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Q21E

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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

A common misconception is that if \(A\) has a strictly dominant eigenvalue, then, for any sufficiently large value of \(k\), the vector \({A^k}{\bf{x}}\) is approximately equal to an eigenvector of \(A\). For the three matrices below, study what happens to \({A^k}{\bf{x}}\) when \({\bf{x = }}\left( {{\bf{.5,}}{\bf{.5}}} \right)\), and try to draw general conclusions (for a \({\bf{2 \times 2}}\) matrix).

a. \(A{\bf{ = }}\left( {\begin{aligned}{ {20}{c}}{{\bf{.8}}}&{\bf{0}}\\{\bf{0}}&{{\bf{.2}}}\end{aligned}} \right)\) b. \(A{\bf{ = }}\left( {\begin{aligned}{ {20}{c}}{\bf{1}}&{\bf{0}}\\{\bf{0}}&{{\bf{.8}}}\end{aligned}} \right)\) c. \(A{\bf{ = }}\left( {\begin{aligned}{ {20}{c}}{\bf{8}}&{\bf{0}}\\{\bf{0}}&{\bf{2}}\end{aligned}} \right)\)

  1. Conclusion: If \({\bf{x}} \ne 0\) and all the eigenvalues of the matrix \(A\) are less than \(1\) in magnitude then \({A^k}{\bf{x}}\) is approximately an eigenvector for the larger \(k\).
  2. Conclusion: If the strictly dominant eigenvalue of \(A\) is \(1\), and if \(x\) has a component in the direction of the corresponding eigenvector, then \({A^k}{\bf{x}}\) will converge to a multiple of that eigenvector.
  3. Conclusion: If the eigenvalues of \(A\) are all greater than \(1\) in magnitude, and if \(x\) is not an eigenvector, then the distance from \({A^k}{\bf{x}}\) to the nearest eigenvector will increase as \(k \to \infty \).
See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

Write about what happened to \({A^k}{\bf{x}}\)

Consider \(A = \left( {\begin{aligned}{ {20}{c}}{.8}&0\\0&{.2}\end{aligned}} \right)\), \({\bf{x}} = \left( {\begin{aligned}{ {20}{c}}{.5}\\{.5}\end{aligned}} \right)\)

Now the sequence for \({A^k}{\bf{x}}\), \(k = 1,...5\).

Therefore,

\(\left( {\begin{aligned}{ {20}{c}}{.4}\\{.1}\end{aligned}} \right),\left( {\begin{aligned}{ {20}{c}}{.32}\\{.02}\end{aligned}} \right),\left( {\begin{aligned}{ {20}{c}}{.256}\\{.004}\end{aligned}} \right),\left( {\begin{aligned}{ {20}{c}}{.2048}\\{.0008}\end{aligned}} \right),\left( {\begin{aligned}{ {20}{c}}{.16384}\\{.00016}\end{aligned}} \right)\)

From the above conclusion, \({A^k}{\bf{x}}\) is approximate \(.8\).

Conclusion: If \({\bf{x}} \ne 0\) and all the eigenvalues of the matrix \(A\) are less than \(1\) in magnitude then \({A^k}{\bf{x}}\) is approximately an eigenvector for the larger \(k\).

Write about what happened to \({A^k}{\bf{x}}\)

Consider \(A = \left( {\begin{aligned}{ {20}{c}}1&0\\0&{.8}\end{aligned}} \right)\), \({\bf{x}} = \left( {\begin{aligned}{ {20}{c}}{.5}\\{.5}\end{aligned}} \right)\)

Now the sequence for \({A^k}{\bf{x}}\), \(k = 1,...5\).

Therefore,

\(\left( {\begin{aligned}{ {20}{c}}{.5}\\{.4}\end{aligned}} \right),\left( {\begin{aligned}{ {20}{c}}{.5}\\{.32}\end{aligned}} \right),\left( {\begin{aligned}{ {20}{c}}{.5}\\{.256}\end{aligned}} \right),\left( {\begin{aligned}{ {20}{c}}{.5}\\{.2048}\end{aligned}} \right),\left( {\begin{aligned}{ {20}{c}}{.5}\\{.16384}\end{aligned}} \right)\)

From the above conclusion \({A^k}{\bf{x}}\) is approximate \(\left( {\begin{aligned}{ {20}{c}}{.5}\\0\end{aligned}} \right)\).

