Question

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

10. \(A = \left( {\begin{aligned}{}{.3}&{}&{.4}\\{ - .3}&{}&{1.1}\end{aligned}} \right)\).

\(\)

Short answer

The direction of greatest attraction and or repulsion is eigenvector \({v_1}\)and.

Here, \({{\rm{v}}_1} = \left( {\begin{aligned}{}2\\1\end{aligned}} \right)\).

Step-by-step solution

Find eigenvalue

\(A = \left( {\begin{aligned}{}{.3}&{}&{ - .4}\\{ - .3}&{}&{1.1}\end{aligned}} \right)\)

For finding eigenvalue,

\(\begin{aligned}{}det\left( {{\rm{A - }}\lambda {\rm{I }}} \right){\rm{ = }}\left( {\left( {{\rm{.3 - }}\lambda } \right)\left( {1.1 - \lambda } \right)} \right) - \left( {\left( { - .3} \right)\left( { - .4} \right)} \right)\\0 = {\lambda ^2}{\rm{ - 1}}{\rm{.4}}\lambda {\rm{ + }}{\rm{.45 }}\end{aligned}\)

From this characteristic equation,

\(\begin{aligned}{}\lambda &= \frac{{1.4 \pm \sqrt {{{1.4}^2} - 4\left( {.45} \right)} }}{2}\\ &= \frac{{1.4 \pm \sqrt {1.6} }}{2}\\ &= \frac{{1.4 \pm .4}}{2}\\ &= 0.5,0.9\end{aligned}\)

The origin is an attractor because both eigenvalues are less than 1 in magnitude.

Find the eigenvector 

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

\(\begin{aligned}{}\left( {A - 2I} \right)x = 0\\\left( {\begin{aligned}{}{ - .2}&{}&{ - .4}&{}&0\\{ - .3}&{}&{ - .6}&{}&0\end{aligned}} \right) \sim \left( {\begin{aligned}{}1&{}&{ - 2}&{}&0\\0&{}&0&{}&0\end{aligned}} \right)\end{aligned}\)

So, \({x_1} = 2{x_2}\).

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

Hence eigenvector \({{\rm{v}}_1} = \left( {\begin{aligned}{}2\\1\end{aligned}} \right)\).