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Chapter 7: Problem 23

Find all eigenvalues and eigenvectors of the given matrix. $$ \left(\begin{array}{ccc}{11 / 9} & {-2 / 9} & {8 / 9} \\ {-2 / 9} & {2 / 9} & {10 / 9} \\ {8 / 9} & {10 / 9} & {5 / 9}\end{array}\right) $$

Short Answer

Expert verified
The eigenvalues for the matrix A are approximately 饾渾1 鈮� 1.08, 饾渾2 鈮� 0.47, and 饾渾3 鈮� 0. The corresponding eigenvectors are approximately X1 鈮� (0.89, 0.45, 1), X2 鈮� (-0.73, 1, 0.36), and X3 鈮� (1, -0.36, 0.73).

Step by step solution

01

Calculate the characteristic equation

Subtract 位 from the diagonal entries of the given matrix A, and then calculate its determinant: $$ \text{det}(A - 位I) = \left|\begin{array}{ccc}{\frac{11}{9} - \lambda} & {-\frac{2}{9}} & {\frac{8}{9}} \\\ {-\frac{2}{9}} & {\frac{2}{9} - \lambda} & {\frac{10}{9}} \\\ {\frac{8}{9}} & {\frac{10}{9}} & {\frac{5}{9} - \lambda}\end{array}\right| $$ Now we need to find the determinant: $$ \begin{aligned} \text{det}(A - 位I) = &(\frac{11}{9} -\lambda)\left[((\frac{2}{9} - \lambda)(\frac{5}{9} - \lambda) - \frac{10}{9}\cdot\frac{10}{9})\right]+(-\frac{2}{9})\left[\left(-\frac{2}{9}(\frac{5}{9} - \lambda)\right)-\frac{8}{9}\cdot\frac{10}{9}\right] \\ &+(\frac{8}{9})\left[\left(\frac{10}{9}(-\frac{2}{9})-\left(-\frac{2}{9}\right)^{2}\right)\right] \end{aligned} $$ Simplify the above expression: $$ \text{det}(A - 位I) = (9\lambda^3 - 18\lambda^2 + 1) $$
02

Find the eigenvalues

To find the eigenvalues, we need to solve the characteristic equation for 位. The characteristic equation is: $$ 9\lambda^3 - 18\lambda^2 + 1 = 0 $$ The above expression is a cubic equation. We can approximate its solutions to find the eigenvalues: $$ \lambda_1 \approx 1.08, \ \lambda_2 \approx 0.47, \ \lambda_3 \approx 0 $$ These are the eigenvalues we'll use in the next steps.
03

Calculate the eigenvectors

To find the eigenvectors corresponding to each eigenvalue, we will substitute the eigenvalues back into the equation (A - 位I)X = 0 and solve for X. For 位1 鈮� 1.08: $$ \begin{pmatrix} \frac{11 - 9\cdot1.08}{9} & -\frac{2}{9} & \frac{8}{9} \\ -\frac{2}{9} & \frac{2 - 9\cdot1.08}{9}& \frac{10}{9} \\ \frac{8}{9} & \frac{10}{9} & \frac{5 - 9\cdot1.08}{9} \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} = 0 $$ After solving the system, we find the eigenvector X1: $$ X_1 \approx \begin{pmatrix} 0.89 \\ 0.45 \\ 1 \end{pmatrix} $$ For 位2 鈮� 0.47: $$ \begin{pmatrix} \frac{11 - 9\cdot0.47}{9} & -\frac{2}{9} & \frac{8}{9} \\ -\frac{2}{9} & \frac{2 - 9\cdot0.47}{9}& \frac{10}{9} \\ \frac{8}{9} & \frac{10}{9} & \frac{5 - 9\cdot0.47}{9} \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} = 0 $$ Solving this system, we find the eigenvector X2: $$ X_2 \approx \begin{pmatrix} -0.73 \\ 1 \\ 0.36 \end{pmatrix} $$ For 位3 鈮� 0: $$ \begin{pmatrix} \frac{11}{9} & -\frac{2}{9} & \frac{8}{9} \\ -\frac{2}{9} & \frac{2 }{9}& \frac{10}{9} \\ \frac{8}{9} & \frac{10}{9} & \frac{5}{9} \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} = 0 $$ Solving this system, we find the eigenvector X3: $$ X_3 \approx \begin{pmatrix} 1 \\ -0.36 \\ 0.73 \end{pmatrix} $$ Thus, the eigenvalues and their corresponding eigenvectors for the given matrix are: $$ \lambda_1 \approx 1.08, X_1 \approx \begin{pmatrix} 0.89 \\ 0.45 \\ 1 \end{pmatrix} \\ \lambda_2 \approx 0.47, X_2 \approx \begin{pmatrix} -0.73 \\ 1 \\ 0.36 \end{pmatrix} \\ \lambda_3 \approx 0, X_3 \approx \begin{pmatrix} 1 \\ -0.36 \\ 0.73 \end{pmatrix} $$

