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14.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.

\(A = \left( {\begin{aligned}{}{1.7}&{}&{.6}\\{ - .4}&{}&{.7}\end{aligned}} \right)\)

Short Answer

Expert verified

The origin is a repellor as both the eigenvalues are greater than one.

The direction of greatest repulsion is through the origin and eigenvector: \({v_1} = \left( {\begin{aligned}{}{ - 3}\\2\end{aligned}} \right)\).

Step by step solution

01

Discrete Dynamical System

For anyDynamical Systemwith matrix \(A\), the eigenvalues of \(A\) will determine the system of:

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

02

Evaluating Eigenvalues of a given matrix

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

\(A = \left( {\begin{aligned}{}{1.7}&{}&{.6}\\{ - .4}&{}&{.7}\end{aligned}} \right)\)

For eigenvalues, we have:

\(\begin{aligned}{}\det \left( {A - \lambda I} \right) = 0\\\left| {\begin{aligned}{}{1.7 - \lambda }&{}&{0.6}\\{ - 0.4}&{}&{0.7 - \lambda }\end{aligned}} \right| = 0\\{\lambda ^2} - 2.4\lambda + 1.43 = 0\\\lambda = 1.1,1.3\end{aligned}\)

Here, both the eigenvalues are greater than 1.

Hence, the origin is a repellor.

03

Eigenvector for eigenvalue 1.3

Now, for eigenvectors, we have:

At \(\lambda = 1.3\).

\(\begin{aligned}{}\left( {A - \left( {1.3} \right)I} \right){v_1} &= 0\\\left( {\begin{aligned}{}{1.7 - 1.3}&{}&{0.6}\\{ - 0.4}&{}&{0.7 - 1.3}\end{aligned}} \right)\left( {\begin{aligned}{}x\\y\end{aligned}} \right) &= 0\\\left( {\begin{aligned}{}{0.4}&{}&{0.6}\\{ - 0.4}&{}&{ - 0.6}\end{aligned}} \right)\left( {\begin{aligned}{}x\\y\end{aligned}} \right) &= 0\end{aligned}\)

Also, the system of equationwill be:

\(\begin{aligned}{}0.4x + 0.6y = 0\\ - 0.4x - 0.6y = 0\end{aligned}\)

For this, the general solution, we have:

\(\begin{aligned}{}{v_1} &= \left( {\begin{aligned}{}x\\y\end{aligned}} \right)\\ &= \left( {\begin{aligned}{}{ - \frac{6}{4}y}\\y\end{aligned}} \right)\\ &= \left( {\begin{aligned}{}{ - 6}\\4\end{aligned}} \right)\\ &= \left( {\begin{aligned}{}{ - 3}\\2\end{aligned}} \right)\end{aligned}\)

Here, the eigenvector is: \({v_1} = \left( {\begin{aligned}{}{ - 3}\\2\end{aligned}} \right)\).

Hence, the direction of greatest repulsion is through the origin and eigenvector: \({v_1} = \left( {\begin{aligned}{}{ - 3}\\2\end{aligned}} \right)\).

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Most popular questions from this chapter

Question: In Exercises 21 and 22, \(A\) and \(B\) are \(n \times n\) matrices. Mark each statement True or False. Justify each answer.

  1. If \(A\) is \(3 \times 3\), with columns \({{\rm{a}}_1}\), \({{\rm{a}}_2}\), and \({{\rm{a}}_3}\), then \(\det A\) equals the volume of the parallelepiped determined by \({{\rm{a}}_1}\), \({{\rm{a}}_2}\), and \({{\rm{a}}_3}\).
  2. \(\det {A^T} = \left( { - 1} \right)\det A\).
  3. The multiplicity of a root \(r\) of the characteristic equation of \(A\) is called the algebraic multiplicity of \(r\) as an eigenvalue of \(A\).
  4. A row replacement operation on \(A\) does not change the eigenvalues.

A common misconception is that if \(A\) has a strictly dominant eigenvalue, then, for any sufficiently large value of \(k\), the vector \({A^k}{\bf{x}}\) is approximately equal to an eigenvector of \(A\). For the three matrices below, study what happens to \({A^k}{\bf{x}}\) when \({\bf{x = }}\left( {{\bf{.5,}}{\bf{.5}}} \right)\), and try to draw general conclusions (for a \({\bf{2 \times 2}}\) matrix).

a. \(A{\bf{ = }}\left( {\begin{aligned}{ {20}{c}}{{\bf{.8}}}&{\bf{0}}\\{\bf{0}}&{{\bf{.2}}}\end{aligned}} \right)\) b. \(A{\bf{ = }}\left( {\begin{aligned}{ {20}{c}}{\bf{1}}&{\bf{0}}\\{\bf{0}}&{{\bf{.8}}}\end{aligned}} \right)\) c. \(A{\bf{ = }}\left( {\begin{aligned}{ {20}{c}}{\bf{8}}&{\bf{0}}\\{\bf{0}}&{\bf{2}}\end{aligned}} \right)\)

Let \(B = \left\{ {{{\bf{b}}_1},{{\bf{b}}_2},{{\bf{b}}_3}} \right\}\)be a basis for a vector space \(V\) and\(T:V \to {\mathbb{R}^2}\) be a linear transformation with the property that

\(T\left( {{x_1}{{\bf{b}}_1} + {x_2}{{\bf{b}}_2} + {x_3}{{\bf{b}}_3}} \right) = \left( {\begin{aligned}{2{x_1} - 4{x_2} + 5{x_3}}\\{ - {x_2} + 3{x_3}}\end{aligned}} \right)\)

Find the matrix for \(T\) relative to \(B\) and the standard basis for \({\mathbb{R}^2}\).

Question 19: Let \(A\) be an \(n \times n\) matrix, and suppose A has \(n\) real eigenvalues, \({\lambda _1},...,{\lambda _n}\), repeated according to multiplicities, so that \(\det \left( {A - \lambda I} \right) = \left( {{\lambda _1} - \lambda } \right)\left( {{\lambda _2} - \lambda } \right) \ldots \left( {{\lambda _n} - \lambda } \right)\) . Explain why \(\det A\) is the product of the n eigenvalues of A. (This result is true for any square matrix when complex eigenvalues are considered.)

Compute the quantities in Exercises 1-8 using the vectors

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

3. \(\frac{1}{{{\mathop{\rm w}\nolimits} \cdot {\mathop{\rm w}\nolimits} }}{\mathop{\rm w}\nolimits} \)

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