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Chapter 1: Q28E (page 1)

Consider a dynamical system x鈫�(t+1)=Ax鈫�(t)with two components. The accompanying sketch shows the initial state vector x鈫�0and two eigenvectors 蠀1鈫�鈥�鈥�and鈥�鈥�蠀2鈫�of A (with eigen values 位1鈫�and位2鈫� respectively). For the given values of 位1鈫�and位2鈫�, draw a rough trajectory. Consider the future and the past of the system.

位1鈫�=1.2,位2鈫�=1.1

Short Answer

Expert verified

So, the required solution is Atx0=1.2t伪蠀1+1.1t尾蠀2.

Step by step solution

01

Define the eigenvector

Eigenvector:An eigenvector ofAis a nonzero vectorvinRnsuch thatAv=位v, for some scalar位.

02

Note the given data

It is given that:

位1鈫�=1.2,位2鈫�=1.1

Given graph is:

03

Calculate the required matrix

We have:

A蠀1=1.2蠀1A蠀2=1.1蠀2

For x0=伪蠀1+尾蠀2,We have:

Ax0=A(伪蠀1+尾蠀2)=伪A蠀1+尾A蠀2=1.2伪蠀1+1.1尾蠀2

Therefore, localid="1668085548202" Atx0=1.2t伪蠀1+1.1t尾蠀2.

Hence, the solution is Atx0=1.2t伪蠀1+1.1t尾蠀2..

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Most popular questions from this chapter

Find all the polynomials of degree鈮�2[a polynomial of the formf(t)=a+bt+ct2] whose graph goes through the points (1,3)and(2,6),such that f'(1)=1[wheref'(t)denotes the derivative].

Question: Determine whether the statements that follow are true or false, and justify your answer.

19. There exits a matrix A such thatA[-12]=[357].

Suppose an experiment leads to the following system of equations:

\(\begin{aligned}{c}{\bf{4}}.{\bf{5}}{x_{\bf{1}}} + {\bf{3}}.{\bf{1}}{x_{\bf{2}}} = {\bf{19}}.{\bf{249}}\\1.6{x_{\bf{1}}} + 1.1{x_{\bf{2}}} = 6.843\end{aligned}\) (3)

  1. Solve system (3), and then solve system (4), below, in which the data on the right have been rounded to two decimal places. In each case, find the exactsolution.

\(\begin{aligned}{c}{\bf{4}}.{\bf{5}}{x_{\bf{1}}} + {\bf{3}}.{\bf{1}}{x_{\bf{2}}} = {\bf{19}}.{\bf{25}}\\1.6{x_{\bf{1}}} + 1.1{x_{\bf{2}}} = 6.8{\bf{4}}\end{aligned}\) (4)

  1. The entries in (4) differ from those in (3) by less than .05%. Find the percentage error when using the solution of (4) as an approximation for the solution of (3).
  1. Use your matrix program to produce the condition number of the coefficient matrix in (3).

Let \({{\mathop{\rm v}\nolimits} _1} = \left[ {\begin{array}{*{20}{c}}1\\0\\{ - 2}\end{array}} \right],{v_2} = \left[ {\begin{array}{*{20}{c}}{ - 3}\\1\\8\end{array}} \right],\) and \({\rm{y = }}\left[ {\begin{array}{*{20}{c}}h\\{ - 5}\\{ - 3}\end{array}} \right]\). For what values(s) of \(h\) is \(y\) in the plane generated by \({{\mathop{\rm v}\nolimits} _1}\) and \({{\mathop{\rm v}\nolimits} _2}\)

In Exercise 2, compute \(u + v\) and \(u - 2v\).

2. \(u = \left[ {\begin{array}{*{20}{c}}3\\2\end{array}} \right]\), \(v = \left[ {\begin{array}{*{20}{c}}2\\{ - 1}\end{array}} \right]\).
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