Problem 28 - 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 \(\mathbf{D}=\mathbf{P}^{T} \mathbf{A P}\) $$ \left(\begin{array}{lll} 3 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 1 \end{array}\right) $$

Step-by-Step Solution

Answer

Orthogonal matrix \( \mathbf{P} \) diagonalizes \( \mathbf{A} \), giving \( \mathbf{D} = \begin{bmatrix} 4 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{bmatrix} \).

Step 1: Define Matrix A and Verify Symmetry

The matrix \( \mathbf{A} \) is given by \( \begin{bmatrix} 3 & 0 & 1 \ 0 & 1 & 0 \ 1 & 0 & 1 \end{bmatrix} \). A symmetric matrix satisfies the condition \( \mathbf{A} = \mathbf{A}^T \). Here, \( \mathbf{A}^T = \begin{bmatrix} 3 & 0 & 1 \ 0 & 1 & 0 \ 1 & 0 & 1 \end{bmatrix} \), confirming that \( \mathbf{A} \) is indeed symmetric.

Step 2: Calculate Eigenvalues of A

To find the eigenvalues, solve the characteristic equation \( \det(\mathbf{A} - \lambda \mathbf{I}) = 0 \). The matrix \( \mathbf{A} - \lambda \mathbf{I} \) is \( \begin{bmatrix} 3-\lambda & 0 & 1 \ 0 & 1-\lambda & 0 \ 1 & 0 & 1-\lambda \end{bmatrix} \). Calculate the determinant and solve:\[ (3-\lambda)((1-\lambda)^2) - ((1-\lambda)(0) - (0)(1-\lambda)(1)) - 1((0)(1-\lambda) - (0)(0)) = 0 \]Simplifying gives the equation \( -\lambda^3 + 5\lambda^2 - 7\lambda + 3 = 0 \). Solve this to find the eigenvalues \( \lambda_1 = 4, \lambda_2 = 1, \lambda_3 = 0 \).

Step 3: Find Eigenvectors for Each Eigenvalue

For \( \lambda_1 = 4 \), solve \( (\mathbf{A} - 4\mathbf{I})\mathbf{x} = 0 \). The matrix becomes \( \begin{bmatrix} -1 & 0 & 1 \ 0 & -3 & 0 \ 1 & 0 & -3 \end{bmatrix} \). The eigenvector for \( \lambda_1 \) is proportional to \( \begin{bmatrix} 1 \ 0 \ 1 \end{bmatrix} \).For \( \lambda_2 = 1 \), perform similar steps with \( \begin{bmatrix} 2 & 0 & 1 \ 0 & 0 & 0 \ 1 & 0 & 0 \end{bmatrix} \), leading to an eigenvector proportional to \( \begin{bmatrix} -1 \ 0 \ 1 \end{bmatrix} \).For \( \lambda_3 = 0 \), use \( \begin{bmatrix} 3 & 0 & 1 \ 0 & 1 & 0 \ 1 & 0 & 1 \end{bmatrix} \) which simplifies to a vector along the \( y \)-axis, giving \( \begin{bmatrix} 0 \ 1 \ 0 \end{bmatrix} \) as the eigenvector.

Step 4: Form Orthogonal Matrix P

The orthogonal matrix \( \mathbf{P} \) consists of the normalized eigenvectors as columns: \[ \mathbf{P} = \begin{bmatrix} \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} & 0 \ 0 & 0 & 1 \ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & 0 \end{bmatrix} \].This matrix is orthogonal as \( \mathbf{P}^T \mathbf{P} = \mathbf{I} \) (identity matrix).

Step 5: Diagonalize Matrix A as D

Matrix \( \mathbf{D} \) is formed as \( \mathbf{D} = \mathbf{P}^T \mathbf{A} \mathbf{P} \). By performing the matrix multiplication, \( \mathbf{D} \) becomes: \[ \mathbf{D} = \begin{bmatrix} 4 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 0 \end{bmatrix} \], reflecting the eigenvalues on the diagonal.

Key Concepts

Symmetric Matrix

A symmetric matrix is a square matrix that is equal to its transpose. In simple terms, this means that the matrix remains the same when its rows are switched with its columns. This property is crucial when dealing with concepts like matrix diagonalization. If a matrix \( \mathbf{A} \) is symmetric, then \( \mathbf{A}^T = \mathbf{A} \). This symmetry makes certain mathematical processes, like finding eigenvalues and eigenvectors, much more straightforward.Symmetric matrices have fascinating properties, such as:

• All eigenvalues are real numbers.
• Eigenvectors corresponding to distinct eigenvalues are orthogonal.

These properties are the reason why symmetric matrices are so important in various fields like physics and engineering. Plus, they help make complex computations more manageable.

Eigenvalues and Eigenvectors

Eigenvalues and eigenvectors are fundamental concepts in linear algebra closely tied to matrices. When you multiply a matrix by a vector, an eigenvector doesn't change direction; it simply gets scaled by a certain factor called the eigenvalue. This relationship is described mathematically by the equation:\[\mathbf{A}\mathbf{v} = \lambda\mathbf{v}\]where \( \mathbf{A} \) is a matrix, \( \mathbf{v} \) is an eigenvector, and \( \lambda \) is the eigenvalue associated with it.To find eigenvalues, you solve the characteristic equation:\[\det(\mathbf{A} - \lambda \mathbf{I}) = 0\]Identifying the eigenvalues is the first step. Once they are found, the next step is to determine the eigenvectors.The importance of eigenvalues and eigenvectors comes into play in diagonalization. When a matrix is diagonalizable, it can be expressed in the form \( \mathbf{A} = \mathbf{PDP}^{-1} \), where \( \mathbf{D} \) is a diagonal matrix containing the eigenvalues, and \( \mathbf{P} \) is formed from the eigenvectors. This transformation simplifies matrix computation and has applications in various areas, including stability analysis and vibration studies.

Orthogonal Matrix

An orthogonal matrix is a special kind of square matrix whose rows and columns are orthogonal unit vectors. What makes orthogonal matrices particularly interesting is their key property: when multiplied by their transpose, the result is the identity matrix.Mathematically, if \( \mathbf{P} \) is an orthogonal matrix, then:\[\mathbf{P}^T\mathbf{P} = \mathbf{I}\]where \( \mathbf{I} \) is the identity matrix. This property significantly simplifies computations, as the inverse of an orthogonal matrix is simply its transpose.Orthogonal matrices are important in diagonalization. When you use an orthogonal matrix to diagonalize a symmetric matrix, you get a diagonal matrix with the eigenvalues on its diagonal, which is a powerful transformation. Moreover, orthogonal matrices preserve length and orthogonality, making them ideal for tasks involving rotations and reflections in geometry.Their ease of invertibility and ability to preserve geometric properties make orthogonal matrices invaluable tools in mathematical computations and practical applications alike.