Math Problem Statement
Find the eigenvalues and eigenvectors of the given matrix 𝐴= 2 −2 2 1 1 1 1 3 −1
Solution
We are given the matrix as follows:
2 & -2 & 2 \\ 1 & 1 & 1 \\ 1 & 3 & -1 \end{bmatrix}$$ To find the **eigenvalues** and **eigenvectors** of this matrix, we proceed through the following steps: ### Step 1: Eigenvalue Calculation The eigenvalues of a matrix $$A$$ are the solutions to the characteristic equation: $$\text{det}(A - \lambda I) = 0$$ Where: - $$\lambda$$ represents the eigenvalue. - $$I$$ is the identity matrix. Let’s construct $$A - \lambda I$$: $$A - \lambda I = \begin{bmatrix} 2 - \lambda & -2 & 2 \\ 1 & 1 - \lambda & 1 \\ 1 & 3 & -1 - \lambda \end{bmatrix}$$ Now, we calculate the determinant of $$A - \lambda I$$: $$\text{det}(A - \lambda I) = \begin{vmatrix} 2 - \lambda & -2 & 2 \\ 1 & 1 - \lambda & 1 \\ 1 & 3 & -1 - \lambda \end{vmatrix}$$ Expanding this determinant will give us a cubic equation in terms of $$\lambda$$, which we can solve for the eigenvalues. Let's calculate it. ### Step 2: Eigenvalue Solution I will now calculate the determinant and solve the characteristic equation for the eigenvalues. sympy as sp # Define the variable and matrix lambda_ = sp.symbols('lambda') A = sp.Matrix([ [2, -2, 2], [1, 1, 1], [1, 3, -1] ]) # Calculate the characteristic equation det(A - lambda*I) I = sp.eye(3) # Identity matrix of size 3x3 A_lambda = A - lambda_ * I determinant = A_lambda.det() # Solve for eigenvalues eigenvalues = sp.solve(determinant, lambda_) eigenvaluesThe eigenvalues of the matrix $$A$$ are $$\lambda_1 = -2$$ and $$\lambda_2 = 2$$. ### Step 3: Eigenvectors Calculation For each eigenvalue, we will now find the corresponding eigenvectors by solving the equation: $$(A - \lambda I) \mathbf{v} = 0$$ where $$\mathbf{v}$$ is the eigenvector. Let's find the eigenvectors for each eigenvalue. I'll compute the eigenvectors corresponding to $$\lambda_1 = -2$$ and $$\lambda_2 = 2$$. Solve for eigenvectors corresponding to each eigenvalue eigenvectors_lambda_1 = A_lambda.subs(lambda_, eigenvalues[0]).nullspace() eigenvectors_lambda_2 = A_lambda.subs(lambda_, eigenvalues[1]).nullspace() # Return the eigenvectors eigenvectors_lambda_1, eigenvectors_lambda_2The eigenvectors corresponding to the eigenvalues are as follows: - For $$\lambda_1 = -2$$, the eigenvector is: $$\mathbf{v}_1 = \begin{bmatrix} -\frac{4}{7} \\ -\frac{1}{7} \\ 1 \end{bmatrix}$$ - For $$\lambda_2 = 2$$, the eigenvector is: $$\mathbf{v}_2 = \begin{bmatrix} 0 \\ 1 \\ 1 \end{bmatrix}$$ ### Recap: - Eigenvalues: $$\lambda_1 = -2$$, $$\lambda_2 = 2$$ - Eigenvectors: - $$\mathbf{v}_1 = \begin{bmatrix} -\frac{4}{7} \\ -\frac{1}{7} \\ 1 \end{bmatrix}$$ - $$\mathbf{v}_2 = \begin{bmatrix} 0 \\ 1 \\ 1 \end{bmatrix}$$ Would you like further explanation or details about the steps? Or do you have any additional questions? Here are some related questions: 1. How do you interpret the eigenvalue-eigenvector relationship in physical systems? 2. What is the significance of the eigenvectors' directionality? 3. How can you use the eigenvalues and eigenvectors in matrix diagonalization? 4. What would happen if the matrix $$A$$ had complex eigenvalues? 5. Can eigenvalues and eigenvectors be used in solving differential equations? **Tip:** Eigenvalues can be thought of as scaling factors for the eigenvectors when the matrix acts on them, which helps in understanding the matrix's behavior in transformations.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Linear Algebra
Eigenvalues
Eigenvectors
Matrix Determinants
Formulas
det(A - λI) = 0 (Characteristic Equation)
(A - λI)v = 0 (Eigenvalue-Eigenvector Equation)
Theorems
Eigenvalue Theorem
Diagonalization of Matrices
Suitable Grade Level
Grades 10-12
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