Problem 16 - 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. $$\mathbf{x}^{T} A \mathbf{x}, \quad A=\left[\begin{array}{rrr} 1 & 1 & -1 \\ 1 & 1 & 1 \\ -1 & 1 & 1 \end{array}\right]$$

Step-by-Step Solution

Answer

The principal axes of the quadratic form are the eigenvectors of the given matrix A: \(v_1 = \frac{1}{\sqrt{3}} \begin{bmatrix}1 \\ -1 \\ 1\end{bmatrix}\), \(v_2 = \frac{1}{\sqrt{2}} \begin{bmatrix}0 \\ 1 \\ 1\end{bmatrix}\), and \(v_3 = \frac{1}{\sqrt{6}} \begin{bmatrix}2 \\ -1 \\ -1\end{bmatrix}\). The quadratic form can be expressed as a sum of squares as follows: \(\mathbf{x}^{T} A \mathbf{x} = 0\left(\frac{1}{\sqrt{3}}(x_1 - x_2 + x_3)\right)^2 + 2\left(\frac{1}{\sqrt{2}}(x_2 + x_3)\right)^2 + 2\left(\frac{1}{\sqrt{6}}(2x_1 - x_2 - x_3)\right)^2\)

Step 1: Find the eigenvalues and eigenvectors of A.

First, we need to find the eigenvalues and eigenvectors of A. To find the eigenvalues, we compute the characteristic equation: \(\det(A - \lambda I) = 0\), where \(\lambda\) represents the eigenvalues and I is the identity matrix. Given matrix A: \(A=\left[\begin{array}{rrr} 1 & 1 & -1 \\ 1 & 1 & 1 \\ -1 & 1 & 1 \end{array}\right]\) Compute the determinant: \(\det(A - \lambda I) = \left|\begin{array}{ccc} 1-\lambda & 1 & -1 \\ 1 & 1-\lambda & 1 \\ -1 & 1 & 1-\lambda \end{array}\right| = -2 \lambda^{3}+6 \lambda^{2}-6 \lambda\) The eigenvalues are the roots of the equation: \(\lambda(\lambda - 2)^{2} = 0\) The eigenvalues are \(\lambda_1 = 0\) and \(\lambda_2 = \lambda_3 = 2\). Now, we will find the eigenvectors associated with each eigenvalue. For \(\lambda_1 = 0\), we solve the equation \((A - 0I)v_1 = 0\): \(\left[\begin{array}{rrr} 1 & 1 & -1 \\ 1 & 1 & 1 \\ -1 & 1 & 1 \end{array}\right] \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}\) The solution of this system is \(v_1 = \frac{1}{\sqrt{3}} \begin{bmatrix}1 \\ -1 \\ 1\end{bmatrix}\). For \(\lambda_2 = \lambda_3 = 2\), we solve the equation \((A - 2I)v_2 = 0\): \(\left[\begin{array}{rrr} -1 & 1 & -1 \\ 1 & -1 & 1 \\ -1 & 1 & -1 \end{array}\right] \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}\) The solutions of this system are \(v_2 = \frac{1}{\sqrt{2}} \begin{bmatrix}0 \\ 1 \\ 1\end{bmatrix}\) and \(v_3 = \frac{1}{\sqrt{6}} \begin{bmatrix}2 \\ -1 \\ -1\end{bmatrix}\).

Step 2: Express the quadratic form as a sum of squares.

Now that we have the eigenvectors (principal axes), we can express the quadratic form as a sum of squares using the following formula: \(\mathbf{x}^{T} A \mathbf{x} = \sum_{i=1}^{3} \lambda_i (\mathbf{v}_i^T\mathbf{x})^2\) where \(\mathbf{v}_i\) are the eigenvectors and \(\lambda_i\) are the eigenvalues. Using our computed eigenvalues and eigenvectors, the quadratic form can be expressed as: \(\mathbf{x}^{T} A \mathbf{x} = 0\left(\frac{1}{\sqrt{3}}(x_1 - x_2 + x_3)\right)^2 + 2\left(\frac{1}{\sqrt{2}}(x_2 + x_3)\right)^2 + 2\left(\frac{1}{\sqrt{6}}(2x_1 - x_2 - x_3)\right)^2\) This is the quadratic form expressed as a sum of squares with respect to its principal axes (eigenvectors).

Key Concepts

Eigenvalues and Eigenvectors

In the world of linear algebra, eigenvalues and eigenvectors are fundamental tools that allow us to simplify complex linear transformations. In essence, an eigenvalue is a scalar that shows how much an eigenvector, which is a special vector, is stretched during a transformation defined by a matrix. To find these, we typically compute the solution to the characteristic equation: \(\det(A - \lambda I) = 0\), where \(A\) is our matrix of interest, \(I\) is the identity matrix, and \(\lambda\) are the eigenvalues.
For the given matrix in the problem:

• We solved \(\lambda(\lambda - 2)^{2} = 0\) and found out the eigenvalues are \(\lambda_1 = 0\) and \(\lambda_2 = \lambda_3 = 2\).
• Next, we determine the eigenvectors corresponding to each eigenvalue.

When the eigenvectors are found, they form a set of vectors that can help in various tasks, such as simplifying transformations and diagonalizing matrices.
Understanding these concepts is critical in reducing complex quadratic forms, just like the task in our exercise.

Principal Axes

Principal axes are related to eigenvectors, but they are specifically used in the context of quadratic forms. In simple terms, a principal axis is a direction along which a quadratic form achieves either a maximum or a minimum value. When we express a quadratic form in terms of its principal axes, it becomes significantly simpler.
For our exercise, after finding the eigenvectors (\(v_1\), \(v_2\), and \(v_3\)), we use them to define these principal directions:

• \(v_1 = \frac{1}{\sqrt{3}} \begin{bmatrix} 1 \ -1 \ 1 \end{bmatrix}\)
• \(v_2 = \frac{1}{\sqrt{2}} \begin{bmatrix} 0 \ 1 \ 1 \end{bmatrix}\)
• \(v_3 = \frac{1}{\sqrt{6}} \begin{bmatrix} 2 \ -1 \ -1 \end{bmatrix}\)

Using these principal axes, we can easily express and work with the quadratic form in its simplest form. This is especially useful in calculations in physics and engineering, where translating effects in one direction to another are crucial.

Sum of Squares

In essence, expressing a quadratic form as a sum of squares is all about simplifying the expression. A quadratic form \(\mathbf{x}^{T}A\mathbf{x}\) can be reduced to a sum of squared terms, which presents a clear and solvable set of equations. The formula we use is:

• \(\mathbf{x}^{T}A\mathbf{x} = \sum_{i=1}^{n} \lambda_i (\mathbf{v}_i^T\mathbf{x})^2\)

where \(\lambda_i\) are the eigenvalues and \(\mathbf{v}_i\) are the eigenvectors. For the exercised problem, after calculating the needed values, we rewrite the form as:

• \(0\left(\frac{1}{\sqrt{3}}(x_1 - x_2 + x_3)\right)^2 + 2\left(\frac{1}{\sqrt{2}}(x_2 + x_3)\right)^2 + 2\left(\frac{1}{\sqrt{6}}(2x_1 - x_2 - x_3)\right)^2\)

The task of converting a quadratic form into this format is helpful in optimization problems, where you need clear and concise equations to find maximum, minimum, or saddle points.