Problem 4 - Advanced Engineering Mathematics

Question

In Problems 1-20, determine whether the given matrix \(\mathbf{A}\) is diagonalizable. If so, find the matrix \(\mathbf{P}\) that diagonalizes \(\mathbf{A}\) and the diagonal matrix \(\mathbf{D}\) such that \(\mathbf{D}=\mathbf{P}^{-1} \mathbf{A P}\). $$ \left(\begin{array}{ll} 0 & 5 \\ 1 & 0 \end{array}\right) $$

Step-by-Step Solution

Answer

The matrix \( \mathbf{A} \) is diagonalizable with \( \mathbf{P} = \begin{pmatrix} 1 & 1 \\ \sqrt{5} & -\sqrt{5} \end{pmatrix} \) and \( \mathbf{D} = \begin{pmatrix} \sqrt{5} & 0 \\ 0 & -\sqrt{5} \end{pmatrix} \).

Step 1: Calculate Eigenvalues

First, find the eigenvalues of the matrix \( \mathbf{A} = \begin{pmatrix} 0 & 5 \ 1 & 0 \end{pmatrix} \). We do this by solving the characteristic equation \( \det(\mathbf{A} - \lambda \mathbf{I}) = 0 \), where \( \lambda \) is an eigenvalue and \( \mathbf{I} \) is the identity matrix.

Step 2: Set up the Characteristic Equation

Calculate \( \mathbf{A} - \lambda \mathbf{I} = \begin{pmatrix} 0 & 5 \ 1 & 0 \end{pmatrix} - \begin{pmatrix} \lambda & 0 \ 0 & \lambda \end{pmatrix} = \begin{pmatrix} -\lambda & 5 \ 1 & -\lambda \end{pmatrix} \). The determinant is \( \det(\mathbf{A} - \lambda \mathbf{I}) = (-\lambda)(-\lambda) - (5)(1) = \lambda^2 - 5 = 0. \)

Step 3: Solve the Characteristic Equation

Solve \( \lambda^2 - 5 = 0 \). This gives \( \lambda = \sqrt{5} \) and \( \lambda = -\sqrt{5} \). These values are the eigenvalues of the matrix \( \mathbf{A} \).

Step 4: Calculate Eigenvectors

For each eigenvalue, compute the corresponding eigenvector. Start with the eigenvalue \( \lambda = \sqrt{5} \) and solve \((\mathbf{A} - \lambda \mathbf{I})\mathbf{v} = \mathbf{0}\). Repeat this process for \( \lambda = -\sqrt{5} \).

Step 5: Eigenvector Calculation for \( \lambda = \sqrt{5} \)

Substitute \( \lambda = \sqrt{5} \) into \( \mathbf{A} - \lambda\mathbf{I} = \begin{pmatrix} -\sqrt{5} & 5 \ 1 & -\sqrt{5} \end{pmatrix} \) and solve \( \begin{pmatrix} -\sqrt{5} & 5 \ 1 & -\sqrt{5} \end{pmatrix} \begin{pmatrix} x \ y \end{pmatrix} = \begin{pmatrix} 0 \ 0 \end{pmatrix} \). This results in the eigenvector \( \mathbf{v_1} = \begin{pmatrix} 1 \ \sqrt{5} \end{pmatrix} \).

Step 6: Eigenvector Calculation for \( \lambda = -\sqrt{5} \)

Substitute \( \lambda = -\sqrt{5} \) into \( \mathbf{A} - \lambda\mathbf{I} = \begin{pmatrix} \sqrt{5} & 5 \ 1 & \sqrt{5} \end{pmatrix} \) and solve \( \begin{pmatrix} \sqrt{5} & 5 \ 1 & \sqrt{5} \end{pmatrix} \begin{pmatrix} x \ y \end{pmatrix} = \begin{pmatrix} 0 \ 0 \end{pmatrix} \). This gives the eigenvector \( \mathbf{v_2} = \begin{pmatrix} 1 \ -\sqrt{5} \end{pmatrix} \).

