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Chapter 5: Q40E (page 267)

[M] In Exercises 37–40, use a matrix program to find the eigenvalues of the matrix. Then use the method of Example 4 with a row reduction routine to produce a basis for each eigenspace.

40. \(\left( {\begin{array}{*{20}{c}}{ - 4}&{ - 4}&{20}&{ - 8}&{ - 1}\\{14}&{12}&{46}&{18}&2\\6&4&{ - 18}&8&1\\{11}&7&{ - 37}&{17}&2\\{18}&{12}&{ - 60}&{24}&5\end{array}} \right)\)

Short Answer

Expert verified

The eigenvalues are \(\lambda = \left( {21.68, - 16.68,3,2,2} \right)\). The basis for eigenspace of \(\lambda = 21.68\) is \(N = \left\{ {\left( {\begin{array}{*{20}{c}}{ - 0.33}\\{2.39}\\{0.33}\\{0.583}\\{1.00}\end{array}} \right)} \right\}\). The basis for eigenspace of \(\lambda = - 16.68\) is \(N = \left\{ {\left( {\begin{array}{*{20}{c}}{ - 0.33}\\{ - 0.81}\\{0.33}\\{0.583}\\{1.00}\end{array}} \right)} \right\}\). The basis for eigenspace of \(\lambda = 2\) is \(N = \left\{ {\left( {\begin{array}{*{20}{c}}{ - 2}\\1\\0\\1\\0\end{array}} \right),\left( {\begin{array}{*{20}{c}}{ - 0.5}\\{0.5}\\0\\0\\1\end{array}} \right)} \right\}\). The basis for eigenspace of \(\lambda = 3\) is \(N = \left\{ {\left( {\begin{array}{*{20}{c}}0\\{ - 2}\\0\\1\\0\end{array}} \right)} \right\}\).

Step by step solution

01

Command for input the matrix in MATLAB

Write the command given below to input the matrix

\({\rm{ > > A = }}\left( {{\rm{ - 4 - 4 20}}\,{\rm{ - 8}}\,{\rm{ - 1 ; 14 12 46 }}\,{\rm{18 2; 6 4 - 18 8 1; 11 7 - 37 17 2; 18 12 - 60 24 5}}} \right)\)

02

Command for finding the eigenvalues of the matrix

\( > > {\rm{ ev}} = {\rm{eig}}\left( {\rm{A}} \right)\)

The output will be\({\rm{ev}} = \left( {21.68, - 16.68,3,2,2} \right)\).

Hence, the eigenvalues are \({\rm{ev}} = \left( {21.68, - 16.68,3,2,2} \right)\).

03

Command for finding the null basis for each eigenvalue

Write the command given below to find the null basis corresponding to \(\lambda = 21.68\):

\( > > {\rm{ N}} = {\rm{A}} - {\rm{ev}}\left( 1 \right)*{\rm{eye}}\left( 5 \right)\)

The output is given below:

\(N = \left( {\begin{array}{*{20}{c}}{ - 0.33}\\{2.39}\\{0.33}\\{0.583}\\{1.00}\end{array}} \right)\)

Hence, the basis for eigenspace of \(\lambda = 21.68\) is \(N = \left\{ {\left( {\begin{array}{*{20}{c}}{ - 0.33}\\{2.39}\\{0.33}\\{0.583}\\{1.00}\end{array}} \right)} \right\}\).

Write the command given below to find the null basis corresponding to \(\lambda = - 16.68\):

\( > > {\rm{ N}} = {\rm{A}} - {\rm{ev}}\left( 2 \right)*{\rm{eye}}\left( 5 \right)\)

The output is given below:

\(N = \left( {\begin{array}{*{20}{c}}{ - 0.33}\\{ - 0.81}\\{0.33}\\{0.583}\\{1.00}\end{array}} \right)\)

Hence, the basis for eigenspace of \(\lambda = - 16.68\) is \(N = \left\{ {\left( {\begin{array}{*{20}{c}}{ - 0.33}\\{ - 0.81}\\{0.33}\\{0.583}\\{1.00}\end{array}} \right)} \right\}\).

Write the command given below to find the null basis corresponding to \(\lambda = 3\):

\( > > {\rm{ N}} = {\rm{A}} - {\rm{ev}}\left( 3 \right)*{\rm{eye}}\left( 5 \right)\)

The output is given below:

\(N = \left( {\begin{array}{*{20}{c}}0\\{ - 2}\\0\\1\\0\end{array}} \right)\)

Hence, the basis for eigenspace of \(\lambda = 3\) is \(N = \left\{ {\left( {\begin{array}{*{20}{c}}0\\{ - 2}\\0\\1\\0\end{array}} \right)} \right\}\).

