Question

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

The eigenvalues for the matrix A are approximately 𝜆1 ≈ 1.08, 𝜆2 ≈ 0.47, and 𝜆3 ≈ 0. The corresponding eigenvectors are approximately X1 ≈ (0.89, 0.45, 1), X2 ≈ (-0.73, 1, 0.36), and X3 ≈ (1, -0.36, 0.73).

Step-by-step solution

Calculate the characteristic equation

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

Find the eigenvalues

To find the eigenvalues, we need to solve the characteristic equation for λ. 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.

Calculate the eigenvectors

To find the eigenvectors corresponding to each eigenvalue, we will substitute the eigenvalues back into the equation (A - λI)X = 0 and solve for X. For λ1 ≈ 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 λ2 ≈ 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 λ3 ≈ 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: $$ \lambda_1 \approx 1.08, X_1 \approx \begin{pmatrix} 0.89 \\ 0.45 \\ 1 \end{pmatrix} \\ \lambda_2 \approx 0.47, X_2 \approx \begin{pmatrix} -0.73 \\ 1 \\ 0.36 \end{pmatrix} \\ \lambda_3 \approx 0, X_3 \approx \begin{pmatrix} 1 \\ -0.36 \\ 0.73 \end{pmatrix} $$

Key concepts

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’s 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.