvector \({{\rm{v}}_1} = \left( {\begin{aligned}{}{ - 1}\\1\end{aligned}} \right)\).

The direction of the greatest attraction is through the origin and the eigenvector \({v_2}\).

\(\begin{aligned}{c}\left( {A - 2I} \right)x = 0\\\left( {\begin{aligned}{}{1.2}&{ - .3}&0\\{ - 1.2}&{.3}&0\end{aligned}} \right) \sim \left( {\begin{aligned}{}1&{}&{ - .25}&{}&0\\0&{}&0&{}&0\end{aligned}} \right)\end{aligned}\)

\({x_1} = - .25{x_2}\)ad \({x_2}\)is free.

Hence eigenvector \({{\rm{v}}_2} = \left( {\begin{aligned}{}1\\4\end{aligned}} \right)\)