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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

Question: Let \(A = \left( {\begin{array}{*{20}{c}}a&b\\c&d\end{array}} \right)\). Use formula (1) for a determinant (given before Example 2) to show that \(\det A = ad - bc\). Consider two cases: \(a \ne 0\) and \(a = 0\).

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

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

Question: Construct a random integer-valued \(4 \times 4\) matrix \(A\), and verify that \(A\) and \({A^T}\) have the same characteristic polynomial (the same eigenvalues with the same multiplicities). Do \(A\) and \({A^T}\) have the same eigenvectors? Make the same analysis of a \(5 \times 5\) matrix. Report the matrices and your conclusions.

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

Mark each statement as True or False. Justify each answer.

a. If \(A\) is invertible and 1 is an eigenvalue for \(A\), then \(1\) is also an eigenvalue of \({A^{ - 1}}\)

b. If \(A\) is row equivalent to the identity matrix \(I\), then \(A\) is diagonalizable.

c. If \(A\) contains a row or column of zeros, then 0 is an eigenvalue of \(A\)

d. Each eigenvalue of \(A\) is also an eigenvalue of \({A^2}\).

e. Each eigenvector of \(A\) is also an eigenvector of \({A^2}\)

f. Each eigenvector of an invertible matrix \(A\) is also an eigenvector of \({A^{ - 1}}\)

g. Eigenvalues must be nonzero scalars.

h. Eigenvectors must be nonzero vectors.

i. Two eigenvectors corresponding to the same eigenvalue are always linearly dependent.

j. Similar matrices always have exactly the same eigenvalues.

k. Similar matrices always have exactly the same eigenvectors.

I. The sum of two eigenvectors of a matrix \(A\) is also an eigenvector of \(A\).

m. The eigenvalues of an upper triangular matrix \(A\) are exactly the nonzero entries on the diagonal of \(A\).

n. The matrices \(A\) and \({A^T}\) have the same eigenvalues, counting multiplicities.

o. If a \(5 \times 5\) matrix \(A\) has fewer than 5 distinct eigenvalues, then \(A\) is not diagonalizable.

p. There exists a \(2 \times 2\) matrix that has no eigenvectors in \({A^2}\)

q. If \(A\) is diagonalizable, then the columns of \(A\) are linearly independent.

r. A nonzero vector cannot correspond to two different eigenvalues of \(A\).

s. A (square) matrix \(A\) is invertible if and only if there is a coordinate system in which the transformation \({\bf{x}} \mapsto A{\bf{x}}\) is represented by a diagonal matrix.

t. If each vector \({{\bf{e}}_j}\) in the standard basis for \({A^n}\) is an eigenvector of \(A\), then \(A\) is a diagonal matrix.

u. If \(A\) is similar to a diagonalizable matrix \(B\), then \(A\) is also diagonalizable.

v. If \(A\) and \(B\) are invertible \(n \times n\) matrices, then \(AB\)is similar to \ (BA\ )

w. An \(n \times n\) matrix with \(n\) linearly independent eigenvectors is invertible.

x. If \(A\) is an \(n \times n\) diagonalizable matrix, then each vector in \({A^n}\) can be written as a linear combination of eigenvectors of \(A\).
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