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Chapter 5: Q5.2-13E (page 267)

Question: Exercises 9-14 require techniques section 3.1. Find the characteristic polynomial of each matrix, using either a cofactor expansion or the special formula for \(3 \times 3\) determinants described prior to Exercise 15-18 in Section 3.1. [Note: Finding the characteristic polynomial of a \(3 \times 3\) matrix is not easy to do with just row operations, because the variable \(\lambda \) is involved.

13. \(\left[ {\begin{array}{*{20}{c}}6&- 2&0\\- 2&9&0\\5&8&3\end{array}} \right]\)

Short Answer

Expert verified

The characteristic polynomial of the matrix is \( - {\lambda ^3} + 18{\lambda ^2} - 95\lambda + 150\).

Step by step solution

01

Definition of the characteristic polynomial

The eigenvalue of an \(n \times n\) matrix \(A\) is a scalar \(\lambda \) such that \(\lambda \) satisfies the characteristic equation \(\det \left( {A - \lambda I} \right) = 0\).

When \(A\) is an \(n \times n\) matrix, \(\det \left( {A - \lambda I} \right)\) is the characteristic polynomial of \(A\), which is the polynomial of degree \(n\).

02

Determine the characteristic polynomial of the matrix

Use the cofactor expression down the third column to obtain the characteristic polynomial of the matrix, as shown below.

\[\begin{array}\det \left( {A - \lambda I} \right) = \det \left[ {\begin{array}{*{20}{c}}{6 - \lambda }&{ - 2}&0\\{ - 2}&{9 - \lambda }&0\\5&8&{3 - \lambda }\end{array}} \right]\\ = \left( {3 - \lambda } \right)\det \left[ {\begin{array}{*{20}{c}}{6 - \lambda }&{ - 2}\\{ - 2}&{9 - \lambda }\end{array}} \right]\\ = \left( {3 - \lambda } \right)\left[ {\left( {6 - \lambda } \right)\left( {9 - \lambda } \right) - \left( { - 2} \right)\left( { - 2} \right)} \right]\\ = \left( {3 - \lambda } \right)\left( {{\lambda ^2} + - 15\lambda + 50} \right)\\ = - {\lambda ^3} + 18{\lambda ^2} - 95\lambda + 150\\ = \left( {3 - \lambda } \right)\left( {\lambda - 5} \right)\left( {\lambda - 10} \right)\end{array}\]

Thus, the characteristic polynomial of the matrix is \( - {\lambda ^3} + 18{\lambda ^2} - 95\lambda + 150\).

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Most popular questions from this chapter

In Exercises 9鈥�16, find a basis for the eigenspace corresponding to each listed eigenvalue.

10. \(A = \left( {\begin{array}{*{20}{c}}{10}&{ - 9}\\4&{ - 2}\end{array}} \right)\), \(\lambda = 4\)

Question 18: It can be shown that the algebraic multiplicity of an eigenvalue \(\lambda \) is always greater than or equal to the dimension of the eigenspace corresponding to \(\lambda \). Find \(h\) in the matrix \(A\) below such that the eigenspace for \(\lambda = 5\) is two-dimensional:

\[A = \left[ {\begin{array}{*{20}{c}}5&{ - 2}&6&{ - 1}\\0&3&h&0\\0&0&5&4\\0&0&0&1\end{array}} \right]\]

Question: Exercises 9-14 require techniques section 3.1. Find the characteristic polynomial of each matrix, using either a cofactor expansion or the special formula for \(3 \times 3\) determinants described prior to Exercise 15-18 in Section 3.1. [Note: Finding the characteristic polynomial of a \(3 \times 3\) matrix is not easy to do with just row operations, because the variable \(\lambda \) is involved.

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

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.

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.)
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