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Q5.2-7E

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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

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

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

Characteristic polynomial: \({\lambda ^2} - 9\lambda + 32\).

\(A\) has no real Eigen values.

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Step by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

Step 1: Find the characteristic polynomial

If is an \(n \times n\) matrix, then \(det\left( {A - \lambda I} \right)\), which is a polynomial of degree \(n\), is called the characteristic polynomial of \(A\).

It is given that \(A = \left[ {\begin{array}{*{20}{c}}5&3\\- 4&4\end{array}} \right]\) and \(I = \left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right]\) is identity matrix. Find the matrix\(\left( {A - \lambda I} \right)\) as shown below:

\[\begin{array}A - \lambda I = \left[ {\begin{array}{*{20}{c}}3&- 4\\4&8\end{array}} \right] - \lambda \left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}{3 - \lambda }&{ - 4}\\4&{8 - \lambda }\end{array}} \right]\end{array}\]

Now, calculate the determinant of the matrix \(\left( {A - \lambda I} \right)\) as shown below:

\[\begin{array}det\left( {A - \lambda I} \right) = \det \left[ {\begin{array}{*{20}{c}}5&3\\{ - 4}&4\end{array}} \right]\\ = \left( {5 - \lambda } \right)\left( {4 - {\rm{\lambda }}} \right) + 12\\ = {\lambda ^2} - 9\lambda + 20 + 12\\ = {\lambda ^2} - 9\lambda + 32\end{array}\]

So, the characteristic polynomial of is \({\lambda ^2} - 9\lambda + 32\).

Step 2: Describe the characteristic equation

To find the eigenvalues of the matrix, we must calculate all the scalars such that \(\left( {A - \lambda I} \right)x = 0\) has a non-trivial solution which is equivalent to finding all such that the matrix \(\left( {A - \lambda I} \right)\)is not invertible, that is, when determinant of \(\left( {A - \lambda I} \right)\)is zero.

Thus, the eigenvalues of \(A\) are the solutions of the characteristic equation\(\det \left( {A - \lambda I} \right) = 0\). So, find the characteristic equation \(\det \left( {A - \lambda I} \right) = 0\).

\[\begin{array}det\left[ {\begin{array}{*{20}{c}}5&3\\- 4&4\end{array}} \right] = 0\\\left( {5 - \lambda } \right)\left( {4 - {\rm{\lambda }}} \right) + 12 = 0\\{\lambda ^2} - 9\lambda + 20 + 12 = 0\\{\lambda ^2} - 9\lambda + 32 = 0\end{array}\]

Step 3: Find roots of characteristic equation

For the quadratic equation, \(a{x^2} + bx + c = 0\) , the general solution is given as\(x = \frac{{ - b \pm \sqrt {{b^2} - 4ac} \;\;}}{{2a}}\).

Thus, the solution of the characteristic equation \({\lambda ^2} - 9\lambda + 32 = 0\) is obtained as follows:

\[\begin{array}{\lambda ^2} - 9\lambda + 32 = 0\\\lambda = \frac{{ - \left( { - 9} \right) \pm \sqrt {{{\left( { - 9} \right)}^2} - 4\left( {32} \right)} }}{2}\\ = \frac{{ - 9 \pm \sqrt { - 47} }}{2}\end{array}\]

The eigenvalues of \(A\) are complex. So, \(A\) has no real Eigen values. This implies that no real vector \(x\)in \({\mathbb{R}^2}\)satisfies the characteristic equation \(\left( {A - \lambda I} \right)x = 0\).

Most popular questions for Math Textbooks

For the matrix A, find real closed formulas for the trajectory x→(t+1)=Ax¯(t)where x→=[01]. Draw a rough sketch

A = [15-27]

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

(M) The MATLAB command roots\(\left( p \right)\) computes the roots of the polynomial equation \(p\left( t \right) = {\bf{0}}\). Read a MATLAB manual, and then describe the basic idea behind the algorithm for the roots command.

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

In Exercises 9–16, find a basis for the eigenspace corresponding to each listed eigenvalue.

10. \(A = \left( {\begin{array}{*{20}{c}}{10}&{ - 9}\\4&{ - 2}\end{array}} \right)\), \(\lambda = 4\)
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