Problem 28 - 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 \(\mathbf{D}=\mathbf{P}^{T} \mathbf{A P}\). $$ \left(\begin{array}{lll} 3 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 1 \end{array}\right) $$

Step-by-Step Solution

Answer

The orthogonal matrix \(\mathbf{P}\) and diagonal matrix \(\mathbf{D}\) respectively are \(\mathbf{P}=\begin{pmatrix} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 1 \end{pmatrix}\) and \(\mathbf{D}=\begin{pmatrix} 3 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}\).

Step 1: Find Eigenvalues

To diagonalize the matrix \(\mathbf{A}\), we first need to find its eigenvalues. Set up the characteristic equation \(\det(\mathbf{A} - \lambda \mathbf{I}) = 0\), where \(\mathbf{I}\) is the identity matrix. For the given matrix, this determinant is:\[\det \left(\begin{array}{ccc} 3-\lambda & 0 & 1 \ 0 & 1-\lambda & 0 \ 1 & 0 & 1-\lambda \end{array}\right) = 0\]Calculate the determinant and solve for \(\lambda\).

Step 2: Calculate Determinant

The determinant of the matrix \(\mathbf{A} - \lambda \mathbf{I}\) is calculated as follows:\[\det \left(\begin{array}{ccc} 3-\lambda & 0 & 1 \ 0 & 1-\lambda & 0 \ 1 & 0 & 1-\lambda \end{array}\right) = (3-\lambda)((1-\lambda)(1-\lambda) - 0) - 0 + 1 \cdot (0 - 0) = (3-\lambda)((1-\lambda)^2)\]This simplifies to:\[(3-\lambda)(\lambda^2 - 2\lambda + 1)\]Solve for roots: \((3-\lambda)(\lambda - 1)^2 = 0\). Hence, \(\lambda = 3\) or \(\lambda = 1\).

Step 3: Find Eigenvectors for \(\lambda = 3\)

Now find the eigenvectors associated with \(\lambda = 3\). Substitute \(\lambda = 3\) into \((\mathbf{A} - 3\mathbf{I})\mathbf{x} = \mathbf{0}\):\[\left(\begin{array}{ccc} 0 & 0 & 1 \ 0 & -2 & 0 \ 1 & 0 & -2 \end{array}\right)\left(\begin{array}{c} x_1 \ x_2 \ x_3 \end{array}\right) = \left(\begin{array}{c} 0 \ 0 \ 0 \end{array}\right)\]Solving this yields \(x_3 = 0\) and \(x_1 = 2x_3 + x_1 = 0\). Choose \(x_1 = 0\), \(x_2 = 1\), \(x_3 = 0\). Thus, an eigenvector is \(\begin{pmatrix} 0 \ 1 \ 0 \end{pmatrix}\).

Step 4: Find Eigenvectors for \(\lambda = 1\)

Next, find the eigenvectors associated with \(\lambda = 1\). Substitute \(\lambda = 1\) into \((\mathbf{A} - \mathbf{I})\mathbf{x} = \mathbf{0}\):\[\left(\begin{array}{ccc} 2 & 0 & 1 \ 0 & 0 & 0 \ 1 & 0 & 0 \end{array}\right)\left(\begin{array}{c} x_1 \ x_2 \ x_3 \end{array}\right) = \left(\begin{array}{c} 0 \ 0 \ 0 \end{array}\right)\]This system of equations gives solutions like \(x_1 = -x_3\), \(x_2\) free, giving us vectors like \(\begin{pmatrix} 1 \ 0 \ -1 \end{pmatrix}\) and another orthogonal to choose \(\begin{pmatrix} 0 \ 0 \ 0 \end{pmatrix}\), similar to completed at Step 5.

Step 5: Construct Diagonal Matrix \(\mathbf{D}\)

The matrix \(\mathbf{D}\) is a diagonal matrix consisting of the eigenvalues of \(\mathbf{A}\). It takes the following form:\[\mathbf{D} = \begin{pmatrix} 3 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{pmatrix}\]This represents the eigenvalues \(\lambda_1 = 3\), \(\lambda_2 = 1\), and '\(\lambda_3 = 1\).

Step 6: Construct Orthogonal Matrix \(\mathbf{P}\)

The orthogonal matrix \(\mathbf{P}\) is formed from the normalized eigenvectors calculated in the previous steps. Organize these as columns in \(\mathbf{P}\):\[\mathbf{P} = \begin{pmatrix} 0 & 1 & 1/\sqrt{2} \ 1 & 0 & 0 \ 0 & -1/\sqrt{2} & 1/\sqrt{2} \end{pmatrix}\]Ensure these vectors are orthogonal and normalized.

Key Concepts

Eigenvalues and Eigenvectors

When you're dealing with matrices, a key concept is finding the eigenvalues and eigenvectors, which are essential for diagonalizing matrices. Think of eigenvalues as special numbers associated with a matrix that show how vectors are stretched. The eigenvector, on the other hand, is the direction that remains unchanged during this transformation, except for the scaling by its corresponding eigenvalue. To find these elements, you must solve the equation \( \det(\mathbf{A} - \lambda \mathbf{I}) = 0\), where \(\mathbf{A}\) is your matrix, \(\lambda\) represents the eigenvalues, and \(\mathbf{I}\) is the identity matrix.

You can follow these steps to find them:

• Set up the characteristic polynomial by calculating the determinant \((\mathbf{A} - \lambda \mathbf{I})\).
• Solve the polynomial equation for \(\lambda\) to get the eigenvalues.
• Substitute each eigenvalue back into the equation \( (\mathbf{A} - \lambda \mathbf{I})\mathbf{x} = \mathbf{0} \) to find eigenvectors.

In practice, these steps help us break complex matrix transformations into simpler, manageable parts.

Orthogonal Matrices

Once you've found the eigenvectors, the next step in matrix diagonalization involves constructing an orthogonal matrix. An orthogonal matrix is powerful in linear algebra because it preserves the length of vectors and the angle between them during transformations. These matrices have unique properties:

• The rows and columns are orthogonal unit vectors, meaning they are perpendicular to each other and have a magnitude of one.
• For any orthogonal matrix \(\mathbf{P}\), the inverse is equal to its transpose, that is \(\mathbf{P}^{-1} = \mathbf{P}^{T}\).

Why are orthogonal matrices useful? They allow for an efficient representation and easy computation of complex transformations, providing stability and accuracy.

In our exercise, you compose \(\mathbf{P}\) by arranging the eigenvectors you found as its columns. Remember to normalize the eigenvectors so each has length one, ensuring the matrix remains orthogonal.

Symmetric Matrices

A symmetric matrix simplifies the process of diagonalization and ensures some beautiful properties in mathematics. These matrices have a unique trait where they remain unchanged when transposed, i.e., \(\mathbf{A} = \mathbf{A}^{T}\). This symmetry brings several advantages:

• A symmetric matrix has real eigenvalues, which makes them more intuitive to work with.
• The eigenvectors of symmetric matrices are orthogonal, meaning you can construct an orthogonal matrix directly from these eigenvectors.

In the context of diagonalization, symmetrical matrices are particularly nice because they allow for a clear decomposition: \(\mathbf{D} = \mathbf{P}^{T}\mathbf{AP}\), where \(\mathbf{D}\) is a diagonal matrix of eigenvalues and \(\mathbf{P}\) is an orthogonal matrix of eigenvectors.

With symmetric matrices, the entire process becomes less computationally intensive and easier to manage, making them a favorite in applications like physics and computer graphics.