Linear Algebra and Its Applications, 5th Edition
Authors: David C. Lay, Steven R. Lay, Judi J. McDonald
ISBN-13: 978-0321982384
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See our solution for Question 12E from Chapter 5.6 from Lay's Linear Algebra and Its Applications, 5th Edition.
Problem 12E
Chapter:
Problem:
In Exercises 9–14, classify the origin as an attractor, repeller, or saddle point of the dynamical system Find the directions of greatest attraction and/or repulsion.
Step-by-Step Solution
Given Information
We are given with following dynamical system \[ \mathbf { x } _ { k + 1 } = A \mathbf { x } _ { k } , \text { where } A = \left[ \begin{array} { c c } { 0.5 } & { 0.6 } \\ { - 0.3 } & { 1.4 } \end{array} \right] \]
Step-1: Eigenvalues of A:
Compute the characteristic polynomial: \[ \begin{aligned} \operatorname { det } ( A - \lambda I ) & = \left| \begin{array} { c c } { 0.5 - \lambda } & { 0.6 } \\ { - 0.3 } & { 1.4 - \lambda } \end{array} \right| \\ & = ( 0.5 - \lambda ) ( 1.4 - \lambda ) + 0.18 \\ & = \lambda ^ { 2 } - 1.9 \lambda + 0.7 + 0.18 \\ & = \lambda ^ { 2 } - 1.9 \lambda + 0.88 \end{aligned} \] The eigenvalues are: \[\begin{array}{l} \lambda = \dfrac{{1.9 \pm \sqrt {{{( - 1.9)}^2} - 4(0.88)} }}{2}\\ = \dfrac{{1.9 \pm \sqrt {3.61 - 3.52} }}{2}\\ = \dfrac{{1.9 \pm \sqrt {0.09} }}{2}\\ = \dfrac{{1.9 \pm 0.3}}{2}\\ = 0.8{\rm{ and }}1.1 \end{array}\]
Step-2:
Since, one of the Eigenvalue 1.1 is greater than 1 in magnitude and an Eigenvalue 0.8 is less than 1 in magnitude, therefore origin is a saddle point for the dynamical system.
The direction of greatest attraction, is given by the the eigenvector whose Eigenvalue is less than 1 in magnitude.
The direction of greatest repulsion, is given by the the eigenvector whose Eigenvalue is more than 1 in magnitude.
Step-3:
Find the eigenvectors for the eigenvalues 1.1
\[ \begin{array} { r } { [ A - ( 1.1 ) I ] \mathbf { v } _ { 1 } = \mathbf { 0 } } \\ { \left[ \begin{array} { c c } { 0.5 - 1.1 } & { 0.6 } \\ { - 0.3 } & { 1.4 - 1.1 } \end{array} \right] \left[ \begin{array} { c } { x _ { 1 } } \\ { x _ { 2 } } \end{array} \right] = 0 } \\ { \left[ \begin{array} { c c c } { - 0.6 } & { 0.6 } \\ { - 0.3 } & { 0.3 } \end{array} \right] \left[ \begin{array} { c } { x _ { 1 } } \\ { x _ { 2 } } \end{array} \right] = 0 } \end{array} \] The system of equations \[\begin{array}{l} - 0.6{x_1} + 0.6{x_2} = 0\\ - 0.3{x_1} + 0.3{x_2} = 0 \end{array}\] Therefore, the eigenvector is:
\[{v_1} = \left[ {\begin{array}{*{20}{c}} 1\\ 1 \end{array}} \right]\]
Step-4:
Find the eigenvectors for the eigenvalues 0.8
\[ \begin{array} { c } { [ A - ( 0.8 ) I ] \mathbf { v } _ { 2 } = \mathbf { 0 } } \\ { \left[ \begin{array} { c c } { 0.5 - 0.8 } & { 0.6 } \\ { - 0.3 } & { 1.4 - 0.8 } \end{array} \right] \left[ \begin{array} { c } { y _ { 1 } } \\ { y _ { 2 } } \end{array} \right] = 0 } \\ { \left[ \begin{array} { c c } { - 0.3 } & { 0.6 } \\ { - 0.3 } & { 0.6 } \end{array} \right] \left[ \begin{array} { c } { y _ { 1 } } \\ { y _ { 2 } } \end{array} \right] = 0 } \end{array} \] The system of equations \[\begin{array}{l} - 0.3{y_1} + 0.6{y_2} = 0\\ - 0.3{y_1} + 0.6{y_2} = 0 \end{array}\] Therefore, the eigenvector is:
