Math Problem Statement

Reduce the qf x1^2+2x2x3 into a conohical formby means of orthogonal transformation.determine it's nature

Solution

To reduce the quadratic form Q(x1,x2,x3)=x12+2x2x3Q(x_1, x_2, x_3) = x_1^2 + 2x_2x_3 into a canonical form using an orthogonal transformation, we can follow these steps:

  1. Identify the quadratic form matrix: We can express the quadratic form in matrix notation as: Q(x)=xTAxQ(\mathbf{x}) = \mathbf{x}^T A \mathbf{x} where x=(x1x2x3)\mathbf{x} = \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} and the matrix AA is given by:

    1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{pmatrix}$$
  2. Find the eigenvalues and eigenvectors of AA:

    • Calculate the characteristic polynomial det(AλI)=0\det(A - \lambda I) = 0:
    1 - \lambda & 0 & 0 \\ 0 & -\lambda & 1 \\ 0 & 1 & -\lambda \end{pmatrix}$$ The determinant is: $$(1 - \lambda)(\lambda^2 - 1) = 0$$ Thus, the eigenvalues are $$\lambda_1 = 1$$ and $$\lambda_2 = 1$$ (with algebraic multiplicity 2, corresponding to the 2x2 block).
  3. Solve for eigenvectors:

    • For λ=1\lambda = 1: (AI)v=0    (000011011)(v1v2v3)=0(A - I)\mathbf{v} = 0 \implies \begin{pmatrix} 0 & 0 & 0 \\ 0 & -1 & 1 \\ 0 & 1 & -1 \end{pmatrix} \begin{pmatrix} v_1 \\ v_2 \\ v_3 \end{pmatrix} = 0 This gives the eigenvector v1=(100)\mathbf{v_1} = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} (for λ=1\lambda = 1), and a basis for the eigenspace corresponding to the remaining eigenvalues can be obtained.
  4. Construct the orthogonal transformation matrix PP: After finding eigenvectors, normalize them to form the orthogonal matrix PP.

  5. Transform the quadratic form: Using Q(y)=yT(PTAP)yQ(\mathbf{y}) = \mathbf{y}^T (P^T A P) \mathbf{y}, where y=PTx\mathbf{y} = P^T \mathbf{x}.

  6. Determine the canonical form: Calculate PTAPP^T A P to find the new quadratic form's matrix.

  7. Classify the nature:

    • The nature of the quadratic form is determined by the signs of the eigenvalues:
      • Positive definite if all eigenvalues are positive,
      • Negative definite if all eigenvalues are negative,
      • Indefinite if there are both positive and negative eigenvalues.

Following through this process would yield the canonical form and its nature. In this case, since there are eigenvalues of 11 (positive) and a block indicating indefinite nature (due to the structure of the form), the quadratic form is classified as indefinite.

If you need detailed calculations or any further explanation, feel free to ask!

Questions for Further Exploration:

  1. What are the specific steps to find eigenvalues for higher-dimensional matrices?
  2. Can you explain the process of diagonalization in more detail?
  3. How do different eigenvalues influence the shape of a quadratic surface?
  4. What are some practical applications of canonical forms in engineering or physics?
  5. How would you handle a quadratic form with more than three variables?

Tip:

When working with quadratic forms, always keep track of the signs of the eigenvalues, as they provide critical information about the nature of the conic sections represented.

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Math Problem Analysis

Mathematical Concepts

Linear Algebra
Quadratic Forms
Eigenvalues and Eigenvectors

Formulas

Q(x) = x^T A x
det(A - λI) = 0

Theorems

Spectral Theorem
Diagonalization

Suitable Grade Level

Grades 11-12