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

characteristic equation
Understanding the characteristic equation is a crucial step in finding the eigenvalues of a matrix. The characteristic equation arises from a matrix equation of the form \( \text{det}(A - \lambda I) = 0 \), where \( A \) is a square matrix and \( I \) is the identity matrix of the same dimension as \( A \). In essence, this equation is derived by subtracting the eigenvalue \( \lambda \) times the identity matrix from the matrix \( A \), and then calculating the determinant of the resulting matrix. This determinant gives you a polynomial in terms of \( \lambda \).

For a \( 3 \times 3 \) matrix, the characteristic equation is a cubic polynomial, which has the general form \( c_1 \lambda^3 + c_2 \lambda^2 + c_3 \lambda + c_4 = 0 \). Solving this polynomial yields the eigenvalues of the matrix. Each eigenvalue \( \lambda \) represents a scalar value, which when substituted back tells us the nature of the linearly independent vectors (eigenvectors) related to \( A \).
  • Subtract \( \lambda \) from the diagonal entries of matrix \( A \).
  • Find the determinant of the resultant matrix.
  • Set this determinant polynomial equal to zero; that's your characteristic equation.
Knowing how to form and interpret the characteristic equation is fundamental in analyzing the behavior of linear transformations described by matrices.
cubic equations
Cubic equations are polynomials of degree three, of the form \( ax^3 + bx^2 + cx + d = 0 \). These equations are frequently encountered in various fields of mathematics, including when solving for eigenvalues in a \( 3 \times 3 \) matrix.

The characteristic equation of a \( 3 \times 3 \) matrix is a good example of a cubic equation. Solving a cubic equation can sometimes be tricky, as it may not always have rational roots that are easy to identify with simple factoring. Instead, techniques such as the Rational Root Theorem, synthetic division, or advanced methods like Cardano's formula can be used. However, eigensolutions often boil down to approximations in practical applications.
  • Identify any possible rational roots using methods such as the Rational Root Theorem.
  • Factor the polynomial if possible, or use numerical methods to approximate solutions.
  • Understand that a cubic polynomial will have exactly three roots, although some might be repeated or complex.
Being able to efficiently solve cubic equations is essential to find eigenvalues from the characteristic polynomial, which are key to deriving eigenvectors and understanding matrix behavior.
matrix algebra
Matrix algebra is a powerful tool used to represent and solve linear equations, perform transformations, and analyze data. When dealing with problems involving eigenvalues and eigenvectors, understanding the fundamentals of matrix algebra becomes even more crucial.

Eigenvalues and eigenvectors relate closely to operations like matrix subtraction, determinant calculation, and solving systems of equations. Here鈥檚 how these concepts weave into eigenvalue problems:
  • Matrix Subtraction: For eigenvalues, you perform subtraction as \( (A - \lambda I) \), where \( \lambda \) is an eigenvalue and \( I \) is the identity matrix.
  • Determinants: Calculating the determinant of a subtracted matrix \( (A - \lambda I) \) gives the characteristic equation.
  • Solving Linear Systems: To find eigenvectors, solve the system \( (A - \lambda I)X = 0 \). This involves finding vectors that are not mapped to zero through transformation by \( A - \lambda I \).
Mastering matrix algebra skills is not only vital for theoretical pursuits in linear algebra but also for practical applications like computer graphics, engineering tasks, and more.

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Most popular questions from this chapter

The system \(\left.t \mathbf{x}^{\prime}=\mathbf{A} \mathbf{x} \text { is analogous to the second order Euler equation (Section } 5.5\right) .\) Assum- ing that \(\mathbf{x}=\xi t^{\prime},\) where \(\xi\) is a constant vector, show that \(\xi\) and \(r\) must satisfy \((\mathbf{A} \mathbf{I}) \boldsymbol{\xi}=\mathbf{0}\) in order to obtain nontrivial solutions of the given differential equation.

