Problem 3 - 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 \(\mathbf{D}=\mathbf{P}^{-1} \mathbf{A} \mathbf{P}\). $$ \left(\begin{array}{rr} 0 & 1 \\ -1 & 2 \end{array}\right) $$

Step-by-Step Solution

Answer

The matrix \( \mathbf{A} \) is not diagonalizable because it lacks enough independent eigenvectors.

Step 1: Find the Eigenvalues of Matrix A

The given matrix \( \mathbf{A} \) is \( \begin{pmatrix} 0 & 1 \ -1 & 2 \end{pmatrix} \). To find the eigenvalues, solve the characteristic equation \( \det(\mathbf{A} - \lambda \mathbf{I}) = 0 \), where \( \mathbf{I} \) is the identity matrix of the same dimension as \( \mathbf{A} \).Calculate \( \mathbf{A} - \lambda \mathbf{I} = \begin{pmatrix} 0 & 1 \ -1 & 2 \end{pmatrix} - \begin{pmatrix} \lambda & 0 \ 0 & \lambda \end{pmatrix} = \begin{pmatrix} -\lambda & 1 \ -1 & 2-\lambda \end{pmatrix} \).Find the determinant: \[det \begin{pmatrix} -\lambda & 1 \ -1 & 2-\lambda \end{pmatrix} = (-\lambda)(2-\lambda) - (1)(-1) = \lambda^2 - 2\lambda + 1 \].Solve \( \lambda^2 - 2\lambda + 1 = 0 \), factor to get \( (\lambda - 1)^2 = 0 \). Therefore, \( \lambda = 1 \) is the only eigenvalue with multiplicity 2.

Step 2: Find the Eigenvectors of Matrix A

For each eigenvalue found in Step 1, solve \( (\mathbf{A} - \lambda \mathbf{I}) \mathbf{x} = \mathbf{0} \) for the eigenvector \( \mathbf{x} \).Using \( \lambda = 1 \), solve \( \begin{pmatrix} -1 & 1 \ -1 & 1 \end{pmatrix} \begin{pmatrix} x_1 \ x_2 \end{pmatrix} = \begin{pmatrix} 0 \ 0 \end{pmatrix} \).The system of equations is \(-x_1 + x_2 = 0\), which simplifies to \(x_2 = x_1\). Let \(x_1 = t\), then \(x_2 = t\) resulting in the eigenvector \( \begin{pmatrix} t \ t \end{pmatrix} = t \begin{pmatrix} 1 \ 1 \end{pmatrix}\).Since we only find one independent eigenvector for the repeated eigenvalue, the matrix \( \mathbf{A} \) is not diagonalizable.

Key Concepts

Eigenvalues

Eigenvalues are essential in understanding the behavior of a square matrix. They are the special numbers associated with a matrix, which are found by solving the characteristic equation. In simpler terms, eigenvalues tell us how much a matrix can stretch or shrink vectors when applied to them.

To find the eigenvalues of a matrix, we subtract a scalar multiplied by an identity matrix from the original matrix and then determine the determinant of the resulting matrix. Setting this determinant equal to zero gives us the characteristic equation, which we solve to find the eigenvalues.

For example, considering the matrix \( \begin{pmatrix} 0 & 1 \ -1 & 2 \end{pmatrix} \), we derive the equation \( \lambda^2 - 2\lambda + 1 = 0 \) by following these steps, and the only solution to this is the eigenvalue \( \lambda = 1 \) with a multiplicity of 2.

Eigenvectors

Once we have the eigenvalues, the next step is finding the eigenvectors, which are non-zero vectors that, when the matrix acts on them, only scale by the eigenvalue and do not change direction.

Given an eigenvalue, we find its respective eigenvector by solving the equation \( (\mathbf{A} - \lambda \mathbf{I}) \mathbf{x} = \mathbf{0} \). This involves substituting the eigenvalue back into the equation and solving for \( \mathbf{x} \).

In our example, substituting \( \lambda = 1 \) into the equation, we find that the eigenvector satisfying \( \begin{pmatrix} -1 & 1 \ -1 & 1 \end{pmatrix} \mathbf{x} = \mathbf{0} \) is \( t \begin{pmatrix} 1 \ 1 \end{pmatrix} \), where \( t \) is a parameter. Here, unfortunately, we obtain only one independent eigenvector, indicating there is not enough to diagonalize the matrix, as diagonalization requires as many independent eigenvectors as the dimension of the matrix.

Characteristic Equation

The characteristic equation is a fundamental aspect of finding eigenvalues, and it is pivotal in linear algebra for matrix analysis. This polynomial equation arises from the determinant of the matrix obtained by subtracting the scalar \( \lambda \) multiplied by the identity matrix from the original matrix.

To elaborate, if \( \mathbf{A} \) is a square matrix, its characteristic equation is given by \( \det(\mathbf{A} - \lambda \mathbf{I}) = 0 \). Solving this equation provides the eigenvalues of the matrix.

In our exercise, the characteristic equation \( \lambda^2 - 2\lambda + 1 = 0 \) was derived from the matrix \( \begin{pmatrix} -\lambda & 1 \ -1 & 2 - \lambda \end{pmatrix} \). After factoring, the equation \( (\lambda - 1)^2 = 0 \) indicated a single eigenvalue, \( \lambda = 1 \), repeated twice. Understanding the roots of this equation is essential in determining whether a matrix is diagonalizable.

Diagonal Matrix

A diagonal matrix plays a significant role in simplifying matrix operations. It is a special type of matrix where all elements off the main diagonal are zero. Such matrices are important because they simplify matrix computations, including powers and exponential matrices.

Matrix diagonalization involves expressing a given matrix as the product of three matrices: \( \mathbf{P}^{-1} \mathbf{A} \mathbf{P} \), where \( \mathbf{P} \) contains the eigenvectors of \( \mathbf{A} \), and \( \mathbf{D} \) is a diagonal matrix containing the eigenvalues.

In our problem, because \( \mathbf{A} \) did not have enough independent eigenvectors, it was not diagonalizable. Normally, if diagonalization is successful, \( \mathbf{D} \) would contain the eigenvalues on the diagonal, arranged in any order. Diagonal matrices simplify many operations, making them very powerful in theoretical and practical applications.