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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

Consider the growth of a lilac bush. The state of this lilac bush for several years (at year’s end) is shown in the accompanying sketch. Let n(t) be the number of new branches (grown in the year t) and a(t) the number of old branches. In the sketch, the new branches are represented by shorter lines. Each old branch will grow two new branches in the following year. We assume that no branches ever die.

(a) Find the matrix A such that [nt+1at+1]=A[ntat]

(b) Verify that [11]and [2-1] are eigenvectors of A. Find the associated eigenvalues.

(c) Find closed formulas for n(t) and a(t).

[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.

20. \[A{\bf{ = }}\left[ {\begin{array}{*{20}{c}}{\bf{1}}&{\bf{2}}&{\bf{3}}&{\bf{2}}\\{\bf{2}}&{{\bf{12}}}&{{\bf{13}}}&{{\bf{11}}}\\{{\bf{ - 2}}}&{\bf{3}}&{\bf{0}}&{\bf{2}}\\{\bf{4}}&{\bf{5}}&{\bf{7}}&{\bf{2}}\end{array}} \right]\]

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.

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

Let \(A{\bf{ = }}\left( {\begin{aligned}{*{20}{c}}{{a_{{\bf{11}}}}}&{{a_{{\bf{12}}}}}\\{{a_{{\bf{21}}}}}&{{a_{{\bf{22}}}}}\end{aligned}} \right)\). Recall from Exercise \({\bf{25}}\) in Section \({\bf{5}}{\bf{.4}}\) that \({\rm{tr}}\;A\) (the trace of \(A\)) is the sum of the diagonal entries in \(A\). Show that the characteristic polynomial of \(A\) is \({\lambda ^2} - \left( {{\rm{tr}}A} \right)\lambda + \det A\). Then show that the eigenvalues of a \({\bf{2 \times 2}}\) matrix \(A\) are both real if and only if \(\det A \le {\left( {\frac{{{\rm{tr}}A}}{2}} \right)^2}\).

Consider an invertiblen × 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-2In)x→(t)
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