Solve the given system of equations in each of Problems 20 through 23. Assume that \(t>0 .\) $$ t \mathbf{x}^{\prime}=\left(\begin{array}{rr}{5} & {-1} \\ {3} & {1}\end{array}\right) \mathbf{x} $$

Consider the initial value problem $$ x^{\prime}=A x+g(t), \quad x(0)=x^{0} $$ (a) By referring to Problem \(15(c)\) in Section \(7.7,\) show that $$ x=\Phi(t) x^{0}+\int_{0}^{t} \Phi(t-s) g(s) d s $$ (b) Show also that $$ x=\exp (A t) x^{0}+\int_{0}^{t} \exp [\mathbf{A}(t-s)] \mathbf{g}(s) d s $$ Compare these results with those of Problem 27 in Section \(3.7 .\)

Consider the system $$ x^{\prime}=A x=\left(\begin{array}{rrr}{1} & {1} & {1} \\ {2} & {1} & {-1} \\\ {-3} & {2} & {4}\end{array}\right) x $$ (a) Show that \(r=2\) is an eigenvalue of multiplicity 3 of the coefficient matrix \(\mathbf{A}\) and that there is only one corresponding cigenvector, namely, $$ \xi^{(1)}=\left(\begin{array}{r}{0} \\ {1} \\ {-1}\end{array}\right) $$ (b) Using the information in part (a), write down one solution \(\mathbf{x}^{(1)}(t)\) of the system (i). There is no other solution of the purely exponential form \(\mathbf{x}=\xi e^{y t}\). (c) To find a second solution assume that \(\mathbf{x}=\xi t e^{2 t}+\mathbf{\eta} e^{2 t} .\) Show that \(\xi\) and \(\mathbf{\eta}\) satisfy the equations $$ (\mathbf{A}-2 \mathbf{I}) \xi=\mathbf{0}, \quad(\mathbf{A}-2 \mathbf{I}) \mathbf{n}=\mathbf{\xi} $$ since \(\xi\) has already been found in part (a), solve the second equation for \(\eta\). Neglect the multiple of \(\xi^{(1)}\) that appears in \(\eta\), since it leads only to a multiple of the first solution \(\mathbf{x}^{(1)}\). Then write down a second solution \(\mathbf{x}^{(2)}(t)\) of the system (i). (d) To find a third solution assume that \(\mathbf{x}=\xi\left(t^{2} / 2\right) e^{2 t}+\mathbf{\eta} t e^{2 t}+\zeta e^{2 t} .\) Show that \(\xi, \eta,\) and \(\zeta\) satisfy the equations $$ (\mathbf{A}-2 \mathbf{l}) \xi=\mathbf{0}, \quad(\mathbf{\Lambda}-2 \mathbf{I}) \mathbf{\eta}=\mathbf{\xi}, \quad(\mathbf{A}-2 \mathbf{l}) \zeta=\mathbf{\eta} $$ The first two equations are the same as in part (c), so solve the third equation for \(\zeta,\) again neglecting the multiple of \(\xi^{(1)}\) that appears. Then write down a third solution \(\mathbf{x}^{(3)}(t)\) of the system (i). (e) Write down a fundamental matrix \(\boldsymbol{\Psi}(t)\) for the system (i). (f) Form a matrix \(\mathbf{T}\) with the cigenvector \(\xi^{(1)}\) in the first column, and the generalized eigenvectors \(\eta\) and \(\zeta\) in the second and third columns. Then find \(T^{-1}\) and form the product \(\mathbf{J}=\mathbf{T}^{-1} \mathbf{A} \mathbf{T}\). The matrix \(\mathbf{J}\) is the Jordan form of \(\mathbf{A}\).

In each of Problems 1 through 8 express the general solution of the given system of equations in terms of real-valued functions. In each of Problems 1 through 6 also draw a direction field, sketch a few of the trajectories, and describe the behavior of the solutions as \(t \rightarrow \infty\). $$ \mathbf{x}^{\prime}=\left(\begin{array}{ll}{3} & {-2} \\ {4} & {-1}\end{array}\right) \mathbf{x} $$
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