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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

A common misconception is that if \(A\) has a strictly dominant eigenvalue, then, for any sufficiently large value of \(k\), the vector \({A^k}{\bf{x}}\) is approximately equal to an eigenvector of \(A\). For the three matrices below, study what happens to \({A^k}{\bf{x}}\) when \({\bf{x = }}\left( {{\bf{.5,}}{\bf{.5}}} \right)\), and try to draw general conclusions (for a \({\bf{2 \times 2}}\) matrix).

a. \(A{\bf{ = }}\left( {\begin{aligned}{ {20}{c}}{{\bf{.8}}}&{\bf{0}}\\{\bf{0}}&{{\bf{.2}}}\end{aligned}} \right)\) b. \(A{\bf{ = }}\left( {\begin{aligned}{ {20}{c}}{\bf{1}}&{\bf{0}}\\{\bf{0}}&{{\bf{.8}}}\end{aligned}} \right)\) c. \(A{\bf{ = }}\left( {\begin{aligned}{ {20}{c}}{\bf{8}}&{\bf{0}}\\{\bf{0}}&{\bf{2}}\end{aligned}} \right)\)

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

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

Let\(\varepsilon = \left\{ {{{\bf{e}}_1},{{\bf{e}}_2},{{\bf{e}}_3}} \right\}\) be the standard basis for \({\mathbb{R}^3}\),\(B = \left\{ {{{\bf{b}}_1},{{\bf{b}}_2},{{\bf{b}}_3}} \right\}\) be a basis for a vector space \(V\) and\(T:{\mathbb{R}^3} \to V\) be a linear transformation with the property that

\(T\left( {{x_1},{x_2},{x_3}} \right) = \left( {{x_3} - {x_2}} \right){{\bf{b}}_1} - \left( {{x_1} - {x_3}} \right){{\bf{b}}_2} + \left( {{x_1} - {x_2}} \right){{\bf{b}}_3}\)

  1. Compute\(T\left( {{{\bf{e}}_1}} \right)\), \(T\left( {{{\bf{e}}_2}} \right)\) and \(T\left( {{{\bf{e}}_3}} \right)\).
  2. Compute \({\left( {T\left( {{{\bf{e}}_1}} \right)} \right)_B}\), \({\left( {T\left( {{{\bf{e}}_2}} \right)} \right)_B}\) and \({\left( {T\left( {{{\bf{e}}_3}} \right)} \right)_B}\).
  3. Find the matrix for \(T\) relative to \(\varepsilon \), and\(B\).

M] In Exercises 19 and 20, find (a) the largest eigenvalue and (b) the eigenvalue closest to zero. In each case, set \[{{\bf{x}}_{\bf{0}}}{\bf{ = }}\left( {{\bf{1,0,0,0}}} \right)\] and carry out approximations until the approximating sequence seems accurate to four decimal places. Include the approximate eigenvector.

19.\[A{\bf{=}}\left[{\begin{array}{*{20}{c}}{{\bf{10}}}&{\bf{7}}&{\bf{8}}&{\bf{7}}\\{\bf{7}}&{\bf{5}}&{\bf{6}}&{\bf{5}}\\{\bf{8}}&{\bf{6}}&{{\bf{10}}}&{\bf{9}}\\{\bf{7}}&{\bf{5}}&{\bf{9}}&{{\bf{10}}}\end{array}} \right]\]

Suppose \(A = PD{P^{ - 1}}\), where \(P\) is \(2 \times 2\) and \(D = \left( {\begin{array}{*{20}{l}}2&0\\0&7\end{array}} \right)\)

a. Let \(B = 5I - 3A + {A^2}\). Show that \(B\) is diagonalizable by finding a suitable factorization of \(B\).

b. Given \(p\left( t \right)\) and \(p\left( A \right)\) as in Exercise 5 , show that \(p\left( A \right)\) is diagonalizable.
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