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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.

\(A = \left( {\begin{aligned}{}{.4}&{}&{.5}\\{ - .4}&{}&{1.3}\end{aligned}} \right)\)

Short Answer

Expert verified

The origin is an attractor as two eigenvalues obtained, that is, \(0.8{\rm{ and }}0.9\) are less than 1. The direction of the greatest attraction passes through the origin, which is \({v_1} = \left( {\begin{aligned}{}5\\4\end{aligned}} \right)\).

Step by step solution

01

Discrete Dynamical System 

For any Dynamical System with matrix \(A\), the eigenvalues of \(A\) will determine the system of:

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

02

Evaluating Eigenvalues of the given matrix 

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

\(A = \left( {\begin{aligned}{}{.4}&{}{.5}\\{ - .4}&{}&{1.3}\end{aligned}} \right)\)

For eigenvalues, we have:

\(\begin{aligned}{}\det \left( {A - \lambda I} \right) = 0\\\left| {\begin{aligned}{}{0.4 - \lambda }&{}&{0.5}\\{ - 0.4}&{}&{1.3 - \lambda }\end{aligned}} \right| = 0\\\left( {0.4 - \lambda } \right)\left( {1.3 - \lambda } \right) + 0.2 = 0\\{\lambda ^2} - 1.7\lambda + 0.72 = 0\\\lambda = 0.8,0.9\end{aligned}\)

Here, both eigenvalues are less than 1.

Hence, the origin is an attractor for the given system

03

Eigenvector for eigenvalue 0.8

Now, for eigenvectors, we have:

At \(\lambda = 0.8\).

\(\begin{aligned}{}\left( {A - \left( {0.8} \right)I} \right){v_1} = 0\\\left( {\begin{aligned}{}{0.4 - 0.8}&{}&{0.5}\\{ - 0.4}&{}&{1.3 - 0.8}\end{aligned}} \right)\left( {\begin{aligned}{}x\\y\end{aligned}} \right) = 0\\\left( {\begin{aligned}{}{ - 0.4}&{}&{0.5}\\{ - 0.4}&{}&{0.5}\end{aligned}} \right)\left( {\begin{aligned}{}x\\y\end{aligned}} \right) = 0\end{aligned}\)

Also, the system of equation will be:

\(\begin{aligned}{} - 0.4x + 0.5y = 0\\ - 0.4x + 0.5y = 0\end{aligned}\)

For this, the general solution, we have:

\(\begin{aligned}{}{v_1} &= \left( {\begin{aligned}{}x\\y\end{aligned}} \right)\\ &= \left( {\begin{aligned}{}{\frac{5}{4}y}\\y\end{aligned}} \right)\\ &= \left( {\begin{aligned}{}5\\4\end{aligned}} \right)\end{aligned}\)

Here, the eigenvector is: \({v_1} = \left( {\begin{aligned}{}5\\4\end{aligned}} \right)\).

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Most popular questions from this chapter

Question: Let \(A = \left( {\begin{array}{*{20}{c}}a&b\\c&d\end{array}} \right)\). Use formula (1) for a determinant (given before Example 2) to show that \(\det A = ad - bc\). Consider two cases: \(a \ne 0\) and \(a = 0\).

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: Exercises 9-14 require techniques section 3.1. Find the characteristic polynomial of each matrix, using either a cofactor expansion or the special formula for \(3 \times 3\) determinants described prior to Exercise 15-18 in Section 3.1. [Note: Finding the characteristic polynomial of a \(3 \times 3\) matrix is not easy to do with just row operations, because the variable \(\lambda \) is involved.

14. \(\left[ {\begin{array}{*{20}{c}}5&- 2&3\\0&1&0\\6&7&- 2\end{array}} \right]\)

Question: For the matrices in Exercises 15-17, list the eigenvalues, repeated according to their multiplicities.

15. \(\left[ {\begin{array}{*{20}{c}}4&- 7&0&2\\0&3&- 4&6\\0&0&3&{ - 8}\\0&0&0&1\end{array}} \right]\)

Let\(\varepsilon = \left\{ {{{\bf{e}}_1},{{\bf{e}}_2},{{\bf{e}}_3}} \right\}\) be the standard basis for \({\mathbb{R}^3}\),\(B = \left\{ {{{\bf{b}}_1},{{\bf{b}}_2},{{\bf{b}}_3}} \right\}\) be a basis for a vector space \(V\) and\(T:{\mathbb{R}^3} \to V\) be a linear transformation with the property that

\(T\left( {{x_1},{x_2},{x_3}} \right) = \left( {{x_3} - {x_2}} \right){{\bf{b}}_1} - \left( {{x_1} - {x_3}} \right){{\bf{b}}_2} + \left( {{x_1} - {x_2}} \right){{\bf{b}}_3}\)

  1. Compute\(T\left( {{{\bf{e}}_1}} \right)\), \(T\left( {{{\bf{e}}_2}} \right)\) and \(T\left( {{{\bf{e}}_3}} \right)\).
  2. Compute \({\left( {T\left( {{{\bf{e}}_1}} \right)} \right)_B}\), \({\left( {T\left( {{{\bf{e}}_2}} \right)} \right)_B}\) and \({\left( {T\left( {{{\bf{e}}_3}} \right)} \right)_B}\).
  3. Find the matrix for \(T\) relative to \(\varepsilon \), and\(B\).
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