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Chapter 5: Problem 17

Diagonalize the quadratic forms by finding an orthogonal matrix \(Q\) such that the change of variable \(\mathbf{x}=\) Qy transforms the given form into one with no cross-product terms. Give Q and the new quadratic form. $$x^{2}+z^{2}-2 x y+2 y z$$

Short Answer

Expert verified
Orthogonal matrix Q is constructed and transforms the form to \( 2y_1^2 - y_2^2 + y_3^2 \).

Step by step solution

01

Identify the Quadratic Form

The given quadratic form is: \( x^{2} + z^{2} - 2xy + 2yz \). This form describes a quadratic expression in terms of the variables \(x, y,\) and \(z\), including cross-product terms \(-2xy\) and \(2yz\).
02

Write the Quadratic Form as a Matrix

We represent the quadratic form as a symmetric matrix expression: \( \mathbf{x}^T A \mathbf{x} \), where \( \mathbf{x} = \begin{bmatrix} x \ y \ z \end{bmatrix} \) and \( A = \begin{bmatrix} 1 & -1 & 0 \ -1 & 0 & 1 \ 0 & 1 & 1 \end{bmatrix} \).
03

Find Eigenvalues of A

Calculate the eigenvalues of matrix \( A \). This involves solving \( \text{det}(A - \lambda I) = 0 \), where \( I \) is the identity matrix. The characteristic equation is \( (\lambda - 1)(\lambda^2 - 1) + (\lambda - 1) = 0 \). Solving this results in eigenvalues \( \lambda_1 = 2, \lambda_2 = -1, \lambda_3 = 1 \).
04

Find Eigenvectors of A

For each eigenvalue, solve \( (A - \lambda I)\mathbf{v} = 0 \) for the eigenvectors. Corresponding to \( \lambda_1 = 2 \), we find \( \mathbf{v}_1 = \frac{1}{\sqrt{2}} \begin{bmatrix} -1 \ 0 \ 1 \end{bmatrix} \). For \( \lambda_2 = -1 \), \( \mathbf{v}_2 = \frac{1}{\sqrt{6}} \begin{bmatrix} 1 \ 2 \ 1 \end{bmatrix} \). And for \( \lambda_3 = 1 \), \( \mathbf{v}_3 = \frac{1}{\sqrt{3}} \begin{bmatrix} 1 \ 0 \ 1 \end{bmatrix} \).
05

Construct the Orthogonal Matrix Q

Construct matrix \( Q \) using the normalized eigenvectors as columns: \( Q = \begin{bmatrix} \frac{-1}{\sqrt{2}} & \frac{1}{\sqrt{6}} & \frac{1}{\sqrt{3}} \ 0 & \frac{2}{\sqrt{6}} & 0 \ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{6}} & \frac{1}{\sqrt{3}} \end{bmatrix} \). This \( Q \) is orthogonal.
06

Transform the Quadratic Form

The transformation \( \mathbf{x} = Q \mathbf{y} \) leads to \( y^T D y \) where \( D = \begin{bmatrix} \lambda_1 & 0 & 0 \ 0 & \lambda_2 & 0 \ 0 & 0 & \lambda_3 \end{bmatrix} = \begin{bmatrix} 2 & 0 & 0 \ 0 & -1 & 0 \ 0 & 0 & 1 \end{bmatrix} \). The new quadratic form is \( 2y_1^2 - y_2^2 + y_3^2 \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Quadratic Forms
In mathematics, a quadratic form represents a homogeneous polynomial of degree two in a number of variables. The standard form is expressed as \[ Q(x) = ax^2 + bxy + cy^2 + ext{other terms} \]where the coefficients determine the curvature and how the variables interact. In this exercise, we started with the quadratic form \[ x^2 + z^2 - 2xy + 2yz, \]which includes cross-product terms like \( -2xy \) and \( 2yz \).

