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Let \(A = \left( {\begin{aligned}{ {20}{r}}{15}&{16}\\{ - 20}&{ - 21}\end{aligned}} \right)\). The vectors \({\bf{x}}, \ldots ,{A^5}{\bf{x}}\) are \(\left( {\begin{aligned}{ {20}{l}}1\\1\end{aligned}} \right)\),

\(\left( {\begin{aligned}{ {20}{r}}{31}\\{ - 41}\end{aligned}} \right),{\rm{ }}\left( {\begin{aligned}{ {20}{r}}{ - 191}\\{241}\end{aligned}} \right),{\rm{ }}\left( {\begin{aligned}{ {20}{r}}{991}\\{ - 1241}\end{aligned}} \right),{\rm{ }}\left( {\begin{aligned}{ {20}{r}}{ - 4991}\\{6241}\end{aligned}} \right),{\rm{ }}\left( {\begin{aligned}{ {20}{r}}{24991}\\{ - 31241}\end{aligned}} \right)\).

Find a vector with a 1 in the second entry that is close to an eigenvector of \(A\). Use four decimal places. Check your estimate, and give an estimate for the dominant eigenvalue of \(A\).

Short Answer

Expert verified

The value is \(\lambda = - 5.0020\).

Step by step solution

01

Definition of Eigenvector

Eigenvectors, also known as characteristic vectors, appropriate vectors, or latent vectors, are a specific collection of vectors associated with a linear system of equations. Each eigenvector is associated with an eigenvalue.

02

Find the Eigenvalue

The normalized form of \({A^5}x = \left( {\begin{aligned}{ {20}{c}}{24991}\\{ - 31241}\end{aligned}} \right)\) is:

\(\begin{aligned}{c}v = - \frac{1}{{31241}}\left( {\begin{aligned}{ {20}{c}}{24991}\\{ - 31241}\end{aligned}} \right)\\ = \left( {\begin{aligned}{ {20}{c}}{ - .7999}\\1\end{aligned}} \right)\end{aligned}\)

This is a vector with 1 in a second entry that is an approximation of eigenvector of \(A\).

To estimate the eigenvalue of \(A\), compute \(Av\):

\(\begin{aligned}{c}Av = \left( {\begin{aligned}{ {20}{c}}{15}&{16}\\{ - 20}&{ - 21}\end{aligned}} \right)\left( {\begin{aligned}{ {20}{c}}{ - .7999}\\1\end{aligned}} \right)\\ = \left( {\begin{aligned}{ {20}{c}}{4.0015}\\{ - 5.002}\end{aligned}} \right)\end{aligned}\)

The largest entity is 5.002. This means eigenvalue is \(\lambda = - 5.002\). The corresponding vector is \(v = \left( {\begin{aligned}{ {20}{c}}{ - .7999}\\1\end{aligned}} \right)\).

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Most popular questions from this chapter

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 \(A\).

Question: Is \(\left( {\begin{array}{*{20}{c}}4\\{ - 3}\\1\end{array}} \right)\) an eigenvector of \(\left( {\begin{array}{*{20}{c}}3&7&9\\{ - 4}&{ - 5}&1\\2&4&4\end{array}} \right)\)? If so, find the eigenvalue.

Show that \(I - A\) is invertible when all the eigenvalues of \(A\) are less than 1 in magnitude. (Hint: What would be true if \(I - A\) were not invertible?)

Suppose \({\bf{x}}\) is an eigenvector of \(A\) corresponding to an eigenvalue \(\lambda \).

a. Show that \(x\) is an eigenvector of \(5I - A\). What is the corresponding eigenvalue?

b. Show that \(x\) is an eigenvector of \(5I - 3A + {A^2}\). What is the corresponding 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. \(\left( {\begin{array}{*{20}{c}}{ - 2}&{12}\\{ - 1}&5\end{array}} \right) = \left( {\begin{array}{*{20}{c}}3&4\\1&1\end{array}} \right)\left( {\begin{array}{*{20}{c}}2&0\\0&1\end{array}} \right)\left( {\begin{array}{*{20}{c}}{ - 1}&4\\1&{ - 3}\end{array}} \right)\)

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