Math Problem Statement

Calculate the eigenvectors of the matrix \( \begin{bmatrix} 6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3 \end{bmatrix} \).

Solution

To calculate the eigenvectors of the given matrix:

6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3 \end{bmatrix},$$ we will follow these steps: ### Step 1: Find the Eigenvalues Solve the characteristic equation $$\det(A - \lambda I) = 0$$ for eigenvalues $$\lambda$$, where $$I$$ is the identity matrix. ### Step 2: Solve $$(A - \lambda I) \mathbf{v} = 0$$ for each eigenvalue $$\lambda$$ For each eigenvalue, solve this equation to find the corresponding eigenvector(s). Let me calculate these values step-by-step for you. numpy as np from sympy import symbols, Matrix, eye, solve, Eq # Define the matrix A and variable lambda A = Matrix([[6, -2, 2], [-2, 3, -1], [2, -1, 3]]) lam = symbols('λ') # Step 1: Find the characteristic polynomial I = eye(3) # 3x3 Identity matrix char_poly = (A - lam * I).det() # Solve for eigenvalues eigenvalues = solve(Eq(char_poly, 0), lam) char_poly, eigenvaluesThe characteristic polynomial of the matrix is: $$-\lambda^3 + 12\lambda^2 - 36\lambda + 32 = 0.$$ The eigenvalues are: $$\lambda = 2 \quad \text{(with multiplicity 2)}, \quad \lambda = 8.$$ Next, let us calculate the eigenvectors for each eigenvalue. Step 2: Calculate eigenvectors for each eigenvalue eigenvectors = {} for eigenvalue in eigenvalues: eig_matrix = A - eigenvalue * I # (A - λI) eig_null_space = eig_matrix.nullspace() # Null space gives eigenvectors eigenvectors[eigenvalue] = eig_null_space eigenvectorsThe eigenvectors for the matrix are: 1. For $$\lambda = 2$$ (with multiplicity 2): - Eigenvector 1: $$\begin{bmatrix} \frac{1}{2} \\ 1 \\ 0 \end{bmatrix}$$ - Eigenvector 2: $$\begin{bmatrix} -\frac{1}{2} \\ 0 \\ 1 \end{bmatrix}$$ 2. For $$\lambda = 8$$: - Eigenvector: $$\begin{bmatrix} 2 \\ -1 \\ 1 \end{bmatrix}$$ Would you like more details on any step? Here are some follow-up topics: 1. What is the null space and how is it used to find eigenvectors? 2. How do we interpret eigenvalues and eigenvectors geometrically? 3. What happens if the eigenvalues have a higher multiplicity? 4. How can eigenvectors be normalized to unit vectors? 5. Can eigenvectors from different eigenvalues ever be collinear? **Tip:** Eigenvectors corresponding to distinct eigenvalues are always linearly independent.

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Math Problem Analysis

Mathematical Concepts

Linear Algebra
Eigenvalues
Eigenvectors

Formulas

\( \det(A - \lambda I) = 0 \)
\( (A - \lambda I) \mathbf{v} = 0 \)

Theorems

Spectral Theorem
Fundamental Theorem of Linear Algebra

Suitable Grade Level

Undergraduate level

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