To better understand and simplify this form, we represented it using a symmetric matrix. This approach helps us to identify the geometric properties of the form, such as axes of symmetry and orientation. By transforming the form with an orthogonal matrix, we can eliminate cross-product terms, making it easier to analyze and solve related problems.
Orthogonal Matrix
An orthogonal matrix is a square matrix whose columns and rows are orthogonal unit vectors. Mathematically, a matrix \( Q \) is orthogonal if \( Q^TQ = I, \)where \( Q^T \) is the transpose of \( Q \) and \( I \) is the identity matrix. The significance of orthogonal matrices in quadratic forms lies in their ability to keep the length and angles between vectors unchanged.

In this problem, we constructed the orthogonal matrix \( Q \) from the eigenvectors of matrix \( A. \)This \( Q \) matrix helps us transform the original quadratic form, such that it no longer has cross-product terms, which simplifies our analysis. This transformation yields a diagonal matrix, where the terms align with the principal axes of the form, making it easier to interpret geometrically and solve practical problems.
Eigenvalues and Eigenvectors
Eigenvalues and eigenvectors are fundamental concepts in linear algebra. They tell us about the properties of linear transformations represented by a matrix. Given a matrix \( A \), an eigenvalue \( \lambda \) and a corresponding eigenvector \( \mathbf{v} \) satisfy the equation\[ A\mathbf{v} = \lambda \mathbf{v}. \]

For this exercise, we found the eigenvalues of matrix \( A \) as \( \lambda_1 = 2, \lambda_2 = -1, \lambda_3 = 1. \)Each eigenvalue tells us about the factor by which the corresponding eigenvector is scaled during the transformation.
  • Eigenvalues help in determining the natural modes of variation.
  • Eigenvectors provide the direction in which those variations occur.
By normalizing these eigenvectors, they become the columns of our orthogonal matrix \( Q. \) This matrix enables us to transform the quadratic form into its diagonalized version, making it simpler to analyze and solve.

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Most popular questions from this chapter

Use a translation of axes to put the conic in standard position. Identify the graph, give its equation in the translated coordinate system, and sketch the curve. $$4 x^{2}+2 y^{2}-8 x+12 y+6=0$$

Sometimes the graph of a quadratic equation is a straight line, a pair of straight lines, or a single point. We refer to such a graph as a degenerate conic. It is also possible that the equation is not satisfied for any values of the variables, in which case there is no graph at all and we refer to the conic as an imaginary conic. Identify the conic with the given equation as either degenerate or imaginary and, where possible, sketch the graph. $$x^{2}+2 x y+y^{2}=0$$

Let \(A\) be a real \(2 \times 2\) matrix with complex eigenvalues \(\lambda=a \pm b i\) such that \(b \neq 0\) and \(|\lambda|=1 .\) Prove that every trajectory of the dynamical system \(\mathbf{x}_{k+1}=A \mathbf{x}_{k}\) lies on an ellipse. [Hint: Theorem 4.43 shows that if \(\mathbf{v}\) is an eigenvector corresponding to \(\lambda=a-b i\), then the matrix \(P=[\operatorname{Re} \mathbf{v} \quad \operatorname{Im} \mathbf{v}]\) is invertible and \(A=P\left[\begin{array}{cc}a & -b \\ b & a\end{array}\right] P^{-1} .\) Set \(B=\left(P P^{T}\right)^{-1} .\) Show that the quadratic \(\mathbf{x}^{T} B \mathbf{x}=k\) defines an ellipse for all \(k>0\) and prove that if \(x \text { lies on this ellipse, so does } A x .]\)

Find the orthogonal projection of v onto the subspace \(W\) spanned by the vectors \(\mathbf{u}_{\mathbf{r}}\) (You may assume that the vectors \(\mathbf{u}_{i}\) are orthogonal. ) $$\mathbf{v}=\left[\begin{array}{r} 3 \\ 1 \\ -2 \end{array}\right], \mathbf{u}_{1}=\left[\begin{array}{l} 1 \\ 1 \\ 1 \end{array}\right], \mathbf{u}_{2}=\left[\begin{array}{r} 1 \\ -1 \\ 0 \end{array}\right]$$

Let \(W\) be a subspace of \(\mathbb{R}^{n},\) and let \(\mathbf{x}\) be a vector in \(\mathbb{R}^{n}\). Prove that \(x\) is orthogonal to \(W\) if and only if \(\operatorname{proj}_{w}(\mathbf{x})=0\).
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