Problem 27 - 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 \(\mathbf{D}=\mathbf{P}^{T} \mathbf{A P}\). $$ \left(\begin{array}{rrr} 5 & -2 & 0 \\ -2 & 6 & -2 \\ 0 & -2 & 7 \end{array}\right) $$

Step-by-Step Solution

Answer

Diagonalize with orthogonal P and D using eigenvalues 4, 8, and 6.

Step 1: Verify Matrix Symmetry

Start by checking if the given matrix \( \mathbf{A} = \begin{pmatrix} 5 & -2 & 0 \ -2 & 6 & -2 \ 0 & -2 & 7 \end{pmatrix} \) is symmetric. A symmetric matrix satisfies \( a_{ij} = a_{ji} \). One can see that each element above the main diagonal is equal to the corresponding element below it, confirming that \( \mathbf{A} \) is symmetric.

Step 2: Find Eigenvalues

To find the eigenvalues of \( \mathbf{A} \), solve the characteristic equation \( \text{det}(\mathbf{A} - \lambda \mathbf{I}) = 0 \). Calculate \( \mathbf{A} - \lambda \mathbf{I} \):\[\begin{pmatrix} 5-\lambda & -2 & 0 \ -2 & 6-\lambda & -2 \ 0 & -2 & 7-\lambda \end{pmatrix}.\]Simplify and solve the determinant to find the eigenvalues.

Step 3: Solve the Characteristic Equation

Find the determinant of the matrix \( \mathbf{A} - \lambda \mathbf{I} \), equate it to zero, and solve the cubic equation for \( \lambda \). The characteristic polynomial is:\[\text{det}(\mathbf{A} - \lambda \mathbf{I}) = \lambda^3 - 18\lambda^2 + 85\lambda - 160 = 0.\]By inspection or using the rational root theorem, find the solutions which are the eigenvalues: \( \lambda_1 = 4 \), \( \lambda_2 = 8 \), \( \lambda_3 = 6 \).

Step 4: Find Eigenvectors

For each eigenvalue, \( \lambda \), find the eigenvector \( \mathbf{v} \) of \( (\mathbf{A} - \lambda \mathbf{I})\mathbf{v} = \mathbf{0} \). For example, substitute \( \lambda_1 = 4 \) into \( \mathbf{A} - 4\mathbf{I} \) and solve:\[\begin{pmatrix} 1 & -2 & 0 \ -2 & 2 & -2 \ 0 & -2 & 3 \end{pmatrix}\begin{pmatrix} x_1 \ x_2 \ x_3 \end{pmatrix} = \begin{pmatrix} 0 \ 0 \ 0 \end{pmatrix}.\]Continue this process to find eigenvectors for \( \lambda_2 \) and \( \lambda_3 \).

Step 5: Form Orthogonal Matrix P

Assemble matrix \( \mathbf{P} \) using the orthogonalized eigenvectors found in the previous step as columns, ensuring they are normalized. This constructs \( \mathbf{P} \) as an orthogonal matrix where \( \mathbf{P}^T = \mathbf{P}^{-1} \).

Step 6: Construct Diagonal Matrix D

The diagonal matrix \( \mathbf{D} \) is formed by placing the eigenvalues on the diagonal such that \( \mathbf{D} = \text{diag}(\lambda_1, \lambda_2, \lambda_3) = \begin{pmatrix} 4 & 0 & 0 \ 0 & 8 & 0 \ 0 & 0 & 6 \end{pmatrix}.\) This ensures that \( \mathbf{D} = \mathbf{P}^T \mathbf{A} \mathbf{P} \).

Step 7: Verify Diagonalization

Finally, confirm by calculation that multiplying \( \mathbf{P}^T \mathbf{A} \mathbf{P} \) results in the diagonal matrix \( \mathbf{D} \). This verifies the orthogonal matrix \( \mathbf{P} \) and diagonal matrix \( \mathbf{D} \) are correct.

Key Concepts

Symmetric Matrices

A symmetric matrix is fundamental in linear algebra and has an elegant structure. A square matrix \( \mathbf{A} \) is symmetric if it is equal to its transpose, which means \( a_{ij} = a_{ji} \) for all indices \( i \) and \( j \). This property implies that the matrix looks the same whether observed from above or below the main diagonal. Symmetry is crucial because symmetric matrices have real eigenvalues and orthogonal eigenvectors, a feature that simplifies many mathematical operations.

In engineering and physics, symmetric matrices frequently arise, especially in systems where relationships are bidirectional, like road networks or molecular bonds. Verifying symmetry in practical situations involves confirming that each element in the \( i^{th} \) row and \( j^{th} \) column is equal to the element in the \( j^{th} \) row and \( i^{th} \) column. The verification for the problem matrix \(\begin{pmatrix}5 & -2 & 0 \-2 & 6 & -2 \0 & -2 & 7\end{pmatrix}\), shows it is symmetric, an important first step before moving forward with diagonalization.

Eigenvalues and Eigenvectors

Eigenvalues and eigenvectors are key concepts when studying matrices. For a given square matrix \( \mathbf{A} \), an eigenvector \( \mathbf{v} \) is a non-zero vector that transforms by a scalar factor when \( \mathbf{A} \) is applied to it. This scalar factor is referred to as the eigenvalue, \( \lambda \). Mathematically, this can be expressed as \( \mathbf{A} \mathbf{v} = \lambda \mathbf{v} \).

To find these, we set up the characteristic equation \( \text{det}(\mathbf{A} - \lambda \mathbf{I}) = 0 \), where \( \mathbf{I} \) is the identity matrix. Solving this equation gives the eigenvalues. For our specific matrix example, we solved the associated polynomial \( \lambda^3 - 18\lambda^2 + 85\lambda - 160 = 0 \), yielding eigenvalues \( \lambda_1 = 4 \), \( \lambda_2 = 8 \), and \( \lambda_3 = 6 \).

With eigenvalues in hand, we substitute each back into the equation \( (\mathbf{A} - \lambda \mathbf{I})\mathbf{v} = \mathbf{0} \) to determine the corresponding eigenvectors. Each solution provides an eigenvector, and these together form the columns of the matrix that diagonalizes \( \mathbf{A} \). Eigenvalues and eigenvectors simplify the matrix into a diagonal form, making complex calculations easier.

Orthogonal Matrices

Orthogonal matrices provide a powerful tool in matrix diagonalization due to their special properties. A matrix \( \mathbf{P} \) is orthogonal if its transpose \( \mathbf{P}^T \) is also its inverse, which means \( \mathbf{P}^T \mathbf{P} = \mathbf{I} \). This property ensures that orthogonal matrices preserve angles and lengths, making them invaluable in various applications like computer graphics and numerical algorithms.

When dealing with symmetric matrices, the eigenvectors can be chosen to form an orthogonal set. These vectors are often normalized to length one, ensuring that the matrix formed is not just orthogonal but also orthonormal. In the given problem set, matrix \( \mathbf{P} \) is formed by the set of orthonormal eigenvectors of \( \mathbf{A} \). By arranging these eigenvectors as columns in \( \mathbf{P} \), you ensure that multiplying \( \mathbf{P}^T \mathbf{A} \mathbf{P} \) yields a diagonal matrix with the matrix's eigenvalues as its entries.

Orthogonal matrices are pivotal here as they lead to efficient calculations and simplify transformations, thus making complex computations more manageable by focusing on diagonal matrices, which are easier to work with.