Step 7: Form the Matrix \( \mathbf{P} \)

Form the matrix \( \mathbf{P} \) using the eigenvectors \( \begin{pmatrix} 1 & 1 \ \sqrt{5} & -\sqrt{5} \end{pmatrix} \).

Step 8: Diagonal Matrix \( \mathbf{D} \)

The diagonal matrix \( \mathbf{D} \) is formed using the eigenvalues. Thus, \( \mathbf{D} = \begin{pmatrix} \sqrt{5} & 0 \ 0 & -\sqrt{5} \end{pmatrix} \).

Step 9: Verify Diagonalization

To verify, check that \( \mathbf{P}^{-1} \mathbf{A} \mathbf{P} = \mathbf{D} \). Calculate \( \mathbf{P}^{-1} \) and ensure the product results in the diagonal matrix \( \mathbf{D} \).

Key Concepts

Eigenvalues

Eigenvalues are a fundamental concept in linear algebra, especially when working with matrices. They are scalar values associated with a matrix. To find eigenvalues, we solve the equation \( \det(\mathbf{A} - \lambda \mathbf{I}) = 0 \), where \( \mathbf{A} \) is the matrix, \( \lambda \) is the eigenvalue, and \( \mathbf{I} \) is the identity matrix. In this exercise, we found two eigenvalues for the matrix \( \mathbf{A} = \begin{pmatrix} 0 & 5 \ 1 & 0 \end{pmatrix} \) using this method. The eigenvalues are \( \lambda = \sqrt{5} \) and \( \lambda = -\sqrt{5} \).
These eigenvalues help determine if the matrix \( \mathbf{A} \) is diagonalizable. A matrix is diagonalizable if it has enough linearly independent eigenvectors to form the diagonalizing matrix, \( \mathbf{P} \).

Eigenvectors

Eigenvectors correspond to eigenvalues and are vectors that describe the directions of stretching induced by a matrix. Once you determine the eigenvalues, you can find the eigenvectors by solving the equation \( (\mathbf{A} - \lambda \mathbf{I}) \mathbf{v} = \mathbf{0} \) for each eigenvalue \( \lambda \).
For our exercise, we calculated the eigenvectors for \( \lambda = \sqrt{5} \) and \( \lambda = -\sqrt{5} \). They are \( \mathbf{v_1} = \begin{pmatrix} 1 \ \sqrt{5} \end{pmatrix} \) and \( \mathbf{v_2} = \begin{pmatrix} 1 \ -\sqrt{5} \end{pmatrix} \), respectively. These vectors form the matrix \( \mathbf{P} \), which is used to transform \( \mathbf{A} \) into a diagonal matrix.

Characteristic Equation

The characteristic equation is crucial for finding the eigenvalues of a matrix. It's derived from the determinant equation \( \det(\mathbf{A} - \lambda \mathbf{I}) = 0 \). In essence, this equation measures how a matrix respects its identity by offsetting it with a scalar, \( \lambda \). Solving this equation gives us the eigenvalues that are needed to understand the transformation properties of a matrix.
In our exercise, the characteristic equation was \( \lambda^2 - 5 = 0 \), solved by factoring into \( \lambda = \sqrt{5} \) and \( \lambda = -\sqrt{5} \). Understanding the characteristic equation is key to unlocking the behavior of a matrix under linear transformations.

Linear Algebra

Linear algebra involves the study of vectors, vector spaces, linear transformations, and matrices. It forms the backbone of many fields, including computer science, physics, and engineering. In the context of matrix diagonalization, linear algebra provides the tools to determine if a matrix can be transformed into a simpler, diagonal form.
This process involves finding eigenvalues and eigenvectors, and using these to construct a diagonal matrix and a transforming matrix. Matrix diagonalization is a powerful tool because it simplifies complex matrix operations, making calculations more manageable and often providing deeper insights into the system or process being analyzed.

• Matrix diagonalization allows for easier computation of matrix powers and exponentials.
• It simplifies the understanding of dynamic systems described by matrices.
• Linear algebra concepts like eigenvalues and eigenvectors are key for performing principal component analysis (PCA) in data science.

Using the principles of linear algebra, we can effectively understand and solve many real-world problems.