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Chapter 5: Q1E (page 267)

Question: Is \(\lambda = 2\) an eigenvalue of \(\left( {\begin{array}{*{20}{c}}3&2\\3&8\end{array}} \right)\)? Why or why not?

Short Answer

Expert verified

Yes, \(\lambda = 2\) is the eigenvalue of the given matrix \(\left( {\begin{array}{*{20}{c}}3&2\\3&8\end{array}} \right)\), because there exists a nontrivial solution of \(A{\bf{x}} = 2{\bf{x}}\) and columns of the matrix \(\left( {A - 2I} \right)\) are linearly dependent.

Step by step solution

01

Definition of eigenvalue

Let \(\lambda \) is a scaler, \(A\) is an \(n \times n\) matrix and \({\bf{x}}\) is an eigenvector corresponding to \(\lambda \), \(\lambda \) is said to an eigenvalue of the matrix \(A\) if there exists a nontrivial solution \({\bf{x}}\) of \(A{\bf{x}} = \lambda {\bf{x}}\).

02

Determine whether \(\lambda  = 2\) is the eigenvalue of the given matrix

Denote the given matrix by \(A\).

\(A = \left( {\begin{array}{*{20}{c}}3&2\\3&8\end{array}} \right)\)

According to the definition of eigenvalue, \(\lambda = 2\) is the eigenvalue of the matrix \(A\), if satisfies the equation \(A{\bf{x}} = \lambda {\bf{x}}\).

Substitute \(\lambda = 2\) into \(A{\bf{x}} = \lambda {\bf{x}}\).

\(A{\bf{x}} = 2{\bf{x}}\)

The obtained equation is equivalent to \(\left( {A - 2I} \right){\bf{x}} = 0\), which is a homogeneous equation.

So, first, solve the matrix \(\left( {A - 2I} \right)\) by using \(A = \left( {\begin{array}{*{20}{c}}3&2\\3&8\end{array}} \right)\).

\(\begin{array}{c}\left( {A - 2I} \right) = \left( {\begin{array}{*{20}{c}}3&2\\3&8\end{array}} \right) - 2\left( {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}3&2\\3&8\end{array}} \right) - \left( {\begin{array}{*{20}{c}}2&0\\0&2\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}1&2\\3&6\end{array}} \right)\end{array}\)

It can be observed that the columns of the obtained matrix are linearly dependent, where elements of the second column twice the multiple of elements of the first column, which can be written as:

\(\left( {A - 2I} \right) = \left( {\begin{array}{*{20}{c}}1&{2\left( 1 \right)}\\3&{2\left( 3 \right)}\end{array}} \right)\)

Hence, \(\left( {A - 2I} \right){\bf{x}} = 0\) has a nontrivial solution, so \(\lambda = 2\) is an eigenvalue of the given matrix \(\left( {\begin{array}{*{20}{c}}3&2\\3&8\end{array}} \right)\).

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Most popular questions from this chapter

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

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

Question: Is \(\lambda = - 2\) an eigenvalue of \(\left( {\begin{array}{*{20}{c}}7&3\\3&{ - 1}\end{array}} \right)\)? Why or why not?

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

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

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 \(p\): \({C_p} = \left( {\begin{aligned}{*{20}{c}}{\bf{0}}&{\bf{1}}&{\bf{0}}&{...}&{\bf{0}}\\{\bf{0}}&{\bf{0}}&{\bf{1}}&{}&{\bf{0}}\\:&{}&{}&{}&:\\{\bf{0}}&{\bf{0}}&{\bf{0}}&{}&{\bf{1}}\\{ - {a_{\bf{0}}}}&{ - {a_{\bf{1}}}}&{ - {a_{\bf{2}}}}&{...}&{ - {a_{n - {\bf{1}}}}}\end{aligned}} \right)\).

23. Let \(p\) be the polynomial in Exercise \({\bf{22}}\), and suppose the equation \(p\left( t \right) = {\bf{0}}\) has distinct roots \({\lambda _{\bf{1}}},{\lambda _{\bf{2}}},{\lambda _{\bf{3}}}\). Let \(V\) be the Vandermonde matrix

\(V{\bf{ = }}\left( {\begin{aligned}{*{20}{c}}{\bf{1}}&{\bf{1}}&{\bf{1}}\\{{\lambda _{\bf{1}}}}&{{\lambda _{\bf{2}}}}&{{\lambda _{\bf{3}}}}\\{\lambda _{\bf{1}}^{\bf{2}}}&{\lambda _{\bf{2}}^{\bf{2}}}&{\lambda _{\bf{3}}^{\bf{2}}}\end{aligned}} \right)\)

(The transpose of \(V\) was considered in Supplementary Exercise \({\bf{11}}\) in Chapter \({\bf{2}}\).) Use Exercise \({\bf{22}}\) and a theorem from this chapter to deduce that \(V\) is invertible (but do not compute \({V^{{\bf{ - 1}}}}\)). Then explain why \({V^{{\bf{ - 1}}}}{C_p}V\) is a diagonal matrix.

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

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