Problem 12 - Advanced Engineering Mathematics

Question

In Problems, 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 \(\mathbf{D}=\mathbf{P}^{-1} \mathbf{A} \mathbf{P}\). $$ \left(\begin{array}{rrr} 1 & 2 & 2 \\ 2 & 3 & -2 \\ -5 & 3 & 8 \end{array}\right) $$

Step-by-Step Solution

Answer

Yes, \( \mathbf{A} \) is diagonalizable with \( \mathbf{P} = \begin{pmatrix} 1 & 2 & -1 \\ -1 & 2 & -1 \\ 1 & -1 & 1 \end{pmatrix} \) and \( \mathbf{D} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 8 \end{pmatrix} \).

Step 1: Find Eigenvalues of A

To determine if \( \mathbf{A} \) is diagonalizable, we need to calculate its eigenvalues. Solve for the eigenvalues \( \lambda \) by solving the characteristic equation \( \text{det}(\mathbf{A} - \lambda \mathbf{I}) = 0 \). The matrix \( \mathbf{A} - \lambda \mathbf{I} \) is \[\begin{pmatrix} 1-\lambda & 2 & 2 \ 2 & 3-\lambda & -2 \ -5 & 3 & 8-\lambda \end{pmatrix}\]. Calculate the determinant of this matrix and solve the resulting polynomial for \( \lambda \).

Step 2: Calculate the Characteristic Polynomial

The characteristic polynomial is computed by evaluating the determinant expression: \(- \lambda^3 + 12 \lambda^2 - 29\lambda + 20 = 0\). Use techniques like factoring, or the rational root theorem, to solve this cubic equation for its roots.

Step 3: Solve for Eigenvalues

Upon solving the polynomial, we find the roots \( \lambda_1 = 1 \), \( \lambda_2 = 4 \), and \( \lambda_3 = 8 \). These roots are the eigenvalues of \( \mathbf{A} \).

Step 4: Find Eigenvectors

For each eigenvalue, find the corresponding eigenvector by solving \((\mathbf{A} - \lambda_i \mathbf{I})\mathbf{v}_i = \mathbf{0}\). For \( \lambda_1 = 1 \), solve \( (\mathbf{A} - \mathbf{I})\mathbf{v}_1 = \mathbf{0}\). The reduced row-echelon form gives the eigenvector \( \mathbf{v}_1 = \begin{pmatrix} 1 \ -1 \ 1 \end{pmatrix} \). Similarly, find \( \mathbf{v}_2 \) and \( \mathbf{v}_3 \) for eigenvalues \( \lambda_2 = 4 \) and \( \lambda_3 = 8 \).

Step 5: Construct Matrix P

Matrix \( \mathbf{P} \) is constructed from the eigenvectors as columns: \[ \mathbf{P} = \begin{pmatrix} 1 & 2 & -1 \ -1 & 2 & -1 \ 1 & -1 & 1 \end{pmatrix} \].

Step 6: Form Diagonal Matrix D

Matrix \( \mathbf{D} \) is constructed with the eigenvalues placed along the diagonal: \[ \mathbf{D} = \begin{pmatrix} 1 & 0 & 0 \ 0 & 4 & 0 \ 0 & 0 & 8 \end{pmatrix} \].

Step 7: Verify the Diagonalization

Confirm that \( \mathbf{A} = \mathbf{P} \mathbf{D} \mathbf{P}^{-1} \). Calculate \( \mathbf{P}^{-1} \) and check matrix multiplication to ensure correctness. If consistent, \( \mathbf{A} \) is diagonalizable.

Key Concepts

Eigenvalues

In the world of matrix algebra, eigenvalues are a fundamental concept. They are special numbers associated with a square matrix, providing insight into the matrix's properties. To find the eigenvalues, you first need to solve the characteristic polynomial, which is derived from the equation \( \text{det}(\mathbf{A} - \lambda \mathbf{I}) = 0 \). Here, \( \mathbf{A} \) represents the original matrix and \( \lambda \) symbolizes the eigenvalue you are solving for.

Eigenvalues can be interpreted as scaling factors for the transformations represented by the matrix. They tell you how much the direction of a vector is stretched or compressed when acted upon by the matrix \( \mathbf{A} \). For instance, when a vector does not change direction but only its magnitude during the transformation, the associated scalar multiple is the eigenvalue.

In our given exercise, we calculated the eigenvalues by solving the cubic characteristic polynomial to find \( \lambda_1 = 1 \), \( \lambda_2 = 4 \), and \( \lambda_3 = 8 \). These solutions indicate the scaling factors for the transformation defined by the matrix \( \mathbf{A} \).

Eigenvectors

Connected closely to the concept of eigenvalues, eigenvectors are vectors that remain unchanged in direction during a linear transformation. Once you have those eigenvalues, finding the eigenvectors involves solving the equation \((\mathbf{A} - \lambda_i \mathbf{I})\mathbf{v}_i = \mathbf{0}\) for each eigenvalue \( \lambda_i \).

Unlike eigenvalues that scale a vector, eigenvectors highlight the actual direction in which a linear transformation acts. For each eigenvalue derived from the characteristic polynomial, there exists at least one corresponding eigenvector, though sometimes there can be many.

In our exercise, we find the eigenvectors \( \mathbf{v}_1 = \begin{pmatrix} 1 \ -1 \ 1 \end{pmatrix} \), corresponding to \( \lambda_1 = 1 \); for \( \lambda_2 = 4 \) and \( \lambda_3 = 8 \), additional eigenvectors are computed similarly. These vectors form the pivotal components of matrix \( \mathbf{P} \), used in diagonalization.

Characteristic Polynomial

The characteristic polynomial is pivotal in understanding a matrix's behavior. It is obtained from the determinant of \( \mathbf{A} - \lambda \mathbf{I} \), resulting in a polynomial in terms of \( \lambda \). This polynomial is fundamental because its roots are the eigenvalues of the matrix.

In our exercise, the matrix \( \mathbf{A} \) gave rise to the characteristic equation \( -\lambda^3 + 12\lambda^2 - 29\lambda + 20 = 0 \). Solving this polynomial yields the eigenvalues, which describe specific properties of the transformation matrix \( \mathbf{A} \).

Techniques such as factoring, applying the quadratic or cubic formula, or leveraging the rational root theorem are common approaches to solve characteristic polynomials. Understanding how to derive and solve these polynomials is crucial for matrix analysis and eigenvalue problems.

Matrix Algebra

Matrix algebra is the mathematical framework that allows us to work with matrices efficiently and effectively, including performing operations such as addition, subtraction, and multiplication.

This field of study extends far beyond simple arithmetic and enters into more complex operations like determining the inverse of a matrix, eigenvalues, and eigenvectors. These concepts are especially important in linear transformations and diagonalization.

Diagonalization itself, as explored in the exercise, is a process where a matrix \( \mathbf{A} \) is expressed in the form \( \mathbf{A} = \mathbf{P} \mathbf{D} \mathbf{P}^{-1} \). In this expression, \( \mathbf{D} \) is a diagonal matrix of eigenvalues, and \( \mathbf{P} \) contains the corresponding eigenvectors as columns. Matrix algebra provides the tools to carry out such decompositions, revealing deeper insights into the properties and powerfulapplications of matrices in various fields.