Problem 15 - Advanced Engineering Mathematics

Question

In Problems 11-18, proceed as in Example 3 to construct an orthogonal matrix from the eigenvectors of the given symmetric matrix. (The answers are not unique.) $$ \left(\begin{array}{lll} 1 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 1 \end{array}\right) $$

Step-by-Step Solution

Answer

Orthogonal matrix is constructed using eigenvectors: \( Q = \begin{pmatrix} \frac{1}{\sqrt{2}} & 0 & \frac{1}{\sqrt{2}} \\ 0 & 1 & 0 \\ -\frac{1}{\sqrt{2}} & 0 & \frac{1}{\sqrt{2}} \end{pmatrix} \).

Step 1: Identify the Matrix

The given symmetric matrix is \( A = \begin{pmatrix} 1 & 0 & 1 \ 0 & 1 & 0 \ 1 & 0 & 1 \end{pmatrix} \). Our task is to construct an orthogonal matrix from its eigenvectors.

Step 2: Find Eigenvalues

Calculate the eigenvalues of matrix \( A \), by solving the characteristic equation \( \text{det}(A - \lambda I) = 0 \). For this matrix, the characteristic polynomial is \((1-\lambda)^2(1-\lambda) - 0 + (1-\lambda) \cdot 1 \cdot (1-\lambda) = (1-\lambda)^2((1-\lambda) - 1) = (1 - \lambda)^2 (1-2\lambda) \). Solving \((1-\lambda)^2 (1-2\lambda) = 0\), we find eigenvalues: \(\lambda_1 = 0\) and \(\lambda_2 = \lambda_3 = 2\).

Step 3: Find Eigenvectors

For \(\lambda_1 = 0\), solve \( (A - 0I)x = 0 \). The system becomes \( \begin{pmatrix} 1 & 0 & 1 \ 0 & 1 & 0 \ 1 & 0 & 1 \end{pmatrix} \begin{pmatrix} x_1 \ x_2 \ x_3 \end{pmatrix} = \begin{pmatrix} 0 \ 0 \ 0 \end{pmatrix} \). We get eigenvector \( v_1 = \begin{pmatrix} 1 \ 0 \ -1 \end{pmatrix} \). For \(\lambda_2 = 2\), solve \( (A - 2I)x = 0 \), which gives eigenvectors \( v_2 = \begin{pmatrix} 0 \ 1 \ 0 \end{pmatrix} \) and \( v_3 = \begin{pmatrix} 1 \ 0 \ 1 \end{pmatrix} \).

Step 4: Normalize the Eigenvectors

Normalize the eigenvectors to make sure they are unit vectors. For \(v_1 = \begin{pmatrix} 1 \ 0 \ -1 \end{pmatrix}\), normalize to get \( \frac{1}{\sqrt{2}} \begin{pmatrix} 1 \ 0 \ -1 \end{pmatrix} \). For \(v_2 = \begin{pmatrix} 0 \ 1 \ 0 \end{pmatrix}\), it is already a unit vector: \( \begin{pmatrix} 0 \ 1 \ 0 \end{pmatrix} \). For \(v_3 = \begin{pmatrix} 1 \ 0 \ 1 \end{pmatrix}\), normalize to get \( \frac{1}{\sqrt{2}} \begin{pmatrix} 1 \ 0 \ 1 \end{pmatrix} \).

Step 5: Construct the Orthogonal Matrix

The orthogonal matrix \( Q \) is constructed by using the normalized eigenvectors as columns: \( Q = \begin{pmatrix} \frac{1}{\sqrt{2}} & 0 & \frac{1}{\sqrt{2}} \ 0 & 1 & 0 \ -\frac{1}{\sqrt{2}} & 0 & \frac{1}{\sqrt{2}} \end{pmatrix} \). This matrix is orthogonal because it consists of orthonormal vectors, meaning \( Q^T Q = I \).

Key Concepts

Eigenvalues and Eigenvectors

Understanding eigenvalues and eigenvectors is crucial when dealing with matrices, especially symmetric ones. An eigenvalue is a scalar that indicates how much the direction of the eigenvector is stretched or compressed. To find the eigenvalues of a matrix, we solve the characteristic equation \( \text{det}(A - \lambda I) = 0 \), where \( A \) is the original matrix and \( \lambda \) represents the eigenvalues.

Once the eigenvalues are obtained, eigenvectors are found by solving the equation \( (A - \lambda I)x = 0 \). These eigenvectors provide essential directions in the transformation represented by the matrix.

For a symmetric matrix like the one in our example, the eigenvectors turn out to be orthogonal, simplifying the creation of an orthogonal matrix. In this specific case, the symmetric matrix \( A \) yields eigenvalues \( \lambda_1 = 0 \), and \( \lambda_2 = \lambda_3 = 2 \), with corresponding eigenvectors which are fundamental for further steps.

Matrix Normalization

Normalization is the process of adjusting the length of vectors to be unit length, which means they all have a magnitude of 1. It's important because it ensures that vectors are easy to compare and maintain their direction while simplifying computations.

For eigenvectors:

• Normalize by dividing each component of the vector by its magnitude.
• The magnitude (or length) of a vector \( v = \begin{pmatrix} x \ y \ z \end{pmatrix} \) is calculated using \( \sqrt{x^2 + y^2 + z^2} \).

Taking the matrix from our example, we see normalization applied to vectors like \( \begin{pmatrix} 1 \ 0 \ -1 \end{pmatrix} \), resulting in an easier to handle unit vector \( \frac{1}{\sqrt{2}} \begin{pmatrix} 1 \ 0 \ -1 \end{pmatrix} \).

This step ensures that all the eigenvectors placed into our orthogonal matrix are orthonormal, which is essential for achieving the property where \( Q^T Q = I \).

Characteristic Polynomial

The characteristic polynomial helps us determine the eigenvalues of a matrix. For matrix \( A \), the polynomial is derived from \( \text{det}(A - \lambda I) \). This process involves subtracting \( \lambda \) along the main diagonal of the matrix \( A \), leading to a new matrix whose determinant must be calculated.

In simpler terms, the characteristic polynomial is a mathematical expression that represents how the values of \( \lambda \) interact with \( A \). Solving it involves finding the roots, which correspond to the eigenvalues of the matrix.

In our exercise's symmetric matrix, the characteristic polynomial becomes \( (1-\lambda)^2(1-2\lambda) \) and solving \( (1-\lambda)^2(1-2\lambda) = 0 \) gives us the eigenvalues \( \lambda_1 = 0 \) and \( \lambda_2 = \lambda_3 = 2 \). These results are primal for finding the subsequent eigenvectors and effectively utilizing them.

Symmetric Matrix

A symmetric matrix is characterized by having equal elements mirrored across its diagonal. Mathematically, a matrix \( A \) is symmetric if \( A = A^T \). Symmetric matrices have the special property that all its eigenvalues are real, and the resulting eigenvectors are orthogonal.

These properties simplify several computations, including the transformation into an orthogonal matrix. In our example, the original matrix is symmetric:

• Its diagonal elements are equal across its main axis.
• Due to the matrix's symmetry, finding eigenvectors becomes straightforward, with the guarantee of orthogonality simplifying normalization and orthogonal matrix construction.

Recognizing symmetric matrices accelerates problem-solving in linear algebra, particularly in constructing orthogonal matrices and in many applications like principal component analysis.