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Chapter 5: Q5.2-14E (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.

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

Short Answer

Expert verified

The characteristic polynomial of the matrix is \( - {\lambda ^3} + 4{\lambda ^2} + 25\lambda - 28\).

Step by step solution

01

Definition of the characteristic polynomial

The eigen value 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 along the second row to obtain the characteristic polynomial of the matrix, as shown below.

\[\begin{array}\det \left( {A - \lambda I} \right) = \det \left[ {\begin{array}{*{20}{c}}{5 - \lambda }&{ - 2}&3\\0&{1 - \lambda }&0\\6&7&{ - 2 - \lambda }\end{array}} \right]\\ = \left( {1 - \lambda } \right)\det \left[ {\begin{array}{*{20}{c}}{5 - \lambda }&3\\6&{ - 2 - \lambda }\end{array}} \right]\\ = \left( {1 - \lambda } \right)\left[ {\left( {5 - \lambda } \right)\left( { - 2 - \lambda } \right) - \left( 3 \right)\left( 6 \right)} \right]\\ = \left( {1 - \lambda } \right)\left( {{\lambda ^2} - 3\lambda - 28} \right)\\ = - {\lambda ^3} + 4{\lambda ^2} + 25\lambda - 28\,\\ = \left( {1 - \lambda } \right)\left( {\lambda - 7} \right)\left( {\lambda + 4} \right)\end{array}\]

Thus, the characteristic polynomial of the matrix is \( - {\lambda ^3} + 4{\lambda ^2} + 25\lambda - 28\).

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Most popular questions from this chapter

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

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

Let \({\bf{u}}\) be an eigenvector of \(A\) corresponding to an eigenvalue \(\lambda \), and let \(H\) be the line in \({\mathbb{R}^{\bf{n}}}\) through \({\bf{u}}\) and the origin.

  1. Explain why \(H\) is invariant under \(A\) in the sense that \(A{\bf{x}}\) is in \(H\) whenever \({\bf{x}}\) is in \(H\).
  2. Let \(K\) be a one-dimensional subspace of \({\mathbb{R}^{\bf{n}}}\) that is invariant under \(A\). Explain why \(K\) contains an eigenvector of \(A\).

Suppose \(A\) is diagonalizable and \(p\left( t \right)\) is the characteristic polynomial of \(A\). Define \(p\left( A \right)\) as in Exercise 5, and show that \(p\left( A \right)\) is the zero matrix. This fact, which is also true for any square matrix, is called the Cayley-Hamilton theorem.

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