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Chapter 5: Q5.6-9E (page 267)

In Exercises 9鈥�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

Expert verified

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

01

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.

02

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

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Most popular questions from this chapter

Question: Is \(\left( {\begin{array}{*{20}{c}}1\\4\end{array}} \right)\) an eigenvalue of \(\left( {\begin{array}{*{20}{c}}{ - 3}&1\\{ - 3}&8\end{array}} \right)\)? If so, find the eigenvalue.

Question: Diagonalize the matrices in Exercises \({\bf{7--20}}\), if possible. The eigenvalues for Exercises \({\bf{11--16}}\) are as follows:\(\left( {{\bf{11}}} \right)\lambda {\bf{ = 1,2,3}}\); \(\left( {{\bf{12}}} \right)\lambda {\bf{ = 2,8}}\); \(\left( {{\bf{13}}} \right)\lambda {\bf{ = 5,1}}\); \(\left( {{\bf{14}}} \right)\lambda {\bf{ = 5,4}}\); \(\left( {{\bf{15}}} \right)\lambda {\bf{ = 3,1}}\); \(\left( {{\bf{16}}} \right)\lambda {\bf{ = 2,1}}\). For exercise \({\bf{18}}\), one eigenvalue is \(\lambda {\bf{ = 5}}\) and one eigenvector is \(\left( {{\bf{ - 2,}}\;{\bf{1,}}\;{\bf{2}}} \right)\).

9. \(\left( {\begin{array}{*{20}{c}}3&{ - 1}\\1&5\end{array}} \right)\)

Let \({\bf{u}}\) be an eigenvector of \(A\) corresponding to an eigenvalue \(\lambda \), and let \(H\) be the line in \({\mathbb{R}^{\bf{n}}}\) through \({\bf{u}}\) and the origin.

  1. Explain why \(H\) is invariant under \(A\) in the sense that \(A{\bf{x}}\) is in \(H\) whenever \({\bf{x}}\) is in \(H\).
  2. Let \(K\) be a one-dimensional subspace of \({\mathbb{R}^{\bf{n}}}\) that is invariant under \(A\). Explain why \(K\) contains an eigenvector of \(A\).

If \(p\left( t \right) = {c_0} + {c_1}t + {c_2}{t^2} + ...... + {c_n}{t^n}\), define \(p\left( A \right)\) to be the matrix formed by replacing each power of \(t\) in \(p\left( t \right)\)by the corresponding power of \(A\) (with \({A^0} = I\) ). That is,

\(p\left( t \right) = {c_0} + {c_1}I + {c_2}{I^2} + ...... + {c_n}{I^n}\)

Show that if \(\lambda \) is an eigenvalue of A, then one eigenvalue of \(p\left( A \right)\) is\(p\left( \lambda \right)\).

Compute the quantities in Exercises 1-8 using the vectors

\({\mathop{\rm u}\nolimits} = \left( {\begin{aligned}{*{20}{c}}{ - 1}\\2\end{aligned}} \right),{\rm{ }}{\mathop{\rm v}\nolimits} = \left( {\begin{aligned}{*{20}{c}}4\\6\end{aligned}} \right),{\rm{ }}{\mathop{\rm w}\nolimits} = \left( {\begin{aligned}{*{20}{c}}3\\{ - 1}\\{ - 5}\end{aligned}} \right),{\rm{ }}{\mathop{\rm x}\nolimits} = \left( {\begin{aligned}{*{20}{c}}6\\{ - 2}\\3\end{aligned}} \right)\)

2. \({\mathop{\rm w}\nolimits} \cdot {\mathop{\rm w}\nolimits} ,{\mathop{\rm x}\nolimits} \cdot {\mathop{\rm w}\nolimits} ,\,\,{\mathop{\rm and}\nolimits} \,\,\frac{{{\mathop{\rm x}\nolimits} \cdot {\mathop{\rm w}\nolimits} }}{{{\mathop{\rm w}\nolimits} \cdot {\mathop{\rm w}\nolimits} }}\)
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