\[{v_2} = \left[ {\begin{array}{*{20}{c}} 2\\ 1 \end{array}} \right]\]
Step-4: The directions
The greatest attraction is through the origin and the Eigenvector $v_2$ and
The greatest repulsion is through the origin and the Eigenvector $v_1$ and
We are given with following dynamical system \[ \mathbf { x } _ { k + 1 } = A \mathbf { x } _ { k } , \text { where } A = \left[ \begin{array} { c c } { 0.5 } & { 0.6 } \\ { - 0.3 } & { 1.4 } \end{array} \right] \]
Step-1: Eigenvalues of A:
Compute the characteristic polynomial: \[ \begin{aligned} \operatorname { det } ( A - \lambda I ) & = \left| \begin{array} { c c } { 0.5 - \lambda } & { 0.6 } \\ { - 0.3 } & { 1.4 - \lambda } \end{array} \right| \\ & = ( 0.5 - \lambda ) ( 1.4 - \lambda ) + 0.18 \\ & = \lambda ^ { 2 } - 1.9 \lambda + 0.7 + 0.18 \\ & = \lambda ^ { 2 } - 1.9 \lambda + 0.88 \end{aligned} \] The eigenvalues are: \[\begin{array}{l} \lambda = \dfrac{{1.9 \pm \sqrt {{{( - 1.9)}^2} - 4(0.88)} }}{2}\\ = \dfrac{{1.9 \pm \sqrt {3.61 - 3.52} }}{2}\\ = \dfrac{{1.9 \pm \sqrt {0.09} }}{2}\\ = \dfrac{{1.9 \pm 0.3}}{2}\\ = 0.8{\rm{ and }}1.1 \end{array}\]
Step-2:
Since, one of the Eigenvalue 1.1 is greater than 1 in magnitude and an Eigenvalue 0.8 is less than 1 in magnitude, therefore origin is a saddle point for the dynamical system.
The direction of greatest attraction, is given by the the eigenvector whose Eigenvalue is less than 1 in magnitude.
The direction of greatest repulsion, is given by the the eigenvector whose Eigenvalue is more than 1 in magnitude.
Step-3:
Find the eigenvectors for the eigenvalues 1.1
\[ \begin{array} { r } { [ A - ( 1.1 ) I ] \mathbf { v } _ { 1 } = \mathbf { 0 } } \\ { \left[ \begin{array} { c c } { 0.5 - 1.1 } & { 0.6 } \\ { - 0.3 } & { 1.4 - 1.1 } \end{array} \right] \left[ \begin{array} { c } { x _ { 1 } } \\ { x _ { 2 } } \end{array} \right] = 0 } \\ { \left[ \begin{array} { c c c } { - 0.6 } & { 0.6 } \\ { - 0.3 } & { 0.3 } \end{array} \right] \left[ \begin{array} { c } { x _ { 1 } } \\ { x _ { 2 } } \end{array} \right] = 0 } \end{array} \] The system of equations \[\begin{array}{l} - 0.6{x_1} + 0.6{x_2} = 0\\ - 0.3{x_1} + 0.3{x_2} = 0 \end{array}\] Therefore, the eigenvector is:
\[{v_1} = \left[ {\begin{array}{*{20}{c}} 1\\ 1 \end{array}} \right]\]
Step-4:
Find the eigenvectors for the eigenvalues 0.8
\[ \begin{array} { c } { [ A - ( 0.8 ) I ] \mathbf { v } _ { 2 } = \mathbf { 0 } } \\ { \left[ \begin{array} { c c } { 0.5 - 0.8 } & { 0.6 } \\ { - 0.3 } & { 1.4 - 0.8 } \end{array} \right] \left[ \begin{array} { c } { y _ { 1 } } \\ { y _ { 2 } } \end{array} \right] = 0 } \\ { \left[ \begin{array} { c c } { - 0.3 } & { 0.6 } \\ { - 0.3 } & { 0.6 } \end{array} \right] \left[ \begin{array} { c } { y _ { 1 } } \\ { y _ { 2 } } \end{array} \right] = 0 } \end{array} \] The system of equations \[\begin{array}{l} - 0.3{y_1} + 0.6{y_2} = 0\\ - 0.3{y_1} + 0.6{y_2} = 0 \end{array}\] Therefore, the eigenvector is:
\[{v_2} = \left[ {\begin{array}{*{20}{c}} 2\\ 1 \end{array}} \right]\]
Step-4: The directions
The greatest attraction is through the origin and the Eigenvector $v_2$ and
The greatest repulsion is through the origin and the Eigenvector $v_1$ and