Math Problem Statement

Solve it

Compute the canonical form which transforms the quadratic form Q = 3x_1^2 + 3x_3^2 + 4x_1x_2 + 8x_1x_3 + 8x_2x_3.

Solution

To solve this, we will compute the canonical form of the quadratic form Q=3x12+3x32+4x1x2+8x1x3+8x2x3Q = 3x_1^2 + 3x_3^2 + 4x_1x_2 + 8x_1x_3 + 8x_2x_3. The canonical form involves finding the principal axes by diagonalizing the symmetric matrix representing the quadratic form.

Here are the steps:

1. Write the matrix of the quadratic form

The general quadratic form can be written as: Q=xTAxQ = \mathbf{x}^T \mathbf{A} \mathbf{x} where x=[x1x2x3]\mathbf{x} = \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} and A\mathbf{A} is the symmetric matrix associated with the quadratic form. The coefficients of the quadratic terms go on the diagonal, and the coefficients of the mixed terms are halved for symmetricity.

Thus, the matrix A\mathbf{A} is:

3 & 2 & 4 \\ 2 & 0 & 4 \\ 4 & 4 & 3 \end{bmatrix}$$ --- ### 2. **Find the eigenvalues and eigenvectors of $$\mathbf{A}$$** The canonical form is obtained by diagonalizing $$\mathbf{A}$$. The eigenvalues give the entries of the diagonal matrix, and the eigenvectors provide the transformation matrix. #### Eigenvalue computation: We solve $$\det(\mathbf{A} - \lambda \mathbf{I}) = 0$$, where $$\mathbf{I}$$ is the identity matrix and $$\lambda$$ is the eigenvalue. $$\det \begin{bmatrix} 3-\lambda & 2 & 4 \\ 2 & -\lambda & 4 \\ 4 & 4 & 3-\lambda \end{bmatrix} = 0$$ Expanding this determinant gives the characteristic polynomial. --- ### 3. **Construct the canonical form** Using the eigenvalues $$\lambda_1, \lambda_2, \lambda_3$$, we write the quadratic form in terms of the new variables $$y_1, y_2, y_3$$: $$Q = \lambda_1 y_1^2 + \lambda_2 y_2^2 + \lambda_3 y_3^2$$ --- Let me compute these steps explicitly. numpy as np # Define the matrix A A = np.array([[3, 2, 4], [2, 0, 4], [4, 4, 3]]) # Compute eigenvalues and eigenvectors eigenvalues, eigenvectors = np.linalg.eig(A) eigenvalues, eigenvectorsThe eigenvalues of the matrix $$\mathbf{A}$$ are: $$\lambda_1 = 9.0368, \quad \lambda_2 = -0.1535, \quad \lambda_3 = -2.8833$$ The corresponding eigenvectors (normalized) are: $$\mathbf{v}_1 = \begin{bmatrix} 0.5932 \\ 0.4320 \\ 0.6793 \end{bmatrix}, \quad \mathbf{v}_2 = \begin{bmatrix} 0.7864 \\ -0.4913 \\ -0.3744 \end{bmatrix}, \quad \mathbf{v}_3 = \begin{bmatrix} -0.1720 \\ -0.7563 \\ 0.6312 \end{bmatrix}$$ ### Canonical Form The quadratic form in canonical form is: $$Q = 9.0368 \, y_1^2 - 0.1535 \, y_2^2 - 2.8833 \, y_3^2$$ where $$y_1, y_2, y_3$$ are the transformed variables obtained using the eigenvectors as the basis. Do you want me to explain any steps in detail or compute anything further? ### Related Questions 1. How are eigenvalues used to classify the quadratic form? 2. What is the role of eigenvectors in diagonalization? 3. Can the negative eigenvalues indicate something geometrically? 4. How do we verify that the transformation preserves the quadratic nature? 5. What happens if $$\mathbf{A}$$ is not symmetric? ### Tip Always verify if the matrix is symmetric before applying diagonalization; it ensures that the eigenvalues are real, simplifying interpretation!

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Math Problem Analysis

Mathematical Concepts

Matrix Algebra
Eigenvalues and Eigenvectors
Diagonalization of Matrices
Quadratic Forms

Formulas

Q = x^T A x (Quadratic form in matrix notation)
det(A - λI) = 0 (Characteristic polynomial for eigenvalues)

Theorems

Spectral Theorem for symmetric matrices

Suitable Grade Level

Undergraduate (Linear Algebra level)

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