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Expert-verified Found in: Page 267 ### 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 # 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

## 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 $$k \to \infty$$. ### Want to see more solutions like these? 