Problem 26 - 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 \(\mathbf{D}=\mathbf{P}^{T} \mathbf{A P}\) $$ \left(\begin{array}{rrr} 1 & -2 & 2 \\ -2 & 1 & -2 \\ 2 & -2 & 1 \end{array}\right) $$

Step-by-Step Solution

Answer

Matrix \( \mathbf{P} \) is orthogonal with columns as normalized eigenvectors of \( \mathbf{A} \), and \( \mathbf{D} \) is diagonal with eigenvalues as entries.

Step 1: Verify Symmetry of Matrix A

First, we need to confirm that the matrix \( \mathbf{A} \) is symmetric. A symmetric matrix is equal to its transpose, i.e., \( \mathbf{A} = \mathbf{A}^T \).\[\mathbf{A} = \begin{pmatrix} 1 & -2 & 2 \ -2 & 1 & -2 \ 2 & -2 & 1 \end{pmatrix}\]Since the matrix is equal to its transpose, it is indeed symmetric.

Step 2: Find Eigenvalues of Matrix A

Solve for the eigenvalues by calculating the determinant of \( \mathbf{A} - \lambda \mathbf{I} \) and set it to zero.\[| \mathbf{A} - \lambda \mathbf{I} | = \begin{vmatrix} 1-\lambda & -2 & 2 \ -2 & 1-\lambda & -2 \ 2 & -2 & 1-\lambda \end{vmatrix} = 0\]Simplifying this determinant results in the characteristic polynomial \(-\lambda^3 + 3\lambda^2 - 4\lambda = 0\). Factor to find eigenvalues: \( \lambda (\lambda-1)(\lambda-4) = 0 \), giving \( \lambda_1 = 0, \lambda_2 = 1, \lambda_3 = 4 \).

Step 3: Find Eigenvectors for Each Eigenvalue

For each eigenvalue, find the corresponding eigenvector by solving \( (\mathbf{A} - \lambda \mathbf{I}) \mathbf{v} = 0 \).- For \( \lambda_1 = 0 \): Solve \( \mathbf{A} \mathbf{v} = 0 \) leading to \( \mathbf{v}_1 = c_1 \begin{pmatrix} 1 \cr 1 \cr 1 \end{pmatrix} \).- For \( \lambda_2 = 1 \): Solve \( (\mathbf{A} - \mathbf{I}) \mathbf{v} = 0 \) leading to \( \mathbf{v}_2 = c_2 \begin{pmatrix} 1 \cr 0 \cr -1 \end{pmatrix} \).- For \( \lambda_3 = 4 \): Solve \( (\mathbf{A} - 4\mathbf{I}) \mathbf{v} = 0 \) leading to \( \mathbf{v}_3 = c_3 \begin{pmatrix} 1 \cr -2 \cr 1 \end{pmatrix} \).

Step 4: Normalize the Eigenvectors

To find the orthogonal matrix \( \mathbf{P} \), normalize each eigenvector.- For \( \mathbf{v}_1 = \begin{pmatrix} 1 \cr 1 \cr 1 \end{pmatrix} \), normalize to \( \frac{1}{\sqrt{3}} \begin{pmatrix} 1 \cr 1 \cr 1 \end{pmatrix} \).- For \( \mathbf{v}_2 = \begin{pmatrix} 1 \cr 0 \cr -1 \end{pmatrix} \), normalize to \( \frac{1}{\sqrt{2}} \begin{pmatrix} 1 \cr 0 \cr -1 \end{pmatrix} \).- For \( \mathbf{v}_3 = \begin{pmatrix} 1 \cr -2 \cr 1 \end{pmatrix} \), normalize to \( \frac{1}{\sqrt{6}} \begin{pmatrix} 1 \cr -2 \cr 1 \end{pmatrix} \).

Step 5: Construct the Orthogonal Matrix P

Arrange the normalized eigenvectors as the columns of the orthogonal matrix \( \mathbf{P} \):\[\mathbf{P} = \begin{pmatrix} \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{6}} \ \frac{1}{\sqrt{3}} & 0 & \frac{-2}{\sqrt{6}} \ \frac{1}{\sqrt{3}} & \frac{-1}{\sqrt{2}} & \frac{1}{\sqrt{6}} \end{pmatrix}\]This is orthogonal since its columns are orthonormal.

Step 6: Form the Diagonal Matrix D

The diagonal matrix \( \mathbf{D} \) contains the eigenvalues on its diagonal:\[\mathbf{D} = \begin{pmatrix} 0 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 4 \end{pmatrix}\]This matrix is obtained by ensuring \( \mathbf{D} = \mathbf{P}^T \mathbf{A} \mathbf{P} \).

Key Concepts

Symmetric Matrix

In linear algebra, a symmetric matrix is one that is equal to its transpose. This means that the matrix looks the same when flipped over its main diagonal. A key property of symmetric matrices is that they always have real eigenvalues. This makes them particularly suitable for various applications, such as in physics and engineering.
To check if a given matrix is symmetric, you simply compare it to its transpose. If they are identical, then the matrix is symmetric. For example, consider a matrix \( \mathbf{A} \):

• Matrix \( \mathbf{A} \) is symmetric if \( \mathbf{A} = \mathbf{A}^T \).

This characteristic is important because symmetric matrices can be diagonalized using orthogonal matrices. This property streamlines calculations and simplifies complex problems.

Eigenvalues

Eigenvalues are special numbers associated with a matrix, born from the characteristic equation of the matrix. They are crucial because they can provide insight into the matrix's properties, such as stability and transformation behavior.
To find the eigenvalues of a matrix \( \mathbf{A} \), you have to solve the equation \(| \mathbf{A} - \lambda \mathbf{I} | = 0 \), where \( \lambda \) represents the eigenvalues and \( \mathbf{I} \) is the identity matrix.
Key steps involve:

• Calculating the determinant of \( \mathbf{A} - \lambda \mathbf{I} \).
• Setting the determinant equation to zero to solve for \( \lambda \).

Finding these values is a fundamental step for matrix diagonalization and understanding the matrix's behavior.

Eigenvectors

Eigenvectors accompany eigenvalues and describe the direction and impact of a matrix transformation. After finding the eigenvalues, each eigenvector corresponds to one eigenvalue, forming a complete set of vectors that describe the matrix's action.
To find eigenvectors, solve the equation \( (\mathbf{A} - \lambda \mathbf{I}) \mathbf{v} = 0 \) for each eigenvalue \( \lambda \).

• The resulting vectors from this calculation are not just any vectors, they are special directions that remain constant under the transformation defined by the matrix.

Once eigenvectors are determined, they can be normalized to unit vectors to form an orthogonal matrix. This orthogonality facilitates easier calculations and helps in matrix diagonalization.

Orthogonal Matrix

Orthogonal matrices are matrices whose columns and rows are orthogonal unit vectors, meaning they are perpendicular and have a magnitude of one. The significance of orthogonal matrices in mathematics stems from their property of preserving lengths and angles during transformations.
An orthogonal matrix \( \mathbf{P} \) fulfills the condition \( \mathbf{P}^T \mathbf{P} = \mathbf{I} \), where \( \mathbf{I} \) is the identity matrix.

• Orthogonal matrices simplify complex matrix operations, such as matrix diagonalization.
• They allow easy transformations without altering the inherent geometrical properties.

In the context of diagonalization, using an orthogonal matrix ensures that the transformation maintains the symmetry and essential properties of the original matrix.