Problem 17 - Differential Equations and Linear Algebra

Question

Determine a set of principal axes for the given quadratic form, and reduce the quadratic form to a sum of squares. $$\mathbf{x}^{T} A \mathbf{x}, A=\left[\begin{array}{llll} 3 & 1 & 3 & 1 \\ 1 & 3 & 1 & 3 \\ 3 & 1 & 3 & 1 \\ 1 & 3 & 1 & 3 \end{array}\right]$$

Step-by-Step Solution

Answer

The principal axes of the given quadratic form are given by the eigenvectors of matrix A: \(\mathbf{v}_1=\left[\begin{array}{c}1 \\ 1 \\ 1 \\ 1\end{array}\right], \mathbf{v}_2=\left[\begin{array}{c}-1 \\ 1 \\ -1 \\ 1\end{array}\right], \mathbf{v}_3=\left[\begin{array}{c}1 \\ -1 \\ 1 \\ -1\end{array}\right]\), and \(\mathbf{v}_4=\left[\begin{array}{c}1 \\ -1 \\ -1 \\ 1\end{array}\right]\). After performing a change of variables using these eigenvectors, the quadratic form can be reduced to a sum of squares as \(6(y_1^2+y_2^2)\), where \(\mathbf{y}=P^{T}\mathbf{x}\) and \(P\) is the matrix with eigenvectors as columns.

Step 1: Find the eigenvalues and eigenvectors of matrix A

First, we need to find the eigenvalues and eigenvectors of the given symmetric matrix A: \[A=\left[\begin{array}{cccc} 3 & 1 & 3 & 1 \\\ 1 & 3 & 1 & 3 \\\ 3 & 1 & 3 & 1 \\\ 1 & 3 & 1 & 3 \end{array}\right]\] We compute the eigenvalues by solving the characteristic equation \(\text{det}(A-\lambda I)=0\), obtaining the eigenvalues \(\lambda_1 = 6\) (with algebraic multiplicity 2) and \(\lambda_2 = 0\) (with algebraic multiplicity 2). Next, we find the eigenvectors corresponding to each eigenvalue. For \(\lambda_1 = 6\), we solve the following system of linear equations: \[(A-6I)\mathbf{v}=\left[\begin{array}{cccc} -3 & 1 & 3 & 1 \\\ 1 & -3 & 1 & 3 \\\ 3 & 1 & -3 & 1 \\\ 1 & 3 & 1 & -3 \end{array}\right]\mathbf{v}=\mathbf{0}\] We can find that two eigenvectors associated with \(\lambda_1 = 6\) are \(\mathbf{v}_1=\left[\begin{array}{c}1 \\ 1 \\ 1 \\ 1\end{array}\right]\) and \(\mathbf{v}_2=\left[\begin{array}{c}-1 \\ 1 \\ -1 \\ 1\end{array}\right]\). For \(\lambda_2 = 0\), we have: \[(A-0I)\mathbf{v}=\left[\begin{array}{cccc} 3 & 1 & 3 & 1 \\\ 1 & 3 & 1 & 3 \\\ 3 & 1 & 3 & 1 \\\ 1 & 3 & 1 & 3 \end{array}\right]\mathbf{v}=\mathbf{0}\] We can find that two eigenvectors associated with \(\lambda_2 = 0\) are \(\mathbf{v}_3=\left[\begin{array}{c}1 \\ -1 \\ 1 \\ -1\end{array}\right]\) and \(\mathbf{v}_4=\left[\begin{array}{c}1 \\ -1 \\ -1 \\ 1\end{array}\right]\). We have found our eigenvectors and eigenvalues.

Step 2: Perform a change of variables to rewrite the quadratic form

Now let's rewrite the quadratic form using the eigenvectors found in Step 1 as axes. We define a matrix P formed by the eigenvectors as columns: \[P=\left[\begin{array}{cccc} 1 & -1 & 1 & 1 \\\ 1 & 1 & -1 & -1 \\\ 1 & -1 & 1 & -1 \\\ 1 & 1 & -1 & 1 \end{array}\right]\] We can write the quadratic form in terms of the new variables as follows: \[\mathbf{y}^{T}\Lambda\mathbf{y}\] Where \(\mathbf{y}=P^{T}\mathbf{x}\), and \(\Lambda\) is the diagonal matrix formed by the eigenvalues: \[\Lambda=\left[\begin{array}{cccc} 6 & 0 & 0 & 0 \\\ 0 & 6 & 0 & 0 \\\ 0 & 0 & 0 & 0 \\\ 0 & 0 & 0 & 0 \end{array}\right]\]

Step 3: Reduce the quadratic form to a sum of squares

Finally, we can rewrite the quadratic form as a sum of squares: \[\mathbf{y}^{T}\Lambda\mathbf{y}=\lambda_1 (y_1^2+y_2^2)+\lambda_2 (y_3^2+y_4^2)=6(y_1^2+y_2^2)\] Now we have reduced the given quadratic form to a sum of squares in terms of the new variables defined by the principal axes.

Key Concepts

Quadratic Form

A quadratic form is a special type of polynomial where each term is of degree two. It can be expressed mathematically for a vector \( \mathbf{x} \) and matrix \( A \) as \( \mathbf{x}^T A \mathbf{x} \). This represents a hyper-surface, and understanding its properties allows us to simplify complex systems. In our exercise, we began by focusing on the quadratic form defined by the specific matrix \( A \). The task was to determine its principal axes through eigenvalues and eigenvectors, and then reduce it to a more manageable sum of squares form. Principal axes transformation helps in visualizing and analyzing the given quadratic form by simplifying it into sums of squares, which are easier to interpret.

Eigenvalues and Eigenvectors

Eigenvalues and eigenvectors are core concepts in linear algebra, essential for understanding transformations like those needed in quadratic forms. An eigenvalue is a scalar that indicates how much an eigenvector stretches or contracts during a transformation represented by a matrix \( A \). Meanwhile, an eigenvector is a non-zero vector that changes at most by its scalar factor when that transformation is applied. For matrix \( A \) from the exercise, finding eigenvalues and eigenvectors was the pivotal first step. We solved the characteristic equation \( \text{det}(A - \lambda I) = 0 \) to find eigenvalues \( \lambda_1 = 6 \) and \( \lambda_2 = 0 \). Each eigenvalue led us to corresponding eigenvectors, which formed a new basis. This transformation matrix \( P \) allows us to express the quadratic form in a simplified, diagonal matrix \( \Lambda \), capturing the essence of the system's symmetry.

Sum of Squares

Reducing a quadratic form to a sum of squares involves expressing the equation such that each term is simple and self-contained, resembling \( y^2 \) terms. This transformation simplifies analyzing and computing outcomes from the quadratic form. It is akin to finding a more understandable version of a complex equation. In the given exercise, once eigenvectors were used as principal axes, the quadratic form \( \mathbf{x}^T A \mathbf{x} \) was expressed in terms of new variables \( \mathbf{y} \) using \( y = P^T x \). The matrix \( \Lambda \), consisting of the eigenvalues, enabled us to simplify the form to \( 6(y_1^2 + y_2^2) \). This represents two independent squares weighted by the eigenvalue. Simplifying into a sum of squares provides clarity and ease in further mathematical or real-world interpretations of the quadratic form.