Problem 3 - Advanced Engineering Mathematics

Question

In Problems \(1-4\), (a) verify that the indicated column vectors are eigenvectors of the given symmetric matrix, (b) identify the corresponding eigenvalues, and (c) verify that the column vectors are orthogonal. $$ \begin{aligned} &\left(\begin{array}{rrr} 5 & 13 & 0 \\ 13 & 5 & 0 \\ 0 & 0 & -8 \end{array}\right)\\\ &\left(\begin{array}{l} \frac{\sqrt{2}}{2} \\ \frac{\sqrt{2}}{2} \\ 0 \end{array}\right),\left(\begin{array}{c} \frac{\sqrt{3}}{3} \\ -\frac{\sqrt{3}}{3} \\ \frac{\sqrt{3}}{3} \end{array}\right), \quad\left(\begin{array}{c} \frac{\sqrt{6}}{6} \\ -\frac{\sqrt{6}}{6} \\ -\frac{\sqrt{6}}{3} \end{array}\right) \end{aligned} $$

Step-by-Step Solution

Answer

Eigenvectors are: \(9\), \(-8\), \(-2\); they are orthogonal.

Step 1: Verify Eigenvector 1

Multiply the matrix \( \begin{pmatrix} 5 & 13 & 0 \ 13 & 5 & 0 \ 0 & 0 & -8 \end{pmatrix} \) by the vector \( \begin{pmatrix} \frac{\sqrt{2}}{2} \ \frac{\sqrt{2}}{2} \ 0 \end{pmatrix} \). Calculate the result: \( \begin{pmatrix} 5 \cdot \frac{\sqrt{2}}{2} + 13 \cdot \frac{\sqrt{2}}{2} \ 13 \cdot \frac{\sqrt{2}}{2} + 5 \cdot \frac{\sqrt{2}}{2} \ 0 \end{pmatrix} = 9 \cdot \begin{pmatrix} \frac{\sqrt{2}}{2} \ \frac{\sqrt{2}}{2} \ 0 \end{pmatrix} \). Thus, this vector is an eigenvector with eigenvalue \(9\).

Step 2: Verify Eigenvector 2

Multiply the matrix by the vector \( \begin{pmatrix} \frac{\sqrt{3}}{3} \ -\frac{\sqrt{3}}{3} \ \frac{\sqrt{3}}{3} \end{pmatrix} \). Calculate: \( \begin{pmatrix} 5 \cdot \frac{\sqrt{3}}{3} + 13 \cdot -\frac{\sqrt{3}}{3} \ 13 \cdot \frac{\sqrt{3}}{3} + 5 \cdot -\frac{\sqrt{3}}{3} \ 0 + 0 + (-8) \cdot \frac{\sqrt{3}}{3} \end{pmatrix} = \begin{pmatrix} -\frac{8\sqrt{3}}{3} \ -\frac{8\sqrt{3}}{3} \ -\frac{8\sqrt{3}}{3} \end{pmatrix} \). This confirms the vector is an eigenvector with eigenvalue \(-8\).

Step 3: Verify Eigenvector 3

Multiply the matrix by the vector \( \begin{pmatrix} \frac{\sqrt{6}}{6} \ -\frac{\sqrt{6}}{6} \ -\frac{\sqrt{6}}{3} \end{pmatrix} \). Calculate: \( \begin{pmatrix} 5 \cdot \frac{\sqrt{6}}{6} + 13 \cdot -\frac{\sqrt{6}}{6} \cdot 0 \end{pmatrix} = \begin{pmatrix} -2 \cdot \frac{\sqrt{6}}{6} \ -2 \cdot \frac{\sqrt{6}}{6} \ \frac{8\sqrt{6}}{3} \end{pmatrix} \). Simplifying gives \(-2\begin{pmatrix} \frac{\sqrt{6}}{6} \ -\frac{\sqrt{6}}{6} \ -\frac{\sqrt{6}}{3} \end{pmatrix} \), confirming it is an eigenvector with eigenvalue \(-2\).

Step 4: Verify Orthogonality

Check the dot product between each pair of vectors. For vectors 1 and 2: \( \left( \frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}, 0 \right) \cdot \left( \frac{\sqrt{3}}{3}, -\frac{\sqrt{3}}{3}, \frac{\sqrt{3}}{3} \right) = 0 \). For vectors 1 and 3: \( \left( \frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}, 0 \right) \cdot \left( \frac{\sqrt{6}}{6}, -\frac{\sqrt{6}}{6}, -\frac{\sqrt{6}}{3} \right) = 0 \). For vectors 2 and 3: \( \left( \frac{\sqrt{3}}{3}, -\frac{\sqrt{3}}{3}, \frac{\sqrt{3}}{3} \right) \cdot \left( \frac{\sqrt{6}}{6}, -\frac{\sqrt{6}}{6}, -\frac{\sqrt{6}}{3} \right) = 0 \). All dot products are zero, thus confirming orthogonality.

Key Concepts

Eigenvalues

Eigenvalues are special numbers that provide insight into the properties of a matrix. When a matrix acts on its eigenvector, the output is simply a scaled version of that eigenvector. This scaling factor is the eigenvalue. For example, in the exercise provided, when the matrix \( \begin{pmatrix} 5 & 13 & 0 \ 13 & 5 & 0 \ 0 & 0 & -8 \end{pmatrix} \) was multiplied by the vector \( \begin{pmatrix} \frac{\sqrt{2}}{2} \ \frac{\sqrt{2}}{2} \ 0 \end{pmatrix} \), the result was 9 times the original vector. This means that the eigenvalue for this eigenvector is 9.

Identifying eigenvalues helps in various applications like analyzing stability of systems or transforming data. Each distinct eigenvalue can provide different insights, often representing unique modes of action within the system described by the matrix.

Orthogonality

Orthogonality is a concept that refers to the relationship between vectors, where two vectors are orthogonal if their dot product is zero. This means they are perpendicular to each other in a geometrical sense. In the provided exercise, the vectors \( \begin{pmatrix} \frac{\sqrt{2}}{2} \ \frac{\sqrt{2}}{2} \ 0 \end{pmatrix} \), \( \begin{pmatrix} \frac{\sqrt{3}}{3} \ -\frac{\sqrt{3}}{3} \ \frac{\sqrt{3}}{3} \end{pmatrix} \), and \( \begin{pmatrix} \frac{\sqrt{6}}{6} \ -\frac{\sqrt{6}}{6} \ -\frac{\sqrt{6}}{3} \end{pmatrix} \) were shown to be orthogonal, as the dot products between each pair were zero.

Orthogonality is important in various mathematical and engineering applications because it often simplifies calculations. When working with orthogonal vectors, especially in higher dimensions, problems become more manageable, and solution methods become more apparent. It’s particularly significant in fields like computer graphics, signal processing, and systems engineering.

Symmetric Matrix

A symmetric matrix is one that is equal to its transpose. In other words, the elements are mirrored around the main diagonal. A symmetric matrix, such as \( \begin{pmatrix} 5 & 13 & 0 \ 13 & 5 & 0 \ 0 & 0 & -8 \end{pmatrix} \), has many significant properties.

- Eigenvalues of symmetric matrices are always real numbers, which simplifies computational processes.
- Symmetric matrices have orthogonal eigenvectors, which means that the eigenvectors are perpendicular to each other. This is why in the exercise, checking the orthogonality of given vectors confirmed they are eigenvectors of the symmetric matrix.

Symmetric matrices often pop up in practical scenarios, such as in the representation of quadratic forms, in physics for simplifying systems, and in statistics when working with covariance matrices. Understanding their properties can simplify otherwise complex mathematical and computational problems.