Problem 22 - 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 \(\mathbf{D}=\mathbf{P}^{T} \mathbf{A P}\). $$ \left(\begin{array}{ll} 3 & 2 \\ 2 & 0 \end{array}\right) $$

Step-by-Step Solution

Answer

Orthogonal matrix \( \mathbf{P} = \begin{pmatrix} \frac{2}{\sqrt{5}} & \frac{1}{\sqrt{5}} \\ \frac{1}{\sqrt{5}} & -\frac{2}{\sqrt{5}} \end{pmatrix} \), diagonal matrix \( \mathbf{D} = \begin{pmatrix} 4 & 0 \\ 0 & -1 \end{pmatrix} \).

Step 1: Verify Symmetry

We first confirm that matrix \( \mathbf{A} = \begin{pmatrix} 3 & 2 \ 2 & 0 \end{pmatrix} \) is symmetric. A matrix is symmetric if \( A^T = A \). Since transposing \( \mathbf{A} \) gives the same matrix, it is symmetric.

Step 2: Find Eigenvalues

To find the eigenvalues, we solve the characteristic equation \( \text{det}(\mathbf{A} - \lambda \mathbf{I}) = 0 \). For \( \mathbf{A} = \begin{pmatrix} 3 & 2 \ 2 & 0 \end{pmatrix} \), we calculate: \[ \text{det}(\begin{pmatrix} 3 - \lambda & 2 \ 2 & -\lambda \end{pmatrix}) = (3-\lambda)(-\lambda) - 4 = \lambda^2 - 3\lambda - 4 = 0 \] which solves to \( \lambda_1 = 4 \) and \( \lambda_2 = -1 \).

Step 3: Find Eigenvectors

For each eigenvalue, find a corresponding eigenvector. For \( \lambda_1 = 4 \), solve \( (\mathbf{A} - 4\mathbf{I})\mathbf{v} = \mathbf{0} \) which gives \( \begin{pmatrix} -1 & 2 \ 2 & -4 \end{pmatrix} \begin{pmatrix} x_1 \ x_2 \end{pmatrix} = \begin{pmatrix} 0 \ 0 \end{pmatrix} \). Solving this gives eigenvectors of multiples of \( \begin{pmatrix} 2 \ 1 \end{pmatrix} \). For \( \lambda_2 = -1 \), solve \( (\mathbf{A} + \mathbf{I})\mathbf{v} = \mathbf{0} \) which gives eigenvectors of multiples of \( \begin{pmatrix} 1 \ -2 \end{pmatrix} \).

Step 4: Normalize Eigenvectors

To form an orthogonal matrix, the eigenvectors need to be normalized. Normalize \( \begin{pmatrix} 2 \ 1 \end{pmatrix} \) to \( \begin{pmatrix} \frac{2}{\sqrt{5}} \ \frac{1}{\sqrt{5}} \end{pmatrix} \) and \( \begin{pmatrix} 1 \ -2 \end{pmatrix} \) to \( \begin{pmatrix} \frac{1}{\sqrt{5}} \ -\frac{2}{\sqrt{5}} \end{pmatrix} \).

Step 5: Form Orthogonal Matrix P

The orthogonal matrix \( \mathbf{P} \) is formed by taking the normalized eigenvectors as its columns: \[ \mathbf{P} = \begin{pmatrix} \frac{2}{\sqrt{5}} & \frac{1}{\sqrt{5}} \ \frac{1}{\sqrt{5}} & -\frac{2}{\sqrt{5}} \end{pmatrix} \].

Step 6: Form Diagonal Matrix D

The diagonal matrix \( \mathbf{D} \) is formed using the eigenvalues: \( \mathbf{D} = \begin{pmatrix} 4 & 0 \ 0 & -1 \end{pmatrix} \).

Step 7: Verify the Diagonalization

Check that \( \mathbf{D} = \mathbf{P}^T \mathbf{A} \mathbf{P} \). First, compute \( \mathbf{P}^T \), then multiply \( \mathbf{P}^T \mathbf{A} \) and then multiply the result by \( \mathbf{P} \) to ensure it equals \( \mathbf{D} \). The calculations should confirm the correctness.

Key Concepts

Eigenvalues and Eigenvectors

Eigenvalues and eigenvectors are fundamental concepts in linear algebra, particularly when dealing with matrices like the symmetric matrix in our exercise. An eigenvalue, denoted by \( \lambda \), is a scalar that indicates how much a corresponding eigenvector is stretched or shrunk during a linear transformation represented by matrix \( A \). An eigenvector \( \mathbf{v} \) is a nonzero vector that remains parallel to its original direction after the transformation, adhering to the relation \( A\mathbf{v} = \lambda \mathbf{v} \).

