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Chapter 7: Problem 9

Classify the quadratic forms in Exercises \(9-18 .\) Then make a change of variable, \(\mathbf{x}=P \mathbf{y},\) that transforms the quadratic form into one with no cross-product term. Write the new quadratic form. Construct \(P\) using the methods of Section \(7.1 .\) $$ 4 x_{1}^{2}-4 x_{1} x_{2}+4 x_{2}^{2} $$

Short Answer

Expert verified
The quadratic form is positive definite and can be transformed to \(6y_1^2 + 2y_2^2\) with \( P = \begin{bmatrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \end{bmatrix} \).

Step by step solution

01

Express in Matrix Form

We start by rewriting the given quadratic form \(4x_1^2 - 4x_1x_2 + 4x_2^2\) in matrix form. Here, it can be expressed as \( \mathbf{x}^T A \mathbf{x} \), where \( A \) is the symmetric matrix of the quadratic form, and \( \mathbf{x} = [x_1, x_2]^T \).The matrix \( A \) is:\[A = \begin{bmatrix} 4 & -2 \ -2 & 4 \end{bmatrix}.\]
02

Determine Eigenvalues and Eigenvectors of A

To remove cross-product terms, we need to diagonalize matrix \( A \). Start by finding the eigenvalues by solving the characteristic equation, \( |A - \lambda I| = 0 \).The characteristic equation is:\[\begin{vmatrix} 4-\lambda & -2 \ -2 & 4-\lambda \end{vmatrix} = (4-\lambda)^2 - (-2)^2 = \lambda^2 - 8\lambda + 12 = 0.\]Using the quadratic formula, the eigenvalues are:\[\lambda_1 = 6, \quad \lambda_2 = 2.\]
03

Find Eigenvectors for Each Eigenvalue

For \(\lambda_1 = 6\), solve \((A - 6I)\mathbf{v} = \mathbf{0}\):\[\begin{bmatrix} -2 & -2 \ -2 & -2 \end{bmatrix} \begin{bmatrix} v_1 \ v_2 \end{bmatrix} = \begin{bmatrix} 0 \ 0 \end{bmatrix}.\] This simplifies to \(v_1 = -v_2\), leading to the eigenvector \(\mathbf{v}_1 = \begin{bmatrix} 1 \ -1 \end{bmatrix}\).For \(\lambda_2 = 2\), solve \((A - 2I)\mathbf{v} = \mathbf{0}\):\[\begin{bmatrix} 2 & -2 \ -2 & 2 \end{bmatrix} \begin{bmatrix} v_1 \ v_2 \end{bmatrix} = \begin{bmatrix} 0 \ 0 \end{bmatrix}.\]This gives \(v_1 = v_2\), leading to the eigenvector \(\mathbf{v}_2 = \begin{bmatrix} 1 \ 1 \end{bmatrix}\).
04

Construct the Orthogonal Matrix P

Matrix \( P \) is formed by normalizing the eigenvectors and using them as columns. Normalize \(\mathbf{v}_1\) and \(\mathbf{v}_2\):\[\mathbf{u}_1 = \frac{1}{\sqrt{2}} \begin{bmatrix} 1 \ -1 \end{bmatrix}, \quad \mathbf{u}_2 = \frac{1}{\sqrt{2}} \begin{bmatrix} 1 \ 1 \end{bmatrix}.\]Thus, \( P = \begin{bmatrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \ -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \end{bmatrix}\).
05

Transform Quadratic Form to Remove Cross-Product Terms

The new variable \( \mathbf{y} \) is given by \( \mathbf{x} = P \mathbf{y} \). The quadratic form in terms of \( \mathbf{y} \) is \( \mathbf{y}^T D \mathbf{y} \), where \( D \) is the diagonal matrix of eigenvalues:\[D = \begin{bmatrix} 6 & 0 \ 0 & 2 \end{bmatrix}.\]Therefore, the new quadratic form is:\[6y_1^2 + 2y_2^2.\]
06

Classify the Quadratic Form

The quadratic form can be classified based on the signs of its eigenvalues. Both eigenvalues are positive (6 and 2), indicating that the quadratic form is positive definite.

