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Chapter 5: Q40E (page 267)

[M] In Exercises 37鈥�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

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

9. \(\left( {\begin{array}{*{20}{c}}3&{ - 1}\\1&5\end{array}} \right)\)

Question 19: Let \(A\) be an \(n \times n\) matrix, and suppose A has \(n\) real eigenvalues, \({\lambda _1},...,{\lambda _n}\), repeated according to multiplicities, so that \(\det \left( {A - \lambda I} \right) = \left( {{\lambda _1} - \lambda } \right)\left( {{\lambda _2} - \lambda } \right) \ldots \left( {{\lambda _n} - \lambda } \right)\) . Explain why \(\det A\) is the product of the n eigenvalues of A. (This result is true for any square matrix when complex eigenvalues are considered.)

Suppose \(A\) is diagonalizable and \(p\left( t \right)\) is the characteristic polynomial of \(A\). Define \(p\left( A \right)\) as in Exercise 5, and show that \(p\left( A \right)\) is the zero matrix. This fact, which is also true for any square matrix, is called the Cayley-Hamilton theorem.

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

8. \(\left( {\begin{array}{*{20}{c}}{\bf{5}}&{\bf{1}}\\{\bf{0}}&{\bf{5}}\end{array}} \right)\)

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 \({\mathbb{R}^4}\).
See all solutions

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