Math Problem Statement
Find the eigenvalues and eigenvectors for the matrix A:
A = [[2, 1, 1], [1, 2, 1], [0, 0, 1]]
Solution
Let's solve the problem:
Problem:
Find the eigenvalues and eigenvectors for the matrix:
2 & 1 & 1 \\ 1 & 2 & 1 \\ 0 & 0 & 1 \end{bmatrix}.$$ ### Solution: #### Step 1: Find the eigenvalues Eigenvalues are the solutions to the characteristic equation: $$\det(A - \lambda I) = 0$$ 1. Subtract $$\lambda I$$ from $$A$$: $$A - \lambda I = \begin{bmatrix} 2-\lambda & 1 & 1 \\ 1 & 2-\lambda & 1 \\ 0 & 0 & 1-\lambda \end{bmatrix}.$$ 2. Compute the determinant: $$\det(A - \lambda I) = \det \begin{bmatrix} 2-\lambda & 1 & 1 \\ 1 & 2-\lambda & 1 \\ 0 & 0 & 1-\lambda \end{bmatrix}.$$ Expand along the third row: $$\det(A - \lambda I) = (1-\lambda) \cdot \det \begin{bmatrix} 2-\lambda & 1 \\ 1 & 2-\lambda \end{bmatrix}.$$ The determinant of the 2x2 matrix is: $$\det \begin{bmatrix} 2-\lambda & 1 \\ 1 & 2-\lambda \end{bmatrix} = (2-\lambda)(2-\lambda) - (1)(1) = (2-\lambda)^2 - 1.$$ So: $$\det(A - \lambda I) = (1-\lambda)((2-\lambda)^2 - 1).$$ Simplify $$(2-\lambda)^2 - 1$$: $$(2-\lambda)^2 - 1 = 4 - 4\lambda + \lambda^2 - 1 = \lambda^2 - 4\lambda + 3.$$ Thus: $$\det(A - \lambda I) = (1-\lambda)(\lambda^2 - 4\lambda + 3).$$ Factorize $$\lambda^2 - 4\lambda + 3$$: $$\lambda^2 - 4\lambda + 3 = (\lambda - 1)(\lambda - 3).$$ So: $$\det(A - \lambda I) = (1-\lambda)(\lambda - 1)(\lambda - 3).$$ #### Eigenvalues: The eigenvalues are the roots of $$\det(A - \lambda I) = 0$$: $$\lambda = 1, \, 1, \, 3.$$ #### Step 2: Find the eigenvectors For each eigenvalue, solve $$(A - \lambda I)x = 0$$. --- 1. **For $$\lambda = 1$$:** $$A - I = \begin{bmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 0 & 0 & 0 \end{bmatrix}.$$ Solve: $$\begin{bmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 0 & 0 & 0 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = 0.$$ From the first row: $$x_1 + x_2 + x_3 = 0.$$ Let $$x_3 = 1$$, then $$x_1 + x_2 = -1$$. Choose $$x_1 = 1, x_2 = -2$$. Eigenvector: $$v_1 = \begin{bmatrix} 1 \\ -2 \\ 1 \end{bmatrix}.$$ --- 2. **For $$\lambda = 3$$:** $$A - 3I = \begin{bmatrix} -1 & 1 & 1 \\ 1 & -1 & 1 \\ 0 & 0 & -2 \end{bmatrix}.$$ Solve: $$\begin{bmatrix} -1 & 1 & 1 \\ 1 & -1 & 1 \\ 0 & 0 & -2 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = 0.$$ From the third row: $$-2x_3 = 0 \implies x_3 = 0.$$ From the first row: $$-x_1 + x_2 = 0 \implies x_1 = x_2.$$ Eigenvector: $$v_2 = \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix}.$$ --- ### Final Answer: Eigenvalues: $$\lambda = 1, 1, 3$$. Eigenvectors: $$v_1 = \begin{bmatrix} 1 \\ -2 \\ 1 \end{bmatrix}, \quad v_2 = \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix}.$$ --- Would you like a detailed verification or help with similar problems? ### Related Questions: 1. How are eigenvalues used in real-world applications like physics or machine learning? 2. What is the geometric significance of eigenvalues and eigenvectors? 3. How does the multiplicity of eigenvalues affect eigenvectors? 4. Can a matrix have fewer eigenvectors than eigenvalues? Why? 5. How does diagonalization relate to eigenvalues and eigenvectors? ### Tip: When solving for eigenvectors, always ensure your solutions satisfy $$(A - \lambda I)x = 0$$.Ask a new question for Free
By Image
Drop file here or Click Here to upload
Math Problem Analysis
Mathematical Concepts
Linear Algebra
Eigenvalues
Eigenvectors
Formulas
Determinant formula: det(A - λI) = 0
Eigenvector equation: (A - λI)x = 0
Theorems
Cayley-Hamilton theorem
Eigenvalue decomposition
Suitable Grade Level
College/University level
Related Recommendation
Eigenvalues and Eigenvectors of a 3x3 Matrix
Finding Eigenvalues and Eigenvectors for the Matrix A = [[2, -2, 2], [1, 1, 1], [1, 3, -1]]
Eigenvectors of Matrix A = [[1, 1, 1], [-1, -3, -3], [2, 4, 4]]
Eigenvalues and Eigenvectors of Matrix A = [[0, 2, 2], [2, 2, -4], [0, 1, 1]]
Eigenvalues and Eigenvectors of Matrix A = [[0, 1, 1], [1, 2, 1], [3, 5, 2]]