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Key Concepts

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

Eigenvectors
Eigenvectors play a critical role when working with quadratic forms. They are vectors that, when multiplied by a matrix, only scale in magnitude without changing direction. For a given matrix \(A\), an eigenvector \( \mathbf{v} \) satisfies the equation \( A\mathbf{v} = \lambda \mathbf{v} \), where \( \lambda \) is the corresponding eigenvalue.
  • Eigenvectors help diagonalize a matrix, which is crucial in simplifying quadratic forms.
  • They provide a way to express the matrix in a simpler form, identifying key directions (eigenvectors) along which the quadratic form is stretched or compressed (eigenvalues).
You can think of eigenvectors as the "natural" directions in which the action of the matrix is purely scaling.Finding eigenvectors involves solving the system of equations formed by \( (A - \lambda I) \mathbf{v} = \mathbf{0} \). For each eigenvalue, this leads to corresponding eigenvectors that are not solely determined; they typically form a subspace. Normalizing these vectors is often necessary to ease computations and align with orthogonal matrix requirements.
Positive Definite
A quadratic form is classified as positive definite if all its eigenvalues are positive. This property is important for several reasons:
  • Positive definite matrices imply that the quadratic form will always yield positive values for any non-zero vector input.
  • This condition assures that the quadratic form does not create any directions in which it outputs negative values, ensuring stability and robustness in applications such as optimization and stability analysis.
  • Positive definiteness is often required for mathematical programming algorithms, such as those used in least squares problems and convex optimization.
In our solution, the eigenvalues were found to be 6 and 2, both positive, confirming that the quadratic form is positive definite. Knowing this, we can predict that the graph of the quadratic form represents an elliptic paraboloid opening upwards.
Diagonalization
Diagonalization is the process of converting a matrix into a diagonal matrix through a similarity transformation involving its eigenvectors. This transformation is invaluable when dealing with quadratic forms, as it offers a simplified representation.
  • A matrix \( A \) can be diagonalized if there exists a matrix \( P \), composed of its eigenvectors, such that \( P^{-1}AP = D \), where \( D \) is a diagonal matrix with the eigenvalues of \( A \) on the diagonal.
  • For the given quadratic form, we diagonalized the matrix to eliminate cross-product terms, making it easier to analyze or graph the quadratic form.
  • The resulting diagonal matrix \( D \) reflects the quadratic form's principal axes by showing how it scales along these axes.
Diagonalization simplifies many operations, such as finding matrix powers and solving systems of equations, by reducing them to calculations involving the simpler, diagonal matrix.
Orthogonal Matrix
An orthogonal matrix is a square matrix whose columns and rows are orthonormal vectors, meaning they are perpendicular and have unit length. For a matrix \( P \) to be orthogonal, it must satisfy \( P^TP = PP^T = I \), where \( I \) is the identity matrix.
  • Orthogonal matrices simplify mathematical computations, as they preserve lengths and angles.
  • When transforming a quadratic form, an orthogonal matrix \( P \) helps in changing variables in such a way that the new variables are uncorrelated, removing cross-product terms from the form.
  • The orthogonal matrix constructed in the solution is based on normalized eigenvectors of the matrix \( A \), ensuring that transformations are stable and reversible.
In our case, the matrix \( P \) created through eigenvectors normalization leads to a transparent change of variable, making the representation of the quadratic form straightforward and devoid of any cross terms.

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Most popular questions from this chapter

Find an SVD of each matrix [Hint: In Exercise 11, one choice for \(U\) is \(\left[\begin{array}{rrr}{-1 / 3} & {2 / 3} & {2 / 3} \\ {2 / 3} & {-1 / 3} & {2 / 3} \\ {2 / 3} & {2 / 3} & {-1 / 3}\end{array}\right]\) In Exercise \(12,\) one column of \(U\) can be \(\left[\begin{array}{c}{1 / \sqrt{6}} \\ {-2 / \sqrt{6}} \\ {1 / \sqrt{6}}\end{array}\right].\)] \(\left[\begin{array}{ll}{7} & {1} \\ {5} & {5} \\ {0} & {0}\end{array}\right]\)

In Exercises 1 and \(2,\) convert the matrix of observations to meandeviation form, and construct the sample covariance matrix. $$ \left[\begin{array}{cccccc}{1} & {5} & {2} & {6} & {7} & {3} \\ {3} & {11} & {6} & {8} & {15} & {11}\end{array}\right] $$

Find the singular values of the matrices. \(\left[\begin{array}{ll}{2} & {3} \\ {0} & {2}\end{array}\right]\)

Show that if an \(n \times n\) matrix \(A\) is positive definite, then there exists a positive definite matrix \(B\) such that \(A=B^{T} B .[\text { Hint: }\) Write \(A=P D P^{T},\) with \(P^{T}=P^{-1} .\) Produce a diagonal matrix \(C\) such that \(D=C^{T} C,\) and let \(B=P C P^{T} .\) Show that \(B\) works. \(]\)

Compute the quadratic form \(\mathbf{x}^{T} A \mathbf{x},\) for \(A=\left[\begin{array}{ccc}{3} & {2} & {0} \\ {2} & {2} & {1} \\ {0} & {1} & {0}\end{array}\right]\) and $$ \text { a. }\mathbf{x}=\left[\begin{array}{l}{x_{1}} \\ {x_{2}} \\\ {x_{3}}\end{array}\right] \quad \text { b. } \mathbf{x}=\left[\begin{array}{r}{-2} \\ {-1} \\ {5}\end{array}\right] \quad \text { c. } \mathbf{x}=\left[\begin{array}{c}{1 / \sqrt{2}} \\ {1 / \sqrt{2}} \\ {1 / \sqrt{2}}\end{array}\right] $$
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