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Chapter 5: Q21E (page 267)

Question: In Exercises 21 and 22, A is an\(n \times n\)matrix. Mark each statement True or False. Justify each answer.

  1. If\(A{\bf{x}} = \lambda {\bf{x}}\)for some vector x, then\(\lambda \)is an eignvalue of A.
  2. A matrix A is not invertible if and only if 0 is an eigenvalue of A.
  3. A number of c is an eigenvalue of A if and only if the equation\(\left( {A - cI} \right){\bf{x}} = {\bf{0}}\)has a nontrivial solution.
  4. Finding an eigenvector of A may be difficult, but checking whether a given vector is in face an eigenvector is easy.
  5. To find the eigenvalues of A, reduce A to echelon form.

Short Answer

Expert verified

a. The given statement is False.

b. The given statement is True.

c. The given statement is True

d. The given statement is True.

e. The given statement is False.

Step by step solution

01

Find an answer for part (a)

The equation \(A{\bf{x}} = \lambda {\bf{x}}\) essentially has a nontrivial solution, then only \(\lambda \) will be the eigenvalue of A.

Thus, statement (a) is false.

02

Find an answer for part (b)

If A is invertible, then 0 is an eigenvalue of A.

Thus, statement (b) is true.

03

Find an answer for part (c)

If a scalar c is the eigenvalue of A, then;

\(\begin{array}{c}A{\bf{x}} = c{\bf{x}}\\\left( {A - c} \right){\bf{x}} = 0\end{array}\)

04

Find an answer for part (d)

To find an eigenvector of a matrix, first, it needs to determine the eigenvalue and then the eigenvector.

On the other hand, to check the eigenvector, we need to solve the equation \(A{\bf{x}} = \lambda {\bf{x}}\), where x is an eigenvector.

Thus, statement (d) is true.

05

Find the answer for part (e)

The characteristic equation of the matrix A is:

\(\det \left( {A - \lambda I} \right) = 0\)

The eigenvalues cannot be created by reducing in echelon form. The echelon form is used to determine the eigenvectors.

Thus, the statement (e) is false.

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Most popular questions from this chapter

Question: Let \(A = \left( {\begin{array}{*{20}{c}}a&b\\c&d\end{array}} \right)\). Use formula (1) for a determinant (given before Example 2) to show that \(\det A = ad - bc\). Consider two cases: \(a \ne 0\) and \(a = 0\).

Question: For the matrices in Exercises 15-17, list the eigenvalues, repeated according to their multiplicities.

15. \(\left[ {\begin{array}{*{20}{c}}4&- 7&0&2\\0&3&- 4&6\\0&0&3&{ - 8}\\0&0&0&1\end{array}} \right]\)

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.

Compute the quantities in Exercises 1-8 using the vectors

\({\mathop{\rm u}\nolimits} = \left( {\begin{array}{*{20}{c}}{ - 1}\\2\end{array}} \right),{\rm{ }}{\mathop{\rm v}\nolimits} = \left( {\begin{array}{*{20}{c}}4\\6\end{array}} \right),{\rm{ }}{\mathop{\rm w}\nolimits} = \left( {\begin{array}{*{20}{c}}3\\{ - 1}\\{ - 5}\end{array}} \right),{\rm{ }}{\mathop{\rm x}\nolimits} = \left( {\begin{array}{*{20}{c}}6\\{ - 2}\\3\end{array}} \right)\)

3. \(\frac{1}{{{\mathop{\rm w}\nolimits} \cdot {\mathop{\rm w}\nolimits} }}{\mathop{\rm w}\nolimits} \)

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.

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