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Chapter 5: Q33E (page 267)

Question: Let \({\rm{u}}\) and \({\rm{v}}\) be the eigenvectors of a matrix \(A\), with corresponding eigenvalues \(\lambda \) and \(\mu \), and let \({c_1}\) and \({c_2}\) be scalars. Define \({{\rm{x}}_k} = {c_1}{\lambda ^k}{\rm{u}} + {c_2}{\mu ^k}{\rm{v}}\,\,\,\,\,\,\left( {k = 0,1,2,...} \right)\).

  1. What is \({{\rm{x}}_{k + 1}}\), by definition?
  2. Compute \({\rm{A}}{{\rm{x}}_k}\) from the formula for \({{\rm{x}}_k}\), and show that \(A{{\rm{x}}_k} = {{\rm{x}}_{k + 1}}\). This calculation will prove that the sequence \(\left\{ {{{\rm{x}}_k}} \right\}\) defined above satisfies the difference equation \({{\rm{x}}_{k + 1}} = A{{\rm{x}}_k}\,\,\,\,\left( {k = 0,1,2,...} \right)\).

Short Answer

Expert verified
  1. \({x_{k + 1}} = {c_1}{\lambda ^{k + 1}}u + {c_2}{\mu ^{k + 1}}v\)
  2. \(A{x_k} = {x_{k + 1}}\)

Step by step solution

01

The value of \({{\rm{x}}_{k + 1}}\)

(a)

Substitute \(k = k + 1\) in the formula of \({x_k} = {c_1}{\lambda ^k}u + {c_2}{\mu ^k}v\) as follows:

\({x_{k + 1}} = {c_1}{\lambda ^{k + 1}}u + {c_2}{\mu ^{k + 1}}v\)

Thus, \({x_{k + 1}} = {c_1}{\lambda ^{k + 1}}u + {c_2}{\mu ^{k + 1}}v\).

02

Computation of \(A{x_k}\)

(b)

Substitute \({x_k} = {c_1}{\lambda ^k}u + {c_2}{\mu ^k}v\) in \(A{x_k}\) and solve as follows:

\(\begin{gathered} A{x_k} = A\left( {{c_1}{\lambda ^k}u + {c_2}{\mu ^k}v} \right) \\ = {c_1}{\lambda ^k}Au + {c_2}{\mu ^k}Av\,\,\,{\text{(}}\because {\text{by linearity)}} \\ = {c_1}{\lambda ^k}\lambda u + {c_2}{\mu ^k}\mu v\,\,{\text{(}}\because u{\text{ and }}v{\text{ are eigenvectors)}} \\ = {c_1}{\lambda ^{k + 1}}u + {c_2}{\mu ^{k + 1}}v \\ = {x_{k + 1}} \\ \end{gathered} \)

Hence, \(A{x_k} = {x_{k + 1}}\).

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

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

Show that if \({\bf{x}}\) is an eigenvector of the matrix product \(AB\) and \(B{\rm{x}} \ne 0\), then \(B{\rm{x}}\) is an eigenvector of\(BA\).

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

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

M] In Exercises 19 and 20, find (a) the largest eigenvalue and (b) the eigenvalue closest to zero. In each case, set \[{{\bf{x}}_{\bf{0}}}{\bf{ = }}\left( {{\bf{1,0,0,0}}} \right)\] and carry out approximations until the approximating sequence seems accurate to four decimal places. Include the approximate eigenvector.

19.\[A{\bf{=}}\left[{\begin{array}{*{20}{c}}{{\bf{10}}}&{\bf{7}}&{\bf{8}}&{\bf{7}}\\{\bf{7}}&{\bf{5}}&{\bf{6}}&{\bf{5}}\\{\bf{8}}&{\bf{6}}&{{\bf{10}}}&{\bf{9}}\\{\bf{7}}&{\bf{5}}&{\bf{9}}&{{\bf{10}}}\end{array}} \right]\]

Question: In Exercises 21 and 22, \(A\) and \(B\) are \(n \times n\) matrices. Mark each statement True or False. Justify each answer.

  1. The determinant of \(A\) is the product of the diagonal entries in \(A\).
  2. An elementary row operation on \(A\) does not change the determinant.
  3. \(\left( {\det A} \right)\left( {\det B} \right) = \det AB\)
  4. If \(\lambda + 5\) is a factor of the characteristic polynomial of \(A\), then 5 is an eigenvalue of \(A\).
See all solutions

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