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Chapter 5: Q13SE (page 267)

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

13. \(A = \left( {\begin{array}{*{20}{c}}3&{ - 2}&8\\0&5&{ - 2}\\0&{ - 4}&3\end{array}} \right)\)

Short Answer

Expert verified

The eigenvalues of \(A\) are \(1\), \(7\), \(3\).

Step by step solution

01

Step 1: Write the echelon form

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

Then we have,

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

02

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

Find the determinant of\(V - \lambda I\).

\(\begin{aligned}{c}\det \left( {\begin{aligned}{*{20}{c}}{5 - \lambda }&{ - 2}\\{ - 4}&{3 - \lambda }\end{aligned}} \right) &= \left( {5 - \lambda } \right)\left( {3 - \lambda } \right) - 8\\ &= 15 - 5\lambda - 3\lambda + {\lambda ^2} - 8\\ &= {\lambda ^2} - 8\lambda + 7\\ &= \left( {\lambda - 1} \right)\left( {\lambda - 7} \right)\end{aligned}\)

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

\(A = \left( {\lambda - 1} \right)\left( {\lambda - 7} \right)\left( {\lambda - 3} \right)\)

Thus, the eigenvalues of \(A\) are \(1\), \(7\), \(3\).

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Most popular questions from this chapter

Question: Let \(A = \left( {\begin{array}{*{20}{c}}{.6}&{.3}\\{.4}&{.7}\end{array}} \right)\), \({v_1} = \left( {\begin{array}{*{20}{c}}{3/7}\\{4/7}\end{array}} \right)\), \({x_0} = \left( {\begin{array}{*{20}{c}}{.5}\\{.5}\end{array}} \right)\). (Note: \(A\) is the stochastic matrix studied in Example 5 of Section 4.9.)

  1. Find a basic for \({\mathbb{R}^2}\) consisting of \({{\rm{v}}_1}\) and anther eigenvector \({{\rm{v}}_2}\) of \(A\).
  2. Verify that \({{\rm{x}}_0}\) may be written in the form \({{\rm{x}}_0} = {{\rm{v}}_1} + c{{\rm{v}}_2}\).
  3. For \(k = 1,2, \ldots \), define \({x_k} = {A^k}{x_0}\). Compute \({x_1}\) and \({x_2}\), and write a formula for \({x_k}\). Then show that \({{\bf{x}}_k} \to {{\bf{v}}_1}\) as \(k\) increases.

[M]Repeat Exercise 25 for \[A{\bf{ = }}\left[ {\begin{array}{*{20}{c}}{{\bf{ - 8}}}&{\bf{5}}&{{\bf{ - 2}}}&{\bf{0}}\\{{\bf{ - 5}}}&{\bf{2}}&{\bf{1}}&{{\bf{ - 2}}}\\{{\bf{10}}}&{{\bf{ - 8}}}&{\bf{6}}&{{\bf{ - 3}}}\\{\bf{3}}&{{\bf{ - 2}}}&{\bf{1}}&{\bf{0}}\end{array}} \right]\].

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 \({\mathbb{R}^2}\).

Suppose \(A = PD{P^{ - 1}}\), where \(P\) is \(2 \times 2\) and \(D = \left( {\begin{array}{*{20}{l}}2&0\\0&7\end{array}} \right)\)

a. Let \(B = 5I - 3A + {A^2}\). Show that \(B\) is diagonalizable by finding a suitable factorization of \(B\).

b. Given \(p\left( t \right)\) and \(p\left( A \right)\) as in Exercise 5 , show that \(p\left( A \right)\) is diagonalizable.

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