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Q20SE

Expert-verified

Linear Algebra and its Applications

Found in: Page 267

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

Exercises 19–23 concern the polynomial \(p\left( t \right) = {a_{\bf{0}}} + {a_{\bf{1}}}t + ... + {a_{n - {\bf{1}}}}{t^{n - {\bf{1}}}} + {t^n}\) and \(n \times n\) matrix \({C_p}\) called the companion matrix of \(p\): \({C_p} = \left( {\begin{aligned}{*{20}{c}}{\bf{0}}&{\bf{1}}&{\bf{0}}&{...}&{\bf{0}}\\{\bf{0}}&{\bf{0}}&{\bf{1}}&{}&{\bf{0}}\\:&{}&{}&{}&:\\{\bf{0}}&{\bf{0}}&{\bf{0}}&{}&{\bf{1}}\\{ - {a_{\bf{0}}}}&{ - {a_{\bf{1}}}}&{ - {a_{\bf{2}}}}&{...}&{ - {a_{n - {\bf{1}}}}}\end{aligned}} \right)\).

20. Let \(p\left( t \right){\bf{ = }}\left( {t{\bf{ - 2}}} \right)\left( {t{\bf{ - 3}}} \right)\left( {t{\bf{ - 4}}} \right){\bf{ = - 24 + 26}}t{\bf{ - 9}}{t^{\bf{2}}}{\bf{ + }}{t^{\bf{3}}}\). Write the companion matrix for \(p\left( t \right)\), and use techniques from chapter \({\bf{3}}\) to find the characteristic polynomial.

The characteristic polynomial of the matrix \({C_p}\) is \( - p\left( \lambda \right)\).

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Find the companion matrix

Consider the polynomial \(p\left( t \right) = {a_0} + {a_1}t + ... + {a_{n - 1}}{t^{n - 1}} + {t^n}\).

The companion matrix of \(p\) is \({C_p} = \left( {\begin{aligned}{*{20}{c}}0&1&0&{...}&0\\0&0&1&{}&0\\:&{}&{}&{}&:\\0&0&0&{}&1\\{ - {a_0}}&{ - {a_1}}&{ - {a_2}}&{...}&{ - {a_{n - 1}}}\end{aligned}} \right)\).

Find the companion matrix for the given polynomial.

\(p\left( t \right) = - 24 + 26t - 9{t^2} + {t^3}\)

Therefore, on comparison we get,

\({C_p} = \left( {\begin{aligned}{*{20}{c}}0&1&0\\0&0&1\\{24}&{ - 26}&9\end{aligned}} \right)\)

Step 2: Find the characteristic polynomial

\(\begin{aligned}{c}\det \left( {{C_p} - \lambda I} \right) &= \det \left( {\begin{aligned}{*{20}{c}}{ - \lambda }&1&0\\0&{ - \lambda }&1\\{24}&{ - 26}&{9 - \lambda }\end{aligned}} \right)\\ &= \left( { - \lambda } \right)\left( {\left( { - \lambda } \right)\left( {9 - \lambda } \right) + 26} \right) + 24\\ &= \left( { - \lambda } \right)\left( { - 9\lambda + {\lambda ^2} + 26} \right) + 24\\ &= 9{\lambda ^2} - {\lambda ^3} - 26\lambda + 24\\ &= - \left( { - 24 + 26\lambda - 9{\lambda ^2} + {\lambda ^3}} \right)\\ &= - p\left( \lambda \right)\end{aligned}\)

Therefore, the characteristic polynomial of the matrix \({C_p}\) is \( - p\left( \lambda \right)\).

Most popular questions for Math Textbooks

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

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

Apply the results of Exercise \({\bf{15}}\) to find the eigenvalues of the matrices \(\left( {\begin{aligned}{*{20}{c}}{\bf{1}}&{\bf{2}}&{\bf{2}}\\{\bf{2}}&{\bf{1}}&{\bf{2}}\\{\bf{2}}&{\bf{2}}&{\bf{1}}\end{aligned}} \right)\) and \(\left( {\begin{aligned}{*{20}{c}}{\bf{7}}&{\bf{3}}&{\bf{3}}&{\bf{3}}&{\bf{3}}\\{\bf{3}}&{\bf{7}}&{\bf{3}}&{\bf{3}}&{\bf{3}}\\{\bf{3}}&{\bf{3}}&{\bf{7}}&{\bf{3}}&{\bf{3}}\\{\bf{3}}&{\bf{3}}&{\bf{3}}&{\bf{7}}&{\bf{3}}\\{\bf{3}}&{\bf{3}}&{\bf{3}}&{\bf{3}}&{\bf{7}}\end{aligned}} \right)\).

Question: Is \(\left( {\begin{array}{*{20}{c}}{ - 1 + \sqrt 2 }\\1\end{array}} \right)\) an eigenvalue of \(\left( {\begin{array}{*{20}{c}}2&1\\1&4\end{array}} \right)\)? If so, find the eigenvalue.

In Exercises \({\bf{3}}\) and \({\bf{4}}\), use the factorization \(A = PD{P^{ - {\bf{1}}}}\) to compute \({A^k}\) where \(k\) represents an arbitrary positive integer.

4. \(\left( {\begin{array}{*{20}{c}}{ - 2}&{12}\\{ - 1}&5\end{array}} \right) = \left( {\begin{array}{*{20}{c}}3&4\\1&1\end{array}} \right)\left( {\begin{array}{*{20}{c}}2&0\\0&1\end{array}} \right)\left( {\begin{array}{*{20}{c}}{ - 1}&4\\1&{ - 3}\end{array}} \right)\)

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

14. \(\left( {\begin{array}{*{20}{c}}4&0&{ - 2}\\2&5&4\\0&0&5\end{array}} \right)\)

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