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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

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 \(p\): \({C_p} = \left( {\begin{aligned}{*{20}{c}}{\bf{0}}&{\bf{1}}&{\bf{0}}&{...}&{\bf{0}}\\{\bf{0}}&{\bf{0}}&{\bf{1}}&{}&{\bf{0}}\\:&{}&{}&{}&:\\{\bf{0}}&{\bf{0}}&{\bf{0}}&{}&{\bf{1}}\\{ - {a_{\bf{0}}}}&{ - {a_{\bf{1}}}}&{ - {a_{\bf{2}}}}&{...}&{ - {a_{n - {\bf{1}}}}}\end{aligned}} \right)\).

23. Let \(p\) be the polynomial in Exercise \({\bf{22}}\), and suppose the equation \(p\left( t \right) = {\bf{0}}\) has distinct roots \({\lambda _{\bf{1}}},{\lambda _{\bf{2}}},{\lambda _{\bf{3}}}\). Let \(V\) be the Vandermonde matrix

\(V{\bf{ = }}\left( {\begin{aligned}{*{20}{c}}{\bf{1}}&{\bf{1}}&{\bf{1}}\\{{\lambda _{\bf{1}}}}&{{\lambda _{\bf{2}}}}&{{\lambda _{\bf{3}}}}\\{\lambda _{\bf{1}}^{\bf{2}}}&{\lambda _{\bf{2}}^{\bf{2}}}&{\lambda _{\bf{3}}^{\bf{2}}}\end{aligned}} \right)\)

(The transpose of \(V\) was considered in Supplementary Exercise \({\bf{11}}\) in Chapter \({\bf{2}}\).) Use Exercise \({\bf{22}}\) and a theorem from this chapter to deduce that \(V\) is invertible (but do not compute \({V^{{\bf{ - 1}}}}\)). Then explain why \({V^{{\bf{ - 1}}}}{C_p}V\) is a diagonal matrix.

All the columns of \(V\) are linearly independent and \(V\) is invertible. Therefore, the matrix \({C_p}\) is diagonalizable and \({V^{ - 1}}{C_p}V\) is a diagonal matrix.

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Step by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

Step 1: Write the companion matrix for \(p\)

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

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

As \(p\left( t \right) = {a_0} + {a_1}t + {a_2}{t^2} + {t^3}\) and \({\lambda _1},{\lambda _2},{\lambda _3}\) are the distinct roots of the equation \(p\left( t \right) = 0\).

Consider \(V = \left( {\begin{aligned}{*{20}{c}}1&1&1\\{{\lambda _1}}&{{\lambda _2}}&{{\lambda _3}}\\{\lambda _1^2}&{\lambda _2^2}&{\lambda _3^2}\end{aligned}} \right)\).

Therefore, the companion matrix is \({C_p} = \left( {\begin{aligned}{*{20}{c}}0&1&0\\0&0&1\\{ - {a_0}}&{ - {a_1}}&{ - {a_2}}\end{aligned}} \right)\).

As \(\lambda \) is a zero of \(p\left( t \right)\), then we get,

\(\begin{aligned}{c}{a_0} + {a_1}\lambda + {a_2}{\lambda ^2} + {\lambda ^3} = 0\\ - {a_0} - {a_1}\lambda - {a_2}{\lambda ^2} = {\lambda ^3}\end{aligned}\)

Step 2: Check whether \(\left( {{\bf{1}},\lambda ,{\lambda ^{\bf{2}}}} \right)\) is an eigenvector of \({C_p}\)

\(\begin{aligned}{c}{C_p} &= \left( {\begin{aligned}{*{20}{c}}1\\\lambda \\{{\lambda ^2}}\end{aligned}} \right)\\ &= \left( {\begin{aligned}{*{20}{c}}0&1&0\\0&0&1\\{ - {a_0}}&{ - {a_1}}&{ - {a_2}}\end{aligned}} \right)\left( {\begin{aligned}{*{20}{c}}1\\\lambda \\{{\lambda ^2}}\end{aligned}} \right)\\ &= \left( {\begin{aligned}{*{20}{c}}\lambda \\{{\lambda ^2}}\\{ - {a_0} - {a_1}\lambda - {a_2}{\lambda ^2}}\end{aligned}} \right)\\ &= \left( {\begin{aligned}{*{20}{c}}\lambda \\{{\lambda ^2}}\\{{\lambda ^3}}\end{aligned}} \right)\\ &= \lambda \left( {\begin{aligned}{*{20}{c}}1\\\lambda \\{{\lambda ^2}}\end{aligned}} \right)\end{aligned}\)

Thus, \(\left( {1,\lambda ,{\lambda ^2}} \right)\) is an eigenvector of \({C_p}\).

Since all eigenvalues \({\lambda _1},{\lambda _2},{\lambda _3}\) are distinct the eigenvectors of \({C_p}\) are linearly independent.

Thus, all the columns of \(V\) are linearly independent and \(V\) is invertible. Therefore, the matrix \({C_p}\) is diagonalizable and \({V^{ - 1}}{C_p}V\) is a diagonal matrix.

Most popular questions for Math Textbooks

Question: Exercises 9-14 require techniques section 3.1. Find the characteristic polynomial of each matrix, using either a cofactor expansion or the special formula for \(3 \times 3\) determinants described prior to Exercise 15-18 in Section 3.1. [Note: Finding the characteristic polynomial of a \(3 \times 3\) matrix is not easy to do with just row operations, because the variable \(\lambda \) is involved.

13. \(\left[ {\begin{array}{*{20}{c}}6&- 2&0\\- 2&9&0\\5&8&3\end{array}} \right]\)

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

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

Question: 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.

3. \(\left( {\begin{array}{*{20}{c}}a&0\\{3\left( {a - b} \right)}&b\end{array}} \right) = \left( {\begin{array}{*{20}{c}}1&0\\3&1\end{array}} \right)\left( {\begin{array}{*{20}{c}}a&0\\0&b\end{array}} \right)\left( {\begin{array}{*{20}{c}}1&0\\{ - 3}&1\end{array}} \right)\)

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

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

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 \(A\).
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