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Chapter 5: Q5E (page 267)

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

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

15. \(\left( {\begin{array}{*{20}{c}}{\bf{7}}&{\bf{4}}&{{\bf{16}}}\\{\bf{2}}&{\bf{5}}&{\bf{8}}\\{{\bf{ - 2}}}&{{\bf{ - 2}}}&{{\bf{ - 5}}}\end{array}} \right)\)

Question: In Exercises \({\bf{5}}\) and \({\bf{6}}\), the matrix \(A\) is factored in the form \(PD{P^{ - {\bf{1}}}}\). Use the Diagonalization Theorem to find the eigenvalues of \(A\) and a basis for each eigenspace.

6. \(\left( {\begin{array}{*{20}{c}}{\bf{4}}&{\bf{0}}&{{\bf{ - 2}}}\\{\bf{2}}&{\bf{5}}&{\bf{4}}\\{\bf{0}}&{\bf{0}}&{\bf{5}}\end{array}} \right){\bf{ = }}\left( {\begin{array}{*{20}{c}}{{\bf{ - 2}}}&{\bf{0}}&{{\bf{ - 1}}}\\{\bf{0}}&{\bf{1}}&{\bf{2}}\\{\bf{1}}&{\bf{0}}&{\bf{0}}\end{array}} \right)\left( {\begin{array}{*{20}{c}}{\bf{5}}&{\bf{0}}&{\bf{0}}\\{\bf{0}}&{\bf{5}}&{\bf{0}}\\{\bf{0}}&{\bf{0}}&4\end{array}} \right)\left( {\begin{array}{*{20}{c}}{\bf{0}}&{\bf{0}}&{\bf{1}}\\{\bf{2}}&{\bf{1}}&{\bf{4}}\\{{\bf{ - 1}}}&{\bf{0}}&{{\bf{ - 2}}}\end{array}} \right)\)


For the matrix A,find real closed formulas for the trajectory x鈫�(t+1)=Ax炉(t)wherex鈫�=[01]. Draw a rough sketchA=[-0.51.5-0.61.3]

Question 18: It can be shown that the algebraic multiplicity of an eigenvalue \(\lambda \) is always greater than or equal to the dimension of the eigenspace corresponding to \(\lambda \). Find \(h\) in the matrix \(A\) below such that the eigenspace for \(\lambda = 5\) is two-dimensional:

\[A = \left[ {\begin{array}{*{20}{c}}5&{ - 2}&6&{ - 1}\\0&3&h&0\\0&0&5&4\\0&0&0&1\end{array}} \right]\]

Exercises 19鈥�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 \(p\): \({C_p} = \left( {\begin{aligned}{*{20}{c}}{\bf{0}}&{\bf{1}}&{\bf{0}}&{...}&{\bf{0}}\\{\bf{0}}&{\bf{0}}&{\bf{1}}&{}&{\bf{0}}\\:&{}&{}&{}&:\\{\bf{0}}&{\bf{0}}&{\bf{0}}&{}&{\bf{1}}\\{ - {a_{\bf{0}}}}&{ - {a_{\bf{1}}}}&{ - {a_{\bf{2}}}}&{...}&{ - {a_{n - {\bf{1}}}}}\end{aligned}} \right)\).

21. Use mathematical induction to prove that for \(n \ge {\bf{2}}\),\(\begin{aligned}{c}det\left( {{C_p} - \lambda I} \right) = {\left( { - {\bf{1}}} \right)^n}\left( {{a_{\bf{0}}} + {a_{\bf{1}}}\lambda + ... + {a_{n - {\bf{1}}}}{\lambda ^{n - {\bf{1}}}} + {\lambda ^n}} \right)\\ = {\left( { - {\bf{1}}} \right)^n}p\left( \lambda \right)\end{aligned}\)

(Hint: Expanding by cofactors down the first column, show that \(det\left( {{C_p} - \lambda I} \right)\) has the form \(\left( { - \lambda B} \right) + {\left( { - {\bf{1}}} \right)^n}{a_{\bf{0}}}\) where \(B\) is a certain polynomial (by the induction assumption).)
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