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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

Suppose \({\bf{x}}\) is an eigenvector of \(A\) corresponding to an eigenvalue \(\lambda \).

a. Show that \(x\) is an eigenvector of \(5I - A\). What is the corresponding eigenvalue?

b. Show that \(x\) is an eigenvector of \(5I - 3A + {A^2}\). What is the corresponding eigenvalue?

(a) It is proved that \(x\) is an eigenvector of \(5I - A\). Therefore,\(\;5 - \lambda \) is the eigenvalue.

(b) It is proved that \(x\) is an eigenvector of \(5I - 3A + {A^2}\). Therefore, \(5 - 3\lambda + {\lambda ^2}\) is the eigenvalue.

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

Step 1: Definition of matrix and eigenvalue

A matrix (plural matrices) is a rectangular array or table of numbers, symbols, or expressions, arranged in rows and columns, which is used to represent a mathematical object or a property of such an object.

The eigenvalue \(\lambda \) is the real or complex number of a matrix \(A\) which is a square matrix that satisfies the following equation

\(\det \left( {A - \lambda I} \right) = 0\)

.

This equation is called the characteristic equation.

Step 2: Find the eigenvector of matrix product

Let \(\lambda \) be an eigenvalue of the matrix \(A\). Then we have

\(Ax = \lambda x\)

(a)

Simplify \((5I - A)x\).

\(\begin{array}{r}(5I - A)x = 5x - Ax\\ = 5x - \lambda x\\ = (5 - \lambda I)x\\ = (5 - \lambda )x\end{array}\)

Therefore\(\;5 - \lambda \) is the eigenvalue.

b)

Consider,

\(\begin{array}{c}\left( {5I - 3A + {A^2}} \right)x = 5x - 3Ax + {A^2}x\\ = 5x - 3\lambda x + A\lambda x\\ = 5x - 3\lambda x + \lambda Ax\\ = 5x - 3\lambda x + {\lambda ^2}x\\ = \left( {5 - 3\lambda + {\lambda ^2}} \right)x\end{array}\)

Therefore \(5 - 3\lambda + {\lambda ^2}\) is the eigenvalue.

Most popular questions for Math Textbooks

Question: In Exercises \({\bf{1}}\) and \({\bf{2}}\), let \(A = PD{P^{ - {\bf{1}}}}\) and compute \({A^{\bf{4}}}\).

2. \(P{\bf{ = }}\left( {\begin{array}{*{20}{c}}2&{ - 3}\\{ - 3}&5\end{array}} \right)\), \(D{\bf{ = }}\left( {\begin{array}{*{20}{c}}{\bf{1}}&{\bf{0}}\\{\bf{0}}&{\frac{{\bf{1}}}{{\bf{2}}}}\end{array}} \right)\)

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

  1. The determinant of \(A\) is the product of the diagonal entries in \(A\).
  2. An elementary row operation on \(A\) does not change the determinant.
  3. \(\left( {\det A} \right)\left( {\det B} \right) = \det AB\)
  4. If \(\lambda + 5\) is a factor of the characteristic polynomial of \(A\), then 5 is an eigenvalue of \(A\).

Question: Find the characteristic polynomial and the eigenvalues of the matrices in Exercises 1-8.

7. \(\left[ {\begin{array}{*{20}{c}}5&3\\- 4&4\end{array}} \right]\)

[M] In Exercises 19 and 20, find (a) the largest eigenvalue and (b) the eigenvalue closest to zero. In each case, set \[{{\bf{x}}_{\bf{0}}}{\bf{ = }}\left( {{\bf{1,0,0,0}}} \right)\] and carry out approximations until the approximating sequence seems accurate to four decimal places. Include the approximate eigenvector.

20. \[A{\bf{ = }}\left[ {\begin{array}{*{20}{c}}{\bf{1}}&{\bf{2}}&{\bf{3}}&{\bf{2}}\\{\bf{2}}&{{\bf{12}}}&{{\bf{13}}}&{{\bf{11}}}\\{{\bf{ - 2}}}&{\bf{3}}&{\bf{0}}&{\bf{2}}\\{\bf{4}}&{\bf{5}}&{\bf{7}}&{\bf{2}}\end{array}} \right]\]

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

7. \(\left( {\begin{array}{*{20}{c}}{\bf{1}}&{\bf{0}}\\{\bf{6}}&{{\bf{ - 1}}}\end{array}} \right)\)
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