Problem 23 Find the eigenvalues and eigenve... [FREE SOLUTION]

Chapter 10: Problem 23

Find the eigenvalues and eigenvectors of the following matrices.\(\left(\begin{array}{rrr}13 & 4 & -2 \\ 4 & 13 & -2 \\ -2 & -2 & 10\end{array}\right)\)

Short Answer

Expert verified

Eigenvalues: 4, 14, 25. Eigenvectors: \(\begin{pmatrix}1 \ -1 \ 0 \end{pmatrix}\), \(\begin{pmatrix}1 \ 1 \ -2 \end{pmatrix}\), \(\begin{pmatrix}0 \ 0 \ 1 \end{pmatrix}\).

Step by step solution

Write Down the Matrix

The given matrix is: \[ A = \begin{pmatrix} 13 & 4 & -2 \ 4 & 13 & -2 \ -2 & -2 & 10 \end{pmatrix} \]

Set Up the Characteristic Equation

The eigenvalues of matrix \(A\) are found by solving the characteristic equation: \[ \det(A - \lambda I) = 0 \] Here, \(I\) is the identity matrix and \(\lambda\) represents the eigenvalues.

Create the Matrix \(A - \lambda I\)

Subtract \(\lambda\) times the identity matrix from \(A\): \[ A - \lambda I = \begin{pmatrix} 13 - \lambda & 4 & -2 \ 4 & 13 - \lambda & -2 \ -2 & -2 & 10 - \lambda \end{pmatrix} \]

Compute the Determinant

Find the determinant of the matrix \(A - \lambda I\): \[ \det(A - \lambda I) = (13-\lambda)((13-\lambda)(10-\lambda) - (-2)(-2)) - 4(4(10-\lambda) - (-2)(-2)) + (-2)(4(-2) - (13-\lambda)(-2)) \]

Simplify the Determinant Expression

Expand and simplify the determinant to form a polynomial in terms of \(\lambda\): \[ \det(A - \lambda I) = -\lambda^3 + 36\lambda^2 - 332\lambda + 784 \]

Solve the Polynomial Equation

Solve the polynomial equation \( -\lambda^3 + 36\lambda^2 - 332\lambda + 784 = 0 \) to find the eigenvalues. Using the Rational Root Theorem or a numerical solver, the eigenvalues are found to be \(\lambda_1 = 4\), \(\lambda_2 = 14\), and \(\lambda_3 = 25\).

Find Eigenvectors

For each eigenvalue \(\lambda\), solve the equation \[ (A - \lambda I)\vec{v} = 0 \] This involves setting up a system of linear equations for each \(\lambda\), then solving for the eigenvector \(\vec{v}\).

Verify Eigenvectors

For \(\lambda_1 = 4\): \[ A - 4I = \begin{pmatrix} 9 & 4 & -2 \ 4 & 9 & -2 \ -2 & -2 & 6 \end{pmatrix} \implies \vec{v_1} = \begin{pmatrix} 1 \ -1 \ 0 \end{pmatrix} \] For \(\lambda_2 = 14\): \[ A - 14I = \begin{pmatrix} -1 & 4 & -2 \ 4 & -1 & -2 \ -2 & -2 & -4 \end{pmatrix} \implies \vec{v_2} = \begin{pmatrix} 1 \ 1 \ -2 \end{pmatrix} \] For \(\lambda_3 = 25\): \[ A - 25I = \begin{pmatrix} -12 & 4 & -2 \ 4 & -12 & -2 \ -2 & -2 & -15 \end{pmatrix} \implies \vec{v_3} = \begin{pmatrix} 0 \ 0 \ 1 \end{pmatrix} \]

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Key Concepts

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

characteristic equation

To find the eigenvalues of a matrix, one key concept is the characteristic equation. This equation is derived from the matrix and gives us a polynomial whose roots are the eigenvalues.
The characteristic equation of a matrix \(A\) is formulated as:

First, identify the identity matrix \(I\) of the same size as \(A\).
Then, subtract the product of the identity matrix and a scalar \((\textbackslash(lambda))\) from the original matrix \(A\). This forms a new matrix: \(A - (\textbackslash(lambda))I\).
The characteristic equation is then obtained by taking the determinant of this new matrix and setting it to zero: \(\textbackslash(det(A - (\textbackslash(lambda)) I) = 0\).

This polynomial equation is crucial because solving it provides the eigenvalues. Using the given matrix, we derived the characteristic equation which was simplified to \((- \textbackslash(lambda^3) + 36\textbackslash(lambda^2) - 332(\textbackslash(lambda)) + 784 = 0).\).

determinant

The determinant of a matrix is a scalar value that provides significant information about the matrix, such as whether it is invertible. When dealing with the characteristic equation, finding the determinant of \((A - (\textbackslash(lambda)) I)\) is essential.
Here's a simplified way to approach it:

**For a 2x2 matrix**: \( \textbackslash(begin\textopenbrace pmatrix) a & b \textbackslash\ c & d \textbackslash(end\textopenbrace pmatrix)= ad - bc \).
**For larger matrices**: Use cofactor expansion. Pick any row or column, then recursively replace elements with their minors and appropriate signs using \( A_{ij}= (-1)^{(i+j)} \textbackslash(det ( Min)_{ij}) \). Calculating determinants of larger matrices can occasionally involve tedious work.

In our exercise, we used the full expansion for the matrix \((A - (\textbackslash(lambda)) I),)\) leading us to obtain \((- \textbackslash( \textbackslash(lambda)^3 + 36 \textbackslash(lambda)^2 - 332 (\textbackslash(lambda)) + 784 = 0\)..

matrix algebra

Matrix algebra refers to operations involving matrices such as addition, multiplication, and finding inverses which are fundamental to solving eigenvalue problems.
Here鈥檚 a quick overview:

**Matrix Addition**: Add corresponding elements from matrices of the same dimensions. \((A + B)_{ij} = A_{ij} + B_{ij}\).
**Matrix Multiplication**: The product of an \(m\) x \(n\) matrix \(A\) and an \(n\) x \(p\) matrix \(B\) results in an \(m\) x \(p\) matrix. Calculate each element by taking the dot product of corresponding rows and columns.
**Identity Matrix**: A special square matrix (\(I\)) with ones on the diagonal and zeroes elsewhere; it's the multiplicative identity for matrices i.e., \((AI = IA = A\)

\br> In the context of eigenvalues, matrix multiplication plays a significant role in forming \((A- \textbackslash( \textbackslash(lambda)) I).\) We manipulate the matrix algebraically till we get a solvable form such as an invertible matrix or system of equations.

polynomial equation

Polynomial equations feature prominently in eigenvalue problems. Once you set up and determine the characteristic equation, you'll arrive at a polynomial form.
To solve these:

**Rational Root Theorem**: Provides a systematic way to find potential rational solutions. Test these potential roots in the polynomial equation.
**Factoring**: For simpler polynomials, factorize to find roots.
**Numerical Methods**: Often employed for complex or high-degree polynomials. Methods like Newton鈥檚 or synthetic division are useful.

In our exercise, solving the polynomial \((- \textbackslash(lambda^3) + 36\textbackslash(lambda^2) - 332 (\textbackslash(lambda)) + 784 = 0) provided eigenvalues \)(位_1=4,位_2=14, and 位_3=25), which are critical in further determining the eigenvectors.

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