Problem 30 - 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 \(\mathbf{D}=\mathbf{P}^{T} \mathbf{A P}\) $$ \left(\begin{array}{llll} 0 & 1 & 0 & 1 \\ 1 & 0 & 1 & 0 \\ 0 & 1 & 0 & 1 \\ 1 & 0 & 1 & 0 \end{array}\right) $$

Step-by-Step Solution

Answer

\(\mathbf{P}\) is orthogonal and \(\mathbf{D}\) is diagonal with eigenvalues.

Step 1: Identify the Symmetric Matrix

The given matrix \( \mathbf{A} \) is \(\begin{bmatrix}0 & 1 & 0 & 1 \1 & 0 & 1 & 0 \0 & 1 & 0 & 1 \1 & 0 & 1 & 0 \\end{bmatrix} \), which is symmetric because it is equal to its transpose.

Step 2: Compute Eigenvalues of Matrix A

To find the eigenvalues, solve the characteristic equation \( \det(\mathbf{A} - \lambda \mathbf{I}) = 0 \), where \( \mathbf{I} \) is the identity matrix. This yields the polynomial equation \( \lambda^4 - 4\lambda^2 = 0 \). Factoring this equation gives \( \lambda^2(\lambda^2 - 4) = 0 \), resulting in eigenvalues \( \lambda_1 = 2, \lambda_2 = -2, \lambda_3 = 0, \lambda_4 = 0 \).

Step 3: Find Eigenvectors for each Eigenvalue

Determine the eigenvectors by solving \( (\mathbf{A} - \lambda \mathbf{I}) \mathbf{v} = \mathbf{0} \) for each eigenvalue. For \( \lambda = 2 \), solve: \([\mathbf{A} - 2\mathbf{I}]\mathbf{v} = \mathbf{0}\). Repeat this for \( \lambda = -2 \) and \( \lambda = 0 \). After solving, the normalized eigenvectors are \( \mathbf{v}_1 = \frac{1}{2}[1, 0, 1, 0]^T \), \( \mathbf{v}_2 = \frac{1}{2}[0, 1, 0, 1]^T \), \( \mathbf{v}_3 = \frac{1}{2}[1, 0, -1, 0]^T \), and \( \mathbf{v}_4 = \frac{1}{2}[0, 1, 0, -1]^T \).

Step 4: Construct Orthogonal Matrix P

Construct the orthogonal matrix \( \mathbf{P} \) using the eigenvectors as columns: \[ \mathbf{P} = \begin{bmatrix} \frac{1}{2} & 0 & \frac{1}{2} & 0 \0 & \frac{1}{2} & 0 & \frac{1}{2} \\frac{1}{2} & 0 & -\frac{1}{2} & 0 \0 & \frac{1}{2} & 0 & -\frac{1}{2} \end{bmatrix} \].

Step 5: Form the Diagonal Matrix D

Construct the diagonal matrix \( \mathbf{D} \) with the eigenvalues on the diagonal: \[ \mathbf{D} = \begin{bmatrix} 2 & 0 & 0 & 0 \0 & -2 & 0 & 0 \0 & 0 & 0 & 0 \0 & 0 & 0 & 0 \end{bmatrix} \].

Step 6: Verify the Diagonalization

Verify that \( \mathbf{D} = \mathbf{P}^T \mathbf{A} \mathbf{P} \). Compute \( \mathbf{P}^T \mathbf{A} \mathbf{P} \) and confirm it results in the diagonal matrix \( \mathbf{D} \). Both matrices should match if the diagonalization is correct.

Key Concepts

Symmetric Matrices

Symmetric matrices are an important class of matrices that appear frequently in linear algebra. A matrix is symmetric if it is equal to its transpose. This means if you swap the rows and columns of a symmetric matrix, you will still get the same matrix.
In a symmetrical matrix \( \mathbf{A} \), the element \( a_{ij} \) is equal to the element \( a_{ji} \). This property lends symmetric matrices some attractive features, such as always having real eigenvalues. For instance, the matrix given in our exercise is:
\[ \begin{bmatrix}0 & 1 & 0 & 1 \1 & 0 & 1 & 0 \0 & 1 & 0 & 1 \1 & 0 & 1 & 0 \\end{bmatrix}\]
It is symmetric because every element across the diagonal mirrors the counterpart on the other side. This property is crucial for matrix diagonalization, as symmetric matrices are guaranteed to be diagonalizable.

Orthogonal Matrices

Orthogonal matrices play a key role in matrix diagonalization. An orthogonal matrix \( \mathbf{P} \) is a square matrix with the property that its transpose is also its inverse. So, it satisfies \( \mathbf{P}^T \mathbf{P} = \mathbf{I} \), where \( \mathbf{I} \) is the identity matrix.
Here are some important aspects of orthogonal matrices:

• Their rows and columns are orthonormal vectors, which means they are perpendicular (orthogonal) and have unit length (normalized).
• Multiplying a matrix by an orthogonal matrix preserves the inner products, giving it the property of "rotation" or "reflection" without changing the length of vectors.
• The determinant of an orthogonal matrix is always \( \pm 1 \).

In the exercise, we found the orthogonal matrix \( \mathbf{P} \) using the eigenvectors as its columns, ensuring positive properties for diagonalization.

Eigenvectors and Eigenvalues

The concepts of eigenvectors and eigenvalues are central to understanding matrix transformations. They provide insights into the matrix structure and are essential for diagonalization.
An eigenvector of a matrix \( \mathbf{A} \) is a non-zero vector \( \mathbf{v} \) such that multiplication by \( \mathbf{A} \) alters only the scale of \( \mathbf{v} \). Mathematically, this is written as:\[ \mathbf{A} \mathbf{v} = \lambda \mathbf{v}\]where \( \lambda \) is the eigenvalue associated with the eigenvector \( \mathbf{v} \).

• Eigenvectors determine the "directions" in which a linear transformation is applied.
• Eigenvalues describe how much the transformation scales the eigenvectors.

In our example, the eigenvalues were found to be 2, -2, and 0. Each eigenvalue had corresponding eigenvectors, which were used to form the orthogonal matrix \( \mathbf{P} \). These eigenpairs simplify matrix analysis significantly by reducing a matrix to its diagonal form.

Characteristic Equation

The characteristic equation is fundamental in finding eigenvalues of a matrix. To derive it, one must solve the equation:\[ \det(\mathbf{A} - \lambda \mathbf{I}) = 0\]where \( \mathbf{I} \) is the identity matrix of the same dimension as \( \mathbf{A} \).
The solution gives us a polynomial equation in terms of \( \lambda \), whose roots are the eigenvalues of \( \mathbf{A} \). In our particular exercise, solving this characteristic equation produced the eigenvalues 2, -2, and 0.

• This method is crucial because it allows us to determine all eigenvalues, which are needed to fully understand the matrix's behavior.
• It is applicable to any square matrix and is particularly straightforward for symmetric matrices.

Understanding how to set up and solve the characteristic equation will enhance one’s ability to work with different linear transformations.