Write the command given below to find the null basis corresponding to \(\lambda = 2\):

\( > > {\rm{ N}} = {\rm{A}} - {\rm{ev}}\left( 4 \right)*{\rm{eye}}\left( 5 \right)\)

The output is given below:

\(N = \left( {\begin{array}{*{20}{c}}{ - 2}&{ - 0.5}\\1&{0.5}\\0&0\\1&0\\0&1\end{array}} \right)\)

Hence, the basis for eigenspace of \(\lambda = 2\) is \(N = \left\{ {\left( {\begin{array}{*{20}{c}}{ - 2}\\1\\0\\1\\0\end{array}} \right),\left( {\begin{array}{*{20}{c}}{ - 0.5}\\{0.5}\\0\\0\\1\end{array}} \right)} \right\}\).

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Most popular questions from this chapter

In Exercises 9–16, find a basis for the eigenspace corresponding to each listed eigenvalue.

10. \(A = \left( {\begin{array}{*{20}{c}}{10}&{ - 9}\\4&{ - 2}\end{array}} \right)\), \(\lambda = 4\)

Exercises 19–23 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 \(p\): \({C_p} = \left( {\begin{aligned}{*{20}{c}}{\bf{0}}&{\bf{1}}&{\bf{0}}&{...}&{\bf{0}}\\{\bf{0}}&{\bf{0}}&{\bf{1}}&{}&{\bf{0}}\\:&{}&{}&{}&:\\{\bf{0}}&{\bf{0}}&{\bf{0}}&{}&{\bf{1}}\\{ - {a_{\bf{0}}}}&{ - {a_{\bf{1}}}}&{ - {a_{\bf{2}}}}&{...}&{ - {a_{n - {\bf{1}}}}}\end{aligned}} \right)\).

19. Write the companion matrix \({C_p}\) for \(p\left( t \right) = {\bf{6}} - {\bf{5}}t + {t^{\bf{2}}}\), and then find the characteristic polynomial of \({C_p}\).

Define\(T:{{\rm P}_3} \to {\mathbb{R}^4}\)by\(T\left( {\bf{p}} \right) = \left( {\begin{aligned}{{\bf{p}}\left( { - 3} \right)}\\{{\bf{p}}\left( { - 1} \right)}\\{{\bf{p}}\left( 1 \right)}\\{{\bf{p}}\left( 3 \right)}\end{aligned}} \right)\).

  1. Show that \(T\) is a linear transformation.
  2. Find the matrix for \(T\) relative to the basis \(\left\{ {1,t,{t^2},{t^3}} \right\}\)for \({{\rm P}_3}\)and the standard basis for \({\mathbb{R}^4}\).

Consider the growth of a lilac bush. The state of this lilac bush for several years (at year’s end) is shown in the accompanying sketch. Let n(t) be the number of new branches (grown in the year t) and a(t) the number of old branches. In the sketch, the new branches are represented by shorter lines. Each old branch will grow two new branches in the following year. We assume that no branches ever die.

(a) Find the matrix A such that [nt+1at+1]=A[ntat]

(b) Verify that [11]and [2-1] are eigenvectors of A. Find the associated eigenvalues.

(c) Find closed formulas for n(t) and a(t).

Apply the results of Exercise \({\bf{15}}\) to find the eigenvalues of the matrices \(\left( {\begin{aligned}{*{20}{c}}{\bf{1}}&{\bf{2}}&{\bf{2}}\\{\bf{2}}&{\bf{1}}&{\bf{2}}\\{\bf{2}}&{\bf{2}}&{\bf{1}}\end{aligned}} \right)\) and \(\left( {\begin{aligned}{*{20}{c}}{\bf{7}}&{\bf{3}}&{\bf{3}}&{\bf{3}}&{\bf{3}}\\{\bf{3}}&{\bf{7}}&{\bf{3}}&{\bf{3}}&{\bf{3}}\\{\bf{3}}&{\bf{3}}&{\bf{7}}&{\bf{3}}&{\bf{3}}\\{\bf{3}}&{\bf{3}}&{\bf{3}}&{\bf{7}}&{\bf{3}}\\{\bf{3}}&{\bf{3}}&{\bf{3}}&{\bf{3}}&{\bf{7}}\end{aligned}} \right)\).
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