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Chapter 5: Q2E (page 267)

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

Short Answer

Expert verified

Yes, \(\lambda = - 2\) is the eigenvalue of the given matrix \(\left( {\begin{array}{*{20}{c}}7&3\\3&{ - 1}\end{array}} \right)\), 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 given matrix

Denote the given matrix by \(A\).

\(A = \left( {\begin{array}{*{20}{c}}7&3\\3&{ - 1}\end{array}} \right)\)

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 \(A = \left( {\begin{array}{*{20}{c}}7&3\\3&{ - 1}\end{array}} \right)\).

\(\begin{array}{c}\left( {A + 2I} \right) = \left( {\begin{array}{*{20}{c}}7&3\\3&{ - 1}\end{array}} \right) + 2\left( {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}7&3\\3&{ - 1}\end{array}} \right) + \left( {\begin{array}{*{20}{c}}2&0\\0&2\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}9&3\\3&1\end{array}} \right)\end{array}\)

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

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

Hence, \(\left( {A + 2I} \right){\bf{x}} = 0\) has a nontrivial solution, so \(\lambda = - 2\) is an eigenvalue of the given matrix \(\left( {\begin{array}{*{20}{c}}7&3\\3&{ - 1}\end{array}} \right)\).

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Most popular questions from this chapter

Apply the results of Exercise \({\bf{15}}\) to find the eigenvalues of the matrices \(\left( {\begin{aligned}{*{20}{c}}{\bf{1}}&{\bf{2}}&{\bf{2}}\\{\bf{2}}&{\bf{1}}&{\bf{2}}\\{\bf{2}}&{\bf{2}}&{\bf{1}}\end{aligned}} \right)\) and \(\left( {\begin{aligned}{*{20}{c}}{\bf{7}}&{\bf{3}}&{\bf{3}}&{\bf{3}}&{\bf{3}}\\{\bf{3}}&{\bf{7}}&{\bf{3}}&{\bf{3}}&{\bf{3}}\\{\bf{3}}&{\bf{3}}&{\bf{7}}&{\bf{3}}&{\bf{3}}\\{\bf{3}}&{\bf{3}}&{\bf{3}}&{\bf{7}}&{\bf{3}}\\{\bf{3}}&{\bf{3}}&{\bf{3}}&{\bf{3}}&{\bf{7}}\end{aligned}} \right)\).

Let\(B = \left\{ {{{\bf{b}}_1},{{\bf{b}}_2},{{\bf{b}}_3}} \right\}\) be a basis for a vector space\(V\). Find \(T\left( {3{{\bf{b}}_1} - 4{{\bf{b}}_2}} \right)\) when \(T\) isa linear transformation from \(V\) to \(V\) whose matrix relative to \(B\) is

\({\left( T \right)_B} = \left( {\begin{aligned}0&{}&{ - 6}&{}&1\\0&{}&5&{}&{ - 1}\\1&{}&{ - 2}&{}&7\end{aligned}} \right)\)

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

17. \(\left[ {\begin{array}{*{20}{c}}3&0&0&0&0\\- 5&1&0&0&0\\3&8&0&0&0\\0&- 7&2&1&0\\- 4&1&9&- 2&3\end{array}} \right]\)

Let\(G = \left( {\begin{aligned}{*{20}{c}}A&X\\{\bf{0}}&B\end{aligned}} \right)\). Use formula\(\left( {\bf{1}} \right)\)for the determinant in section\({\bf{5}}{\bf{.2}}\)to explain why\(\det G = \left( {\det A} \right)\left( {\det B} \right)\). From this, deduce that the characteristic polynomial of\(G\)is the product of the characteristic polynomials of\(A\)and\(B\).

Consider an invertible n × n matrix A such that the zero state is a stable equilibrium of the dynamical system x→(t+1)=Ax→(t) What can you say about the stability of the systems

x→(t+1)=A-1x→(t)
See all solutions

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