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Q7.6-28E

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Linear Algebra and its Applications
Found in: Page 267
Linear Algebra and its Applications

Linear Algebra and its Applications

Book edition 5th
Author(s) David C. Lay, Steven R. Lay and Judi J. McDonald
Pages 483 pages
ISBN 978-03219822384

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Short Answer

Consider an invertible n × n matrix A such that the zero state is a stable equilibrium of the dynamical system x→(t+1)=Ax→(t) What can you say about the stability of the systems

x→(t+1)=(A-2In)x→(t)

The given value is unstable

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Step by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

define eigenvalue

Eigenvalues can be used to determine whether a fixed point (also known as an equilibrium point) is stable or unstable.

Finding the stability of the given system

Most popular questions for Math Textbooks

[M] Repeat Exercise 25 for \[A{\bf{ = }}\left[ {\begin{array}{*{20}{c}}{{\bf{ - 8}}}&{\bf{5}}&{{\bf{ - 2}}}&{\bf{0}}\\{{\bf{ - 5}}}&{\bf{2}}&{\bf{1}}&{{\bf{ - 2}}}\\{{\bf{10}}}&{{\bf{ - 8}}}&{\bf{6}}&{{\bf{ - 3}}}\\{\bf{3}}&{{\bf{ - 2}}}&{\bf{1}}&{\bf{0}}\end{array}} \right]\].

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 \(\left( {{\bf{ - 2,}}\;{\bf{1,}}\;{\bf{2}}} \right)\).

15. \(\left( {\begin{array}{*{20}{c}}{\bf{7}}&{\bf{4}}&{{\bf{16}}}\\{\bf{2}}&{\bf{5}}&{\bf{8}}\\{{\bf{ - 2}}}&{{\bf{ - 2}}}&{{\bf{ - 5}}}\end{array}} \right)\)

Question 20: Use a property of determinants to show that \(A\) and \({A^T}\) have the same characteristic polynomial.

Question: Construct a random integer-valued \(4 \times 4\) matrix \(A\), and verify that \(A\) and \({A^T}\) have the same characteristic polynomial (the same eigenvalues with the same multiplicities). Do \(A\) and \({A^T}\) have the same eigenvectors? Make the same analysis of a \(5 \times 5\) matrix. Report the matrices and your conclusions.

Suppose \(A\) is diagonalizable and \(p\left( t \right)\) is the characteristic polynomial of \(A\). Define \(p\left( A \right)\) as in Exercise 5, and show that \(p\left( A \right)\) is the zero matrix. This fact, which is also true for any square matrix, is called the Cayley-Hamilton theorem.
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