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Chapter 5: Q14SE (page 267)

Use Exercise 12 to find the eigenvalues of the matrices in Exercises 13 and 14.

14. \(A{\bf{ = }}\left( {\begin{array}{*{20}{c}}{\bf{1}}&{\bf{5}}&{{\bf{ - 6}}}&{{\bf{ - 7}}}\\{\bf{2}}&{\bf{4}}&{\bf{5}}&{\bf{2}}\\{\bf{0}}&{\bf{0}}&{{\bf{ - 7}}}&{{\bf{ - 4}}}\\{\bf{0}}&{\bf{0}}&{\bf{3}}&{\bf{1}}\end{array}} \right)\)

Short Answer

Expert verified

The eigenvalues of \(A\) are \( - 1\), \( - 1\), \( - 5\), \(6\).

Step by step solution

01

Step 1: Find the characteristic polynomial of \(U\)

Assume \(G = \left( {\begin{array}{*{20}{c}}U&X\\0&V\end{array}} \right)\).

Then we have,

\(\begin{aligned}{c}A &= \left( {\begin{aligned}{*{20}{c}}1&5&{ - 6}&{ - 7}\\2&4&5&2\\0&0&{ - 7}&{ - 4}\\0&0&3&1\end{aligned}} \right)\\ &= \left( {\begin{aligned}{*{20}{c}}U&X\\0&V\end{aligned}} \right)\end{aligned}\)

On comparison we get,

\(U = \left( {\begin{aligned}{*{20}{c}}1&5\\2&4\end{aligned}} \right)\),\(V = \left( {\begin{aligned}{*{20}{c}}{ - 7}&{ - 4}\\3&1\end{aligned}} \right)\)

Now characteristic polynomial of \(U\) are as shown below:

\(\begin{aligned}{c}\det \left( {\begin{aligned}{*{20}{c}}{1 - \lambda }&5\\2&{4 - \lambda }\end{aligned}} \right) &= \left( {1 - \lambda } \right)\left( {4 - \lambda } \right) - 10\\ &= {\lambda ^2} - 5\lambda - 6\\ &= \left( {\lambda + 1} \right)\left( {\lambda - 6} \right)\end{aligned}\)

02

Step 2: Find the characteristic polynomial of \(V\)

\(\begin{aligned}{c}\det \left( {\begin{aligned}{*{20}{c}}{ - 7 - \lambda }&{ - 4}\\3&{1 - \lambda }\end{aligned}} \right) &= \left( { - 7 - \lambda } \right)\left( {1 - \lambda } \right) + 12\\ &= {\lambda ^2} + 6\lambda + 5\\ &= \left( {\lambda + 1} \right)\left( {\lambda + 5} \right)\end{aligned}\)

On multiplication we get the characteristic polynomial of \(A\).

\(\begin{aligned}{c}A &= \left( {\lambda + 1} \right)\left( {\lambda - 6} \right)\left( {\lambda + 1} \right)\left( {\lambda + 5} \right)\\ &= {\left( {\lambda + 1} \right)^2}\left( {\lambda + 5} \right)\left( {\lambda - 6} \right)\end{aligned}\)

Thus, the eigenvalues of \(A\) are \( - 1\), \( - 1\), \( - 5\), \(6\).

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Most popular questions from this chapter

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.

5. \(\left( {\begin{array}{*{20}{c}}2&2&1\\1&3&1\\1&2&2\end{array}} \right) = \left( {\begin{array}{*{20}{c}}1&1&2\\1&0&{ - 1}\\1&{ - 1}&0\end{array}} \right)\left( {\begin{array}{*{20}{c}}5&0&0\\0&1&0\\0&0&1\end{array}} \right)\left( {\begin{array}{*{20}{c}}{\frac{1}{4}}&{\frac{1}{2}}&{\frac{1}{4}}\\{\frac{1}{4}}&{\frac{1}{2}}&{ - \frac{3}{4}}\\{\frac{1}{4}}&{ - \frac{1}{2}}&{\frac{1}{4}}\end{array}} \right)\)

(M)The MATLAB command roots\(\left( p \right)\) computes the roots of the polynomial equation \(p\left( t \right) = {\bf{0}}\). Read a MATLAB manual, and then describe the basic idea behind the algorithm for the roots command.

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

10. \(\left( {\begin{array}{*{20}{c}}{\bf{2}}&{\bf{3}}\\{\bf{4}}&{\bf{1}}\end{array}} \right)\)

Question: Let \(A = \left( {\begin{array}{*{20}{c}}{.5}&{.2}&{.3}\\{.3}&{.8}&{.3}\\{.2}&0&{.4}\end{array}} \right)\), \({{\rm{v}}_1} = \left( {\begin{array}{*{20}{c}}{.3}\\{.6}\\{.1}\end{array}} \right)\), \({{\rm{v}}_2} = \left( {\begin{array}{*{20}{c}}1\\{ - 3}\\2\end{array}} \right)\), \({{\rm{v}}_3} = \left( {\begin{array}{*{20}{c}}{ - 1}\\0\\1\end{array}} \right)\) and \({\rm{w}} = \left( {\begin{array}{*{20}{c}}1\\1\\1\end{array}} \right)\).

  1. Show that \({{\rm{v}}_1}\), \({{\rm{v}}_2}\), and \({{\rm{v}}_3}\) are eigenvectors of \(A\). (Note: \(A\) is the stochastic matrix studied in Example 3 of Section 4.9.)
  2. Let \({{\rm{x}}_0}\) be any vector in \({\mathbb{R}^3}\) with non-negative entries whose sum is 1. (In section 4.9, \({{\rm{x}}_0}\) was called a probability vector.) Explain why there are constants \({c_1}\), \({c_2}\), and \({c_3}\) such that \({{\rm{x}}_0} = {c_1}{{\rm{v}}_1} + {c_2}{{\rm{v}}_2} + {c_3}{{\rm{v}}_3}\). Compute \({{\rm{w}}^T}{{\rm{x}}_0}\), and deduce that \({c_1} = 1\).
  3. For \(k = 1,2, \ldots ,\) define \({{\rm{x}}_k} = {A^k}{{\rm{x}}_0}\), with \({{\rm{x}}_0}\) as in part (b). Show that \({{\rm{x}}_k} \to {{\rm{v}}_1}\) as \(k\) increases.

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

12. \(\left( {\begin{array}{*{20}{c}}{\bf{4}}&{\bf{2}}&{\bf{2}}\\{\bf{2}}&{\bf{4}}&{\bf{2}}\\{\bf{2}}&{\bf{2}}&{\bf{4}}\end{array}} \right)\)
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