Conclusion: If the strictly dominant eigenvalue of \(A\) is \(1\), and if \(x\) has a component in the direction of the corresponding eigenvector, then \({A^k}{\bf{x}}\) will converge to a multiple of that eigenvector.

Write about what happened to \({A^k}{\bf{x}}\)

Consider \(A = \left( {\begin{aligned}{ {20}{c}}8&0\\0&2\end{aligned}} \right)\), \({\bf{x}} = \left( {\begin{aligned}{ {20}{c}}{.5}\\{.5}\end{aligned}} \right)\)

Now the sequence for \({A^k}{\bf{x}}\), \(k = 1,...5\).

Therefore,

\(\left( {\begin{aligned}{ {20}{c}}4\\1\end{aligned}} \right),\left( {\begin{aligned}{ {20}{c}}{32}\\2\end{aligned}} \right),\left( {\begin{aligned}{ {20}{c}}{256}\\4\end{aligned}} \right),\left( {\begin{aligned}{ {20}{c}}{2048}\\8\end{aligned}} \right),\left( {\begin{aligned}{ {20}{c}}{16384}\\{16}\end{aligned}} \right)\)

From the above conclusion distance of \({A^k}{\bf{x}}\) distance of from either eigenvector of \(A\) is increasing rapidly as \(k\) increases

Conclusion: If the eigenvalues of \(A\) are all greater than \(1\) in magnitude, and if \(x\) is not an eigenvector, then the distance from \({A^k}{\bf{x}}\) to the nearest eigenvector will increase as \(k \to \infty \).

Most popular questions for Math Textbooks

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

(M) Use a matrix program to diagonalize

\(A = \left( {\begin{aligned}{*{20}{c}}{ - 3}&{ - 2}&0\\{14}&7&{ - 1}\\{ - 6}&{ - 3}&1\end{aligned}} \right)\)

If possible. Use the eigenvalue command to create the diagonal matrix \(D\). If the program has a command that produces eigenvectors, use it to create an invertible matrix \(P\). Then compute \(AP - PD\) and \(PD{P^{{\bf{ - 1}}}}\). Discuss your results.

Question: For the matrices in Exercises 15-17, list the eigenvalues, repeated according to their multiplicities.

17. \(\left[ {\begin{array}{*{20}{c}}3&0&0&0&0\\- 5&1&0&0&0\\3&8&0&0&0\\0&- 7&2&1&0\\- 4&1&9&- 2&3\end{array}} \right]\)

Let \(A{\bf{ = }}\left( {\begin{aligned}{*{20}{c}}{{a_{{\bf{11}}}}}&{{a_{{\bf{12}}}}}\\{{a_{{\bf{21}}}}}&{{a_{{\bf{22}}}}}\end{aligned}} \right)\). Recall from Exercise \({\bf{25}}\) in Section \({\bf{5}}{\bf{.4}}\) that \({\rm{tr}}\;A\) (the trace of \(A\)) is the sum of the diagonal entries in \(A\). Show that the characteristic polynomial of \(A\) is \({\lambda ^2} - \left( {{\rm{tr}}A} \right)\lambda + \det A\). Then show that the eigenvalues of a \({\bf{2 \times 2}}\) matrix \(A\) are both real if and only if \(\det A \le {\left( {\frac{{{\rm{tr}}A}}{2}} \right)^2}\).

a. Let \(A\) be a diagonalizable \(n \times n\) matrix. Show that if the multiplicity of an eigenvalue \(\lambda \) is \(n\), then \(A = \lambda I\).

b. Use part (a) to show that the matrix \(A =\left({\begin{aligned}{*{20}{l}}3&1\\0&3\end{aligned}}\right)\) is not diagonalizable.
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