Math Problem Statement
- 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 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 . 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).Ask a new question for Free
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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
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