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Q5.3-1E

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

1. \(P{\bf{ = }}\left( {\begin{array}{*{20}{c}}{\bf{5}}&{\bf{7}}\\{\bf{2}}&{\bf{3}}\end{array}} \right)\), \(D{\bf{ = }}\left( {\begin{array}{*{20}{c}}{\bf{2}}&{\bf{0}}\\{\bf{0}}&{\bf{1}}\end{array}} \right)\)

The required value is \({A^4} = \left( {\begin{array}{*{20}{c}}{226}&{ - 525}\\{90}&{ - 209}\end{array}} \right)\).

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Step by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

Step 1: Write the Diagonalization Theorem

The Diagonalization Theorem: An \(n \times n\) matrix \(A\) is diagonalizable if and only if \(A\) has \(n\) linearly independent eigenvectors. As \(A = PD{P^{ - 1}}\) which has \(D\) a diagonal matrix if and only if the columns of \(P\) are \(n\) linearly independent eigenvectors of \(A\).

Step 2: Find the inverse of the invertible matrix

Consider the matrix \(P = \left( {\begin{array}{*{20}{c}}5&7\\2&3\end{array}} \right)\) and \(D = \left( {\begin{array}{*{20}{c}}2&0\\0&1\end{array}} \right)\).

\[\begin{array}{P^{ - 1}} = \frac{1}{{5 \times 3 - 7 \times 2}}\left( {\begin{array}{*{20}{c}}3&{ - 7}\\{ - 2}&5\end{array}} \right)\\ = \frac{1}{{15 - 14}}\left( {\begin{array}{*{20}{c}}3&{ - 7}\\{ - 2}&5\end{array}} \right)\\ = \frac{1}{1}\left( {\begin{array}{*{20}{c}}3&{ - 7}\\{ - 2}&5\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}3&{ - 7}\\{ - 2}&5\end{array}} \right)\end{array}\]

Step 3: Find \({{\bf{4}}^{{\bf{th}}}}\)the power of the diagonal matrix

\[\begin{array}{l}D = \left( {\begin{array}{*{20}{c}}2&0\\0&1\end{array}} \right)\\{D^2} = \left( {\begin{array}{*{20}{c}}{{2^2}}&0\\0&{{1^2}}\end{array}} \right)\\{D^3} = \left( {\begin{array}{*{20}{c}}{{2^3}}&0\\0&{{1^3}}\end{array}} \right)\\{D^4} = \left( {\begin{array}{*{20}{c}}{{2^4}}&0\\0&{{1^4}}\end{array}} \right)\\{D^4} = \left( {\begin{array}{*{20}{c}}{16}&0\\0&1\end{array}} \right)\end{array}\]

Step 4: Find \({A^{\bf{4}}}\)

Consider the matrix \(P = \left( {\begin{array}{*{20}{c}}5&7\\2&3\end{array}} \right)\) and \(D = \left( {\begin{array}{*{20}{c}}2&0\\0&1\end{array}} \right)\).

As it is given that \(A = PD{P^{ - 1}}\) than by using the formula for \({n^{th}}\) power we get:

\[{A^n} = P{D^n}{P^{ - 1}}\].

Then we get:

\[\begin{array}{A^4} = P{D^4}{P^{ - 1}}\\ = \left( {\begin{array}{*{20}{c}}5&7\\2&3\end{array}} \right)\left( {\begin{array}{*{20}{c}}{16}&0\\0&1\end{array}} \right)\left( {\begin{array}{*{20}{c}}3&{ - 7}\\{ - 2}&5\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}{80}&7\\{32}&3\end{array}} \right)\left( {\begin{array}{*{20}{c}}3&{ - 7}\\{ - 2}&5\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}{226}&{ - 525}\\{90}&{ - 209}\end{array}} \right)\end{array}\]

Thus, \({A^4} = \left( {\begin{array}{*{20}{c}}{226}&{ - 525}\\{90}&{ - 209}\end{array}} \right)\).

Most popular questions for Math Textbooks

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

22. Let \(p\left( t \right) = {a_{\bf{0}}} + {a_{\bf{1}}}t + {a_{\bf{2}}}{t^{\bf{2}}} + {t^{\bf{3}}}\), and let \(\lambda \) be a zero of \(p\).

  1. Write the companion matrix for \(p\).
  2. Explain why \({\lambda ^{\bf{3}}} = - {a_{\bf{0}}} - {a_{\bf{1}}}\lambda - {a_{\bf{2}}}{\lambda ^{\bf{2}}}\), and show that \(\left( {{\bf{1}},\lambda ,{\lambda ^2}} \right)\) is an eigenvector of the companion matrix for \(p\).

Compute the quantities in Exercises 1-8 using the vectors

\({\mathop{\rm u}\nolimits} = \left( {\begin{aligned}{*{20}{c}}{ - 1}\\2\end{aligned}} \right),{\rm{ }}{\mathop{\rm v}\nolimits} = \left( {\begin{aligned}{*{20}{c}}4\\6\end{aligned}} \right),{\rm{ }}{\mathop{\rm w}\nolimits} = \left( {\begin{aligned}{*{20}{c}}3\\{ - 1}\\{ - 5}\end{aligned}} \right),{\rm{ }}{\mathop{\rm x}\nolimits} = \left( {\begin{aligned}{*{20}{c}}6\\{ - 2}\\3\end{aligned}} \right)\)

2. \({\mathop{\rm w}\nolimits} \cdot {\mathop{\rm w}\nolimits} ,{\mathop{\rm x}\nolimits} \cdot {\mathop{\rm w}\nolimits} ,\,\,{\mathop{\rm and}\nolimits} \,\,\frac{{{\mathop{\rm x}\nolimits} \cdot {\mathop{\rm w}\nolimits} }}{{{\mathop{\rm w}\nolimits} \cdot {\mathop{\rm w}\nolimits} }}\)

Let \({\bf{u}}\) be an eigenvector of \(A\) corresponding to an eigenvalue \(\lambda \), and let \(H\) be the line in \({\mathbb{R}^{\bf{n}}}\) through \({\bf{u}}\) and the origin.

  1. Explain why \(H\) is invariant under \(A\) in the sense that \(A{\bf{x}}\) is in \(H\) whenever \({\bf{x}}\) is in \(H\).
  2. Let \(K\) be a one-dimensional subspace of \({\mathbb{R}^{\bf{n}}}\) that is invariant under \(A\). Explain why \(K\) contains an eigenvector of \(A\).

For the matrices A in Exercises 1 through 12, find closed formulas for , where t is an arbitrary positive integer. Follow the strategy outlined in Theorem 7.4.2 and illustrated in Example 2. In Exercises 9 though 12, feel free to use technology.

1. A=1203

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