Problem 25 - Advanced Engineering Mathematics

Question

In Problems, the given matrix \(\mathbf{A}\) is symmetric. Find an orthogonal matrix \(\mathbf{P}\) that diagonalizes \(\mathbf{A}\) and the diagonal matrix \(\mathbf{D}\) such that \(\mathbf{D}=\mathbf{P}^{T} \mathbf{A P}\) $$ \left(\begin{array}{lll} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{array}\right) $$

Step-by-Step Solution

Answer

\( \mathbf{P} \) and \( \mathbf{D} \) are found; \( \mathbf{P} = \begin{pmatrix} \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} & 0 \\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & 0 \\ 0 & 0 & 1 \end{pmatrix} \) and \( \mathbf{D} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1 \end{pmatrix} \).

Step 1: Identify the Symmetric Matrix

The given matrix \( \mathbf{A} \) is: \[\mathbf{A} = \begin{pmatrix} 0 & 1 & 0 \ 1 & 0 & 0 \ 0 & 0 & 1 \end{pmatrix}.\] This matrix is symmetric since \( a_{ij} = a_{ji} \) for all elements.

Step 2: Find Eigenvalues

To diagonalize \( \mathbf{A} \), first find the eigenvalues by solving the characteristic equation \( \det(\mathbf{A} - \lambda \mathbf{I}) = 0 \). \[\det\begin{pmatrix} -\lambda & 1 & 0 \ 1 & -\lambda & 0 \ 0 & 0 & 1 - \lambda \end{pmatrix} = (\lambda^2 - 1)(1-\lambda) = 0\] Solving this gives the eigenvalues: \( \lambda_1 = 1, \lambda_2 = 1, \lambda_3 = -1 \).

Step 3: Find Eigenvectors

For \( \lambda_1 = 1 \) and \( \lambda_2 = 1 \), solve \( (\mathbf{A} - \mathbf{I})\mathbf{x} = 0 \). \[\begin{pmatrix} -1 & 1 & 0 \ 1 & -1 & 0 \ 0 & 0 & 0 \end{pmatrix}\begin{pmatrix} x_1 \ x_2 \ x_3 \end{pmatrix} = \begin{pmatrix} 0 \ 0 \ 0 \end{pmatrix}\] The solution gives the eigenspace spanned by \( \begin{pmatrix} 1 \ 1 \ 0 \end{pmatrix} \). For \( \lambda_3 = -1 \), solve \( (\mathbf{A} + \mathbf{I})\mathbf{x} = 0 \): \[\begin{pmatrix} 1 & 1 & 0 \ 1 & 1 & 0 \ 0 & 0 & 2 \end{pmatrix}\begin{pmatrix} x_1 \ x_2 \ x_3 \end{pmatrix} = \begin{pmatrix} 0 \ 0 \ 0 \end{pmatrix}\] This provides the eigenspace spanned by \( \begin{pmatrix} -1 \ 1 \ 0 \end{pmatrix} \) and \( \begin{pmatrix} 0 \ 0 \ 1 \end{pmatrix} \).

Step 4: Normalize and Form Matrix P

Normalize the eigenvectors to form an orthogonal matrix \( \mathbf{P} \): \[\mathbf{p}_1 = \frac{1}{\sqrt{2}} \begin{pmatrix} 1 \ 1 \ 0 \end{pmatrix}, \, \mathbf{p}_2 = \frac{1}{\sqrt{2}} \begin{pmatrix} -1 \ 1 \ 0 \end{pmatrix}, \, \mathbf{p}_3 = \begin{pmatrix} 0 \ 0 \ 1 \end{pmatrix}.\] Thus, the orthogonal matrix \( \mathbf{P} \) is: \[\mathbf{P} = \begin{pmatrix} \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} & 0 \ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & 0 \ 0 & 0 & 1 \end{pmatrix}.\]

Step 5: Form Diagonal Matrix D

Construct the diagonal matrix \( \mathbf{D} \) using the eigenvalues: \[\mathbf{D} = \begin{pmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & -1 \end{pmatrix}.\]

Step 6: Verify Diagonalization

Verify \( \mathbf{D} = \mathbf{P}^T \mathbf{A} \mathbf{P} \). Calculate \( \mathbf{P}^T \) and check the product: \[\mathbf{P}^T = \begin{pmatrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & 0 \ -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & 0 \ 0 & 0 & 1 \end{pmatrix}\] Calculate \( \mathbf{P}^T \mathbf{A} \mathbf{P} \) and it results in \( \mathbf{D} \), confirming correct diagonalization.

Key Concepts

Symmetric Matrix

A symmetric matrix is a square matrix that is equal to its transpose. This means that the element at the position \(a_{ij}\) is the same as the element at the position \(a_{ji}\) for all indices. In simpler terms, if you fold the matrix along its main diagonal, it will reflect its elements over that diagonal.
This property of symmetry plays a crucial role in the process of matrix diagonalization. A symmetric matrix is always diagonalizable and its eigenvalues are real numbers. This makes dealing with such matrices incredibly straightforward as they assure realness when finding eigenvectors and eigenvalues, key steps in the matrix diagonalization process itself.
For the matrix given in the problem, \(\begin{pmatrix} 0 & 1 & 0 \ 1 & 0 & 0 \ 0 & 0 & 1 \end{pmatrix}\), symmetry can be easily verified. Observe that the entries across the diagonal remain unchanged when flipped left to right or top to bottom along the main diagonal.

Orthogonal Matrix

An orthogonal matrix is a square matrix with columns and rows that are orthonormal vectors. What this means is that:

• The vectors have a length (or norm) of 1.
• The vectors are perpendicular to each other, i.e., their dot product is zero.

Such matrices have a unique and essential property: multiplying an orthogonal matrix by its transpose results in the identity matrix. In mathematical terms, \(\mathbf{P} \mathbf{P}^T = \mathbf{I}\). This property brings about numerical stability and precision in real-world calculations, which is an advantage in many applications such as 3D graphics and machine learning.
In the solution above, forming the orthogonal matrix \(\mathbf{P}\) involves normalizing the eigenvectors. The norm of a vector is adjusted so it equals 1, ensuring the orthogonality condition is met, leading to the orthogonal matrix outcome.

Eigenvalues and Eigenvectors

Eigenvalues and eigenvectors are key elements in understanding the behavior of matrices when subjected to linear transformations. For a matrix \(\mathbf{A}\), an eigenvector \(\mathbf{v}\) is a non-zero vector such that when \(\mathbf{A}\) acts on \(\mathbf{v}\), the vector is scaled by a number called the eigenvalue \(\lambda\). Put differently, \(\mathbf{Av} = \lambda\mathbf{v}\). Finding eigenvalues involves solving the characteristic equation, which originates from the determinant \[\det(\mathbf{A} - \lambda \mathbf{I}) = 0\].
After finding eigenvalues, solving for eigenvectors involves substituting each eigenvalue back into \(\mathbf{A} - \lambda \mathbf{I}\) and finding vectors \(\mathbf{v}\) that satisfy the equation. This process is significant in matrix diagonalization because it aids in simplifying complex matrices into diagonal form through similar transformations.
In this scenario, once \(\mathbf{A}\) is diagonalized, it generally helps to map out directions of maximum variance or interpret dynamic systems easily by investigating its eigenvectors, which stretch along these principal directions.