Find Eigenvalues and Eigenvectors of Matrix G = A^T A

Math Problem Statement

A= [1,0,1;01,1]. Find the eigenvalues and eigenvectors of the matrix G = A superscript T A, and orthonomalise them. Call them v subscript 1,v subscript 2, v subscript 3 according to the order of h subscript 1>or = h subscript 2>or= h subscript 3 of eigenvalues and G

Solution

To solve the problem, let's break it down step by step.

1. Matrix $A$ :

Given:

1 & 0 & 1 \\ 0 & 1 & 1 \end{bmatrix}$$ ### 2. Calculate $$G = A^T A$$: First, we need to compute the transpose of $$A$$ ($$A^T$$), and then multiply $$A^T$$ with $$A$$ to get the matrix $$G$$. $$A^T = \begin{bmatrix} 1 & 0 \\ 0 & 1 \\ 1 & 1 \end{bmatrix}$$ Now, multiply $$A^T$$ with $$A$$: $$G = A^T A = \begin{bmatrix} 1 & 0 \\ 0 & 1 \\ 1 & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 1 \\ 1 & 1 & 2 \end{bmatrix}$$ So, $$G$$ is: $$G = \begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 1 \\ 1 & 1 & 2 \end{bmatrix}$$ ### 3. Find the Eigenvalues and Eigenvectors of $$G$$: To find the eigenvalues, we solve the characteristic equation $$\text{det}(G - \lambda I) = 0$$, where $$\lambda$$ represents the eigenvalues and $$I$$ is the identity matrix. The characteristic equation is: $$\text{det}\left(\begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 1 \\ 1 & 1 & 2 \end{bmatrix} - \lambda \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}\right) = 0$$ This simplifies to: $$\text{det}\left(\begin{bmatrix} 1-\lambda & 0 & 1 \\ 0 & 1-\lambda & 1 \\ 1 & 1 & 2-\lambda \end{bmatrix}\right) = 0$$ Now we calculate the determinant of this matrix to find the characteristic polynomial. ### 4. Solve the Characteristic Polynomial: Expanding the determinant, we have: $$\text{det}((1-\lambda)(1-\lambda)(2-\lambda) - 1(1) + 0) = 0$$ The characteristic polynomial is: $$\lambda^3 - 4\lambda^2 + 5\lambda - 2 = 0$$ Solving for $$\lambda$$ (eigenvalues), we'll get three eigenvalues. ### 5. Find Eigenvectors: For each eigenvalue $$\lambda_i$$, solve the equation $$(G - \lambda_i I)v_i = 0$$ to find the corresponding eigenvector $$v_i$$. ### 6. Orthonormalize the Eigenvectors: Use the Gram-Schmidt process to orthonormalize the eigenvectors $$v_1, v_2, v_3$$. ### 7. Arrange Eigenvectors: Finally, arrange the eigenvectors $$v_1, v_2, v_3$$ in the order corresponding to the eigenvalues $$h_1 \geq h_2 \geq h_3$$. I will now calculate the eigenvalues and corresponding eigenvectors explicitly.The eigenvalues and the corresponding orthonormal eigenvectors of the matrix $$G = A^T A$$ are as follows: ### Eigenvalues: $$\lambda_1 = 3.0, \quad \lambda_2 = 1.0, \quad \lambda_3 \approx 0$$ ### Orthonormal Eigenvectors: $$v_1 = \begin{bmatrix} -0.4082 \\ -0.4082 \\ -0.8165 \end{bmatrix}, \quad v_2 = \begin{bmatrix} -0.7071 \\ 0.7071 \\ 0 \end{bmatrix}, \quad v_3 = \begin{bmatrix} -0.5774 \\ -0.5774 \\ 0.5774 \end{bmatrix}$$ These eigenvectors $$v_1, v_2, v_3$$ correspond to the eigenvalues $$\lambda_1, \lambda_2, \lambda_3$$ in descending order. Would you like more details or have any questions? Here are 5 related questions to expand your understanding: 1. How do we derive the characteristic polynomial for finding eigenvalues? 2. Why are eigenvectors orthonormalized, and what is the significance? 3. How does the Gram-Schmidt process work for orthonormalization? 4. What is the geometric interpretation of eigenvectors and eigenvalues in linear algebra? 5. How does the eigenvalue decomposition relate to diagonalizing a matrix? **Tip:** Eigenvalues close to zero often indicate directions of minimal variance or influence in the matrix's transformation.

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Math Problem Analysis

Mathematical Concepts

Matrix Transposition
Eigenvalues and Eigenvectors
Gram-Schmidt Process

Formulas

Characteristic Polynomial

Theorems

Suitable Grade Level

Advanced Undergraduate

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