The process to find these begins with solving the characteristic equation \( \text{det}(\mathbf{A} - \lambda \mathbf{I}) = 0 \). This equation arises from subtracting \( \lambda \) times the identity matrix \( \mathbf{I} \) from \( \mathbf{A} \), and then taking the determinant. For example, in our problem with matrix \( \mathbf{A} = \begin{pmatrix} 3 & 2 \ 2 & 0 \end{pmatrix}\), we solved \( \lambda^2 - 3\lambda - 4 = 0 \) to find eigenvalues of 4 and -1.

Once eigenvalues are found, eigenvectors are determined by plugging each \( \lambda \) back into \( (\mathbf{A} - \lambda \mathbf{I})\mathbf{v} = \mathbf{0} \), resulting in a vector that aligns with these eigenvalues. Understanding these concepts helps you transform matrices into simpler forms, facilitating more efficient matrix operations.

Orthogonal Matrix

Orthogonal matrices play a crucial role in simplifying complex matrix operations. An orthogonal matrix \( \mathbf{P} \) is defined such that the transpose of \( \mathbf{P} \) is equal to its inverse: \( \mathbf{P}^T = \mathbf{P}^{-1} \). This property means multiplying an orthogonal matrix by its transpose yields the identity matrix.

Orthogonal matrices are essential when diagonalizing symmetric matrices, as shown in our exercise. By transforming the eigenvectors into an orthogonal set, we can align them as columns in \( \mathbf{P} \), forming the orthogonal matrix:

• Each column is a normalized eigenvector, ensuring orthogonality.
• They maintain directionality under unit transformations.

For instance, in our problem, the matrix \( \mathbf{P} = \begin{pmatrix} \frac{2}{\sqrt{5}} & \frac{1}{\sqrt{5}} \ \frac{1}{\sqrt{5}} & -\frac{2}{\sqrt{5}} \end{pmatrix} \) was formed from the normalized eigenvectors, facilitating easier matrix manipulations and leading directly to diagonalization. Understanding orthogonal matrices aids in grasping their powerful role in avoiding computational errors, given their stability in numerical processes.

Characteristic Equation

The characteristic equation is a polynomial equation derived from a matrix. It is pivotal in determining the eigenvalues, thereby laying the groundwork for anything related to the diagonalization of matrices. The characteristic equation is formed by setting the determinant of \( \mathbf{A} - \lambda \mathbf{I} \) to zero: \( \text{det}(\mathbf{A} - \lambda \mathbf{I}) = 0 \). Here, \( \lambda \) represents the eigenvalue, and \( \mathbf{I} \) stands for the identity matrix of the same dimension as \( \mathbf{A} \).

The determinant forms a polynomial equation, where the degree is the same as that of the matrix. Solving this polynomial yields the eigenvalues. In our solution, for example, we substituted these eigenvalues into equations to find the correct eigenvectors, giving us clues about the matrix's behavior.

Knowing how to form and solve the characteristic equation can drastically simplify understanding the transformation properties of a matrix, particularly in scenarios where diagonalization is required.

Matrix Normalization

Matrix normalization involves scaling vectors such that their length, or norm, becomes 1, enabling easier and more stable computations. When dealing with eigenvectors, as in our exercise, normalization plays a fundamental role in forming the orthogonal matrix \( \mathbf{P} \). Normalized vectors maintain the structure of the original vector but are "stretched" or "shrunk" to unit length.

To normalize an eigenvector, divide each component of the vector by its norm. The norm of a vector \( \begin{pmatrix} x_1 \ x_2 \end{pmatrix} \) is computed as \( \sqrt{x_1^2 + x_2^2} \).

• Ensures each vector in \( \mathbf{P} \) is orthogonal and unit-normed.
• Facilitates the construction of \( \mathbf{P} \) to enable matrix diagonalization.

In our symmetric matrix example, eigenvectors \( \begin{pmatrix} 2 \ 1 \end{pmatrix} \) and \( \begin{pmatrix} 1 \ -2 \end{pmatrix} \) were normalized to \( \begin{pmatrix} \frac{2}{\sqrt{5}} \ \frac{1}{\sqrt{5}} \end{pmatrix} \) and \( \begin{pmatrix} \frac{1}{\sqrt{5}} \ -\frac{2}{\sqrt{5}} \end{pmatrix} \). Being familiar with this process is vital when constructing orthogonal matrices to diagonalize matrices effectively.