Math Problem Statement

  1. Reduce the quadratic form 2x2+2y2+2z2-2xy-2yz-2xz to canonical form by an orthogonal transformation and hence find rank, index and nature of the quadratic form

Solution

To reduce the quadratic form 2x2+2y2+2z22xy2yz2xz2x^2 + 2y^2 + 2z^2 - 2xy - 2yz - 2xz to its canonical form using an orthogonal transformation, we need to follow these steps:

1. Express the quadratic form as a matrix:

We can represent the quadratic form in terms of a symmetric matrix AA. The quadratic form can be written as:

\begin{bmatrix} 2 & -1 & -1 \\ -1 & 2 & -1 \\ -1 & -1 & 2 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix}$$ Thus, the symmetric matrix associated with the quadratic form is: $$A = \begin{bmatrix} 2 & -1 & -1 \\ -1 & 2 & -1 \\ -1 & -1 & 2 \end{bmatrix}$$ ### 2. Find the eigenvalues and eigenvectors of the matrix $$A$$: To reduce the quadratic form to its canonical form, we need to diagonalize the matrix $$A$$. This can be done by finding the eigenvalues and eigenvectors of $$A$$. #### Characteristic equation: The characteristic equation is given by: $$\det(A - \lambda I) = 0$$ where $$I$$ is the identity matrix and $$\lambda$$ represents the eigenvalues. Let's compute the determinant: $$\det \begin{bmatrix} 2 - \lambda & -1 & -1 \\ -1 & 2 - \lambda & -1 \\ -1 & -1 & 2 - \lambda \end{bmatrix} = 0$$ Expanding the determinant, we get the characteristic equation: \[ (2 - \lambda)[(2 - \lambda)^2 - (-1)(-1)] - (-1)[(-1)(2 - \lambda) - (-1)(-1)] = 0 \] This simplifies to: $$(2 - \lambda)((2 - \lambda)^2 - 1) - (-1)((2 - \lambda) - 1) = 0$$ $$(2 - \lambda)((2 - \lambda)^2 - 1) + (2 - \lambda - 1) = 0$$ $$(2 - \lambda)((\lambda - 1)(\lambda - 3)) = 0$$ So, the eigenvalues are: $$\lambda_1 = 0, \quad \lambda_2 = 3, \quad \lambda_3 = 3$$ ### 3. Diagonalization: Since the eigenvalues of the matrix $$A$$ are $$0, 3, 3$$, we can now express the quadratic form in canonical form. The quadratic form is reduced to: $$Q(x, y, z) = \lambda_1 x'^2 + \lambda_2 y'^2 + \lambda_3 z'^2$$ where $$x', y', z'$$ are the new variables after the orthogonal transformation. Thus, the quadratic form in canonical form is: $$0x'^2 + 3y'^2 + 3z'^2$$ or simply: $$3y'^2 + 3z'^2$$ ### 4. Determine the rank, index, and nature of the quadratic form: - **Rank**: The rank of the quadratic form is the number of non-zero eigenvalues. Since there are two non-zero eigenvalues, the rank is **2**. - **Index**: The index is the number of negative eigenvalues. Since all non-zero eigenvalues are positive, the index is **0**. - **Nature**: Since the eigenvalues are non-negative (with two positive eigenvalues and one zero eigenvalue), the quadratic form is **positive semi-definite**. ### Conclusion: - **Canonical form**: $$3y'^2 + 3z'^2$$ - **Rank**: 2 - **Index**: 0 - **Nature**: Positive semi-definite --- Would you like further details or clarifications on any of these steps? ### Related Questions: 1. How do you diagonalize a matrix using orthogonal transformations? 2. What is the geometric interpretation of the canonical form of a quadratic equation? 3. How does the rank of a quadratic form relate to its definiteness? 4. What is the role of eigenvectors in diagonalizing a matrix? 5. How can you compute the index of a quadratic form with multiple variables? **Tip:** When dealing with quadratic forms, finding the eigenvalues is key to determining the nature of the form (definite, semi-definite, or indefinite).

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Math Problem Analysis

Mathematical Concepts

Quadratic Forms
Matrix Diagonalization
Eigenvalues and Eigenvectors

Formulas

Quadratic form matrix representation Q(x, y, z) = [x y z] * A * [x y z]^T
Characteristic equation: det(A - λI) = 0

Theorems

Eigenvalue Theorem
Diagonalization of Symmetric Matrices

Suitable Grade Level

Undergraduate Mathematics