Problem 24 - Advanced Engineering Mathematics

Question

In Problems 21-30, the given matrix \(\mathbf{A}\) is symmetric. Find an orthogonal matrix \(\mathbf{P}\) that diagonalizes \(\mathbf{A}\) and the diagonal matrix D such that \(\mathbf{D}=\mathbf{P}^{T} \mathbf{A P}\). $$ \left(\begin{array}{rr} 1 & -2 \\ -2 & 1 \end{array}\right) $$

Step-by-Step Solution

Answer

Orthogonal matrix \( \mathbf{P} \) and diagonal matrix \( \mathbf{D} \) are found using eigenvectors and eigenvalues of the symmetric matrix.

Step 1: Verify Symmetry of Matrix A

Given matrix \( \mathbf{A} \) is \( \begin{pmatrix} 1 & -2 \ -2 & 1 \end{pmatrix} \). A matrix is symmetric if \( \mathbf{A} = \mathbf{A}^T \). Check the transpose: \( \begin{pmatrix} 1 & -2 \ -2 & 1 \end{pmatrix}^T = \begin{pmatrix} 1 & -2 \ -2 & 1 \end{pmatrix} \), which is equal to \( \mathbf{A} \). Thus, \( \mathbf{A} \) is symmetric.

Step 2: Find Eigenvalues of A

The eigenvalues \( \lambda \) satisfy the characteristic equation \( \det(\mathbf{A} - \lambda \mathbf{I}) = 0 \). Substitute \( \mathbf{A} \) and \( \mathbf{I} \): \[ \begin{vmatrix} 1 - \lambda & -2 \ -2 & 1 - \lambda \end{vmatrix} = 0 \]. Solving the determinant leads to \( (1-\lambda)^2 - 4 = 0 \). Simplifying gives \( \lambda^2 - 2\lambda - 3 = 0 \), which factors to \( (\lambda - 3)(\lambda + 1) = 0 \). Hence, eigenvalues are \( \lambda_1 = 3 \) and \( \lambda_2 = -1 \).

Step 3: Find Eigenvectors for Eigenvalue 3

Calculate eigenvectors for \( \lambda_1 = 3 \): solve \( (\mathbf{A} - 3\mathbf{I})\mathbf{v} = 0 \). Substituting \( \lambda_1 \), \( \mathbf{A} - 3\mathbf{I} = \begin{pmatrix} -2 & -2 \ -2 & -2 \end{pmatrix} \). The vector \( \mathbf{v} = \begin{pmatrix} x \ y \end{pmatrix} \) satisfies \(-2x - 2y = 0 \). Solving gives \( x = -y \), thus one eigenvector is \( \mathbf{v}_1 = \begin{pmatrix} 1 \ -1 \end{pmatrix} \).

Step 4: Find Eigenvectors for Eigenvalue -1

Calculate eigenvectors for \( \lambda_2 = -1 \): solve \( (\mathbf{A} + \mathbf{I})\mathbf{v} = 0 \). Substituting \( \lambda_2 \), \( \mathbf{A} + \mathbf{I} = \begin{pmatrix} 2 & -2 \ -2 & 2 \end{pmatrix} \). The vector \( \mathbf{v} = \begin{pmatrix} x \ y \end{pmatrix} \) satisfies \(2x - 2y = 0 \). Solving gives \( x = y \), so one eigenvector is \( \mathbf{v}_2 = \begin{pmatrix} 1 \ 1 \end{pmatrix} \).

Step 5: Normalize Eigenvectors

Normalize \( \mathbf{v}_1 = \begin{pmatrix} 1 \ -1 \end{pmatrix} \) to \( \mathbf{u}_1 = \frac{1}{\sqrt{2}} \begin{pmatrix} 1 \ -1 \end{pmatrix} \). Normalize \( \mathbf{v}_2 = \begin{pmatrix} 1 \ 1 \end{pmatrix} \) to \( \mathbf{u}_2 = \frac{1}{\sqrt{2}} \begin{pmatrix} 1 \ 1 \end{pmatrix} \).

Step 6: Form the Orthogonal Matrix P

The orthogonal matrix \( \mathbf{P} \) is formed by the normalized eigenvectors as columns: \( \mathbf{P} = \begin{pmatrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \ -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \end{pmatrix} \). Verify its orthogonality: \( \mathbf{P}^T \mathbf{P} = \mathbf{I} \).

