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Chapter 5: Q4E (page 267)

Question: Is \(\left( {\begin{array}{*{20}{c}}{ - 1 + \sqrt 2 }\\1\end{array}} \right)\) an eigenvalue of \(\left( {\begin{array}{*{20}{c}}2&1\\1&4\end{array}} \right)\)? If so, find the eigenvalue.

Short Answer

Expert verified

Yes, \(\left( {\begin{array}{*{20}{c}}{ - 1 + \sqrt 2 }\\1\end{array}} \right)\) is the eigenvector of the given matrix \(\left( {\begin{array}{*{20}{c}}2&1\\1&4\end{array}} \right)\) and the eigenvalue is \(3 + \sqrt 2 \).

Step by step solution

01

Definition of eigenvector

If there exists a non-zero vector \({\bf{x}}\) which satisfies \(A{\bf{x}} = \lambda {\bf{x}}\) for some scaler \(\lambda \), then \({\bf{x}}\) be the eigenvector of an \(n \times n\) matrix \(A\), and if \(A{\bf{x}} = \lambda {\bf{x}}\) exists then scaler \(\lambda \) is the eigenvalue of the matrix.

02

Determine whether \(\left( {\begin{array}{*{20}{c}}{ - 1 + \sqrt 2 }\\1\end{array}} \right)\) is the eigenvector of given matrix

Denote the given matrix by \(A\) and the given vector by \({\bf{x}}\).

\(A = \left( {\begin{array}{*{20}{c}}2&1\\1&4\end{array}} \right)\), \({\bf{x}} = \left( {\begin{array}{*{20}{c}}{ - 1 + \sqrt 2 }\\1\end{array}} \right)\)

According to the definition of eigenvalue, \({\bf{x}} = \left( {\begin{array}{*{20}{c}}{ - 1 + \sqrt 2 }\\1\end{array}} \right)\) is the eigenvector of the matrix \(A\), if \(A{\bf{x}} = \lambda {\bf{x}}\).

Find the product of \(A\), and \({\bf{x}}\).

\(\begin{array}{c}A{\bf{x}} = \left( {\begin{array}{*{20}{c}}2&1\\1&4\end{array}} \right)\left( {\begin{array}{*{20}{c}}{ - 1 + \sqrt 2 }\\1\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}{2\left( { - 1 + \sqrt 2 } \right) + 1\left( 1 \right)}\\{1\left( { - 1 + \sqrt 2 } \right) + 4\left( 1 \right)}\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}{ - 1 + 2\sqrt 2 }\\{3 + \sqrt 2 }\end{array}} \right)\end{array}\)

The obtained matrix in the form of given vector can be written as:

\(A{\bf{x}} = \left( {3 + \sqrt 2 } \right) \cdot \left( {\begin{array}{*{20}{c}}{ - 1 + \sqrt 2 }\\1\end{array}} \right)\)

Because,

\(\begin{array}{c}\left( {\begin{array}{*{20}{c}}{\left( {3 + \sqrt 2 } \right)\left( { - 1 + \sqrt 2 } \right)}\\{\left( {3 + \sqrt 2 } \right)\left( 1 \right)}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}{ - 3 + {{\left( {\sqrt 2 } \right)}^2} + 2\sqrt 2 }\\{3 + \sqrt 2 }\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}{ - 1 + 2\sqrt 2 }\\{3 + \sqrt 2 }\end{array}} \right)\\ = A{\bf{x}}\end{array}\)

So, the vector \(\left( {\begin{array}{*{20}{c}}{ - 1 + \sqrt 2 }\\1\end{array}} \right)\) is the eigenvector of the given matrix \(\left( {\begin{array}{*{20}{c}}2&1\\1&4\end{array}} \right)\).

03

Determine the eigenvalue 

As the given vector satisfies the condition \(A{\bf{x}} = \lambda {\bf{x}}\). Which imply that \(\lambda \) is the eigenvalue of the given matrix. So, \(\lambda = 3 + \sqrt 2 \).

So, \(3 + \sqrt 2 \) is the eigenvalue of the given matrix \(\left( {\begin{array}{*{20}{c}}2&1\\1&4\end{array}} \right)\).

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Most popular questions from this chapter

Exercises 19–23 concern the polynomial \(p\left( t \right) = {a_{\bf{0}}} + {a_{\bf{1}}}t + ... + {a_{n - {\bf{1}}}}{t^{n - {\bf{1}}}} + {t^n}\) and \(n \times n\) matrix \({C_p}\) called the companion matrix of \(p\): \({C_p} = \left( {\begin{aligned}{*{20}{c}}{\bf{0}}&{\bf{1}}&{\bf{0}}&{...}&{\bf{0}}\\{\bf{0}}&{\bf{0}}&{\bf{1}}&{}&{\bf{0}}\\:&{}&{}&{}&:\\{\bf{0}}&{\bf{0}}&{\bf{0}}&{}&{\bf{1}}\\{ - {a_{\bf{0}}}}&{ - {a_{\bf{1}}}}&{ - {a_{\bf{2}}}}&{...}&{ - {a_{n - {\bf{1}}}}}\end{aligned}} \right)\).

19. Write the companion matrix \({C_p}\) for \(p\left( t \right) = {\bf{6}} - {\bf{5}}t + {t^{\bf{2}}}\), and then find the characteristic polynomial of \({C_p}\).

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

For the matrices A in Exercises 1 through 12, find closed formulas for , where t is an arbitrary positive integer. Follow the strategy outlined in Theorem 7.4.2 and illustrated in Example 2. In Exercises 9 though 12, feel free to use technology.

1.A=1203

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

Question: In Exercises \({\bf{1}}\) and \({\bf{2}}\), let \(A = PD{P^{ - {\bf{1}}}}\) and compute \({A^{\bf{4}}}\).

1. \(P{\bf{ = }}\left( {\begin{array}{*{20}{c}}{\bf{5}}&{\bf{7}}\\{\bf{2}}&{\bf{3}}\end{array}} \right)\), \(D{\bf{ = }}\left( {\begin{array}{*{20}{c}}{\bf{2}}&{\bf{0}}\\{\bf{0}}&{\bf{1}}\end{array}} \right)\)
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