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Chapter 5: Q6E (page 267)

Let \(A = \left( {\begin{aligned}{ {20}{r}}{ - 2}&{ - 3}\\6&7\end{aligned}} \right)\). Repeat Exercise 5, using the following sequence \({\bf{x}},A{\bf{x}}, \ldots ,{A^5}{\bf{x}}\).

\(\left( {\begin{aligned}{ {20}{l}}1\\1\end{aligned}} \right),\left( {\begin{aligned}{ {20}{r}}{ - 5}\\{13}\end{aligned}} \right),\left( {\begin{aligned}{ {20}{r}}{ - 29}\\{61}\end{aligned}} \right),\left( {\begin{aligned}{ {20}{r}}{ - 125}\\{253}\end{aligned}} \right),\left( {\begin{aligned}{ {20}{r}}{ - 509}\\{1021}\end{aligned}} \right),\left( {\begin{aligned}{ {20}{r}}{ - 2045}\\{4093}\end{aligned}} \right)\)

Short Answer

Expert verified

The value is \(\lambda = 4.0024\).

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 \({A^5}x = \left( {\begin{aligned}{ {20}{c}}{ - 2045}\\{4093}\end{aligned}} \right)\) is:

\(\begin{aligned}{c}v = \frac{1}{{4093}}\left( {\begin{aligned}{ {20}{c}}{ - 2045}\\{4093}\end{aligned}} \right)\\ = \left( {\begin{aligned}{ {20}{c}}{ - .4996}\\1\end{aligned}} \right)\end{aligned}\)

Now, 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 \(Av\) :

\(\begin{aligned}{c}Av = \left( {\begin{aligned}{ {20}{c}}{ - 2}&{ - 3}\\6&7\end{aligned}} \right)\left( {\begin{aligned}{ {20}{c}}{ - .4996}\\1\end{aligned}} \right)\\ = \left( {\begin{aligned}{ {20}{c}}{ - 2.0008}\\{4.0024}\end{aligned}} \right)\end{aligned}\)

The largest entity is 4.0024. This means eigenvalue is \(\lambda = 4.0024\). The corresponding vector is \(v = \left( {\begin{aligned}{ {20}{c}}{ - .4996}\\1\end{aligned}} \right)\).

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Most popular questions from this chapter

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

7. \(\left( {\begin{array}{*{20}{c}}{\bf{1}}&{\bf{0}}\\{\bf{6}}&{{\bf{ - 1}}}\end{array}} \right)\)

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: Find the characteristic polynomial and the eigenvalues of the matrices in Exercises 1-8.

4. \(\left[ {\begin{array}{*{20}{c}}5&-3\\-4&3\end{array}} \right]\)

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.

14. \(\left[ {\begin{array}{*{20}{c}}5&- 2&3\\0&1&0\\6&7&- 2\end{array}} \right]\)

Let \(B = \left\{ {{{\bf{b}}_1},{{\bf{b}}_2},{{\bf{b}}_3}} \right\}\)be a basis for a vector space \(V\) and\(T:V \to {\mathbb{R}^2}\) be a linear transformation with the property that

\(T\left( {{x_1}{{\bf{b}}_1} + {x_2}{{\bf{b}}_2} + {x_3}{{\bf{b}}_3}} \right) = \left( {\begin{aligned}{2{x_1} - 4{x_2} + 5{x_3}}\\{ - {x_2} + 3{x_3}}\end{aligned}} \right)\)

Find the matrix for \(T\) relative to \(B\) and the standard basis for \({\mathbb{R}^2}\).
See all solutions

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