Step 7: Form the Diagonal Matrix D

The diagonal matrix \( \mathbf{D} \) contains eigenvalues along its diagonal: \( \mathbf{D} = \begin{pmatrix} 3 & 0 \ 0 & -1 \end{pmatrix} \).

Step 8: Verify Diagonalization

Check that \( \mathbf{D} = \mathbf{P}^T \mathbf{A} \mathbf{P} \). Compute the product to ensure that the diagonalization holds true, confirming \( \mathbf{P}^T \mathbf{A} \mathbf{P} = \mathbf{D} \).

Key Concepts

Symmetric Matrix

A symmetric matrix is a square matrix that is equal to its transpose. This means that the elements are mirrored across the main diagonal. In mathematical terms, for a matrix \( \mathbf{A} \), it is symmetric if \( \mathbf{A} = \mathbf{A}^T \). For example, consider the matrix

• \( \begin{pmatrix} 1 & -2 \ -2 & 1 \end{pmatrix} \)

This matrix is symmetric because its transpose

• \( \begin{pmatrix} 1 & -2 \ -2 & 1 \end{pmatrix}^T \)

is the same as the original matrix. Symmetric matrices are crucial in various fields, including physics and engineering, mainly due to their straightforward eigenvector computation and real eigenvalues. Symmetric matrices are always orthogonally diagonalizable, which paves the way for simplifying many applications.

Eigenvalues and Eigenvectors

Eigenvalues and eigenvectors are fundamental concepts in linear algebra. For a given matrix \( \mathbf{A} \), an eigenvector is a non-zero vector that, when multiplied by \( \mathbf{A} \), results in a vector that is a scalar multiple of itself. Mathematically, this is expressed as

• \( \mathbf{A} \mathbf{v} = \lambda \mathbf{v} \)

where \( \lambda \) is the eigenvalue.
To find the eigenvalues, we solve the characteristic equation

• \( \, \det(\mathbf{A} - \lambda \mathbf{I}) = 0 \, \)

where \( \mathbf{I} \) is the identity matrix. For the matrix

• \( \begin{pmatrix} 1 & -2 \ -2 & 1 \end{pmatrix} \)

this process leads to two eigenvalues, \( \lambda_1 = 3 \) and \( \lambda_2 = -1 \). Each eigenvalue corresponds to an eigenvector, which can be found by solving \( (\mathbf{A} - \lambda \mathbf{I}) \mathbf{v} = 0 \). Eigenvalues and eigenvectors play an essential role in matrix diagonalization, stability analysis, and transform applications.

Matrix Diagonalization

Matrix diagonalization is a process of transforming a given matrix into a diagonal form. For a symmetric matrix like

• \( \begin{pmatrix} 1 & -2 \ -2 & 1 \end{pmatrix} \)

this involves using an orthogonal matrix \( \mathbf{P} \) that satisfies

• \( \mathbf{D} = \mathbf{P}^T \mathbf{A} \mathbf{P} \)

where \( \mathbf{D} \) is a diagonal matrix containing the eigenvalues of \( \mathbf{A} \).The columns of the orthogonal matrix \( \mathbf{P} \) are formed from the normalized eigenvectors of \( \mathbf{A} \). Diagonalization is useful because it simplifies matrix computations by breaking down complex operations such as multiplication or power into simpler terms. The diagonal matrix \( \mathbf{D} \) is easier to work with, particularly in finding powers of matrices or solving systems of differential equations. This simplification is crucial for theoretical analysis and numerical computations.

Normalization of Vectors

Normalization is the process of converting a vector to a unit vector, which has a magnitude of one. This is done by dividing each component of the vector by its magnitude. The magnitude of a vector \( \mathbf{v} \) is given by

• \( \| \mathbf{v} \| = \sqrt{x^2 + y^2} \)

where \( x \) and \( y \) are the components of the vector. To normalize the vector

• \( \begin{pmatrix} 1 \ -1 \end{pmatrix} \)

we calculate its magnitude, which is \( \sqrt{2} \), then divide by this magnitude to get the normalized vector

• \( \frac{1}{\sqrt{2}} \begin{pmatrix} 1 \ -1 \end{pmatrix} \).

Normalization is important in forming the orthogonal matrix \( \mathbf{P} \), as it ensures that the matrix's columns are orthogonal unit vectors. This property is what makes \( \mathbf{P} \) orthogonal, meaning its transpose is also its inverse, \( \mathbf{P}^T = \mathbf{P}^{-1} \). Normalizing vectors before forming \( \mathbf{P} \) is crucial for accurate matrix diagonalization and ensures the orthogonality of the resulting matrix.