Problem 10 - 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 \(\mathbf{D}=\mathbf{P}^{-1} \mathbf{A} \mathbf{P}\). $$ \left(\begin{array}{rr} 1 & 2 \\ -\frac{1}{2} & 1 \end{array}\right) $$

Step-by-Step Solution

Answer

The matrix is not diagonalizable because its geometric multiplicity is less than its algebraic multiplicity.

Step 1: Find Eigenvalues

To determine if \( \mathbf{A} \) is diagonalizable, we first find its eigenvalues. For a matrix \( \mathbf{A} \), eigenvalues are found by solving the characteristic equation \( \det(\mathbf{A} - \lambda \mathbf{I}) = 0 \). Here, \( \mathbf{A} = \begin{pmatrix} 1 & 2 \ -\frac{1}{2} & 1 \end{pmatrix} \) and \( \mathbf{I} \) is the identity matrix \( \begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix} \). \( \det(\begin{pmatrix} 1-\lambda & 2 \ -\frac{1}{2} & 1-\lambda \end{pmatrix}) = (1-\lambda)^2 + 1 = 0 \). This simplifies to \( (1-\lambda)^2 = 0 \), giving us one eigenvalue: \( \lambda = 1 \).

Step 2: Check Algebraic and Geometric Multiplicities

The algebraic multiplicity of an eigenvalue is the number of times it appears as a root in the characteristic polynomial. Here, the eigenvalue \( \lambda = 1 \) has algebraic multiplicity 2 because \( (1 - \lambda)^2 = 0 \). To check if the matrix is diagonalizable, its geometric multiplicity should be equal to its algebraic multiplicity.

Step 3: Find Eigenvectors

To find eigenvectors, solve \( (\mathbf{A} - \lambda \mathbf{I})\mathbf{v} = \mathbf{0} \). For \( \lambda = 1 \), solve \( (\begin{pmatrix} 1-1 & 2 \ -\frac{1}{2} & 1-1 \end{pmatrix})\begin{pmatrix} x \ y \end{pmatrix} = \begin{pmatrix} 0 \ 0 \end{pmatrix} \), which simplifies to \( \begin{pmatrix} 0 & 2 \ -\frac{1}{2} & 0 \end{pmatrix}\begin{pmatrix} x \ y \end{pmatrix} = \begin{pmatrix} 0 \ 0 \end{pmatrix} \). This system gives the equations \( 2y = 0 \) and \( -\frac{1}{2}x = 0 \), implying that \( y = 0 \) and \( x = 0 \). Therefore, there is no non-trivial solution, meaning the geometric multiplicity is less than 2.

Step 4: Determine If Diagonalizable

A matrix is diagonalizable if the geometric multiplicity of each eigenvalue equals its algebraic multiplicity. Here, \( \lambda = 1 \) has a geometric multiplicity of 0, which is less than its algebraic multiplicity of 2. Therefore, the matrix is not diagonalizable.

Key Concepts

Diagonalization

Diagonalization is a useful process in linear algebra for simplifying the manipulation of matrices. It involves transforming a given matrix into a diagonal matrix, which can make many mathematical operations easier. A matrix \( \mathbf{A} \) is said to be diagonalizable if there exists a matrix \( \mathbf{P} \) such that \( \mathbf{D} = \mathbf{P}^{-1} \mathbf{A} \mathbf{P} \) is a diagonal matrix. However, not all matrices can be diagonalized. For a matrix to be diagonalizable, it must have enough linearly independent eigenvectors to form the columns of \( \mathbf{P} \). This often relates to the concept of eigenvalues and eigenvectors, which we will explore next.
One important note is that diagonalization is only possible if the geometric multiplicity of each eigenvalue matches its algebraic multiplicity. Otherwise, as in the exercise, the matrix is not diagonalizable.

Eigenvalues

Eigenvalues are scalar values associated with a matrix that provide significant insight into its properties. They are derived from the characteristic equation \( \det(\mathbf{A} - \lambda \mathbf{I}) = 0 \), where \( \lambda \) represents the eigenvalue, \( \mathbf{A} \) is the matrix, and \( \mathbf{I} \) is the identity matrix of the same dimensions as \( \mathbf{A} \). These values tell us about the scaling factors of the eigenvectors when the matrix is applied to them.
In the exercise, we determined that the matrix \( \mathbf{A} \) has one eigenvalue \( \lambda = 1 \). This eigenvalue has an algebraic multiplicity of 2, which typically correlates to the number of times it appears as a root in the characteristic polynomial. Finding eigenvalues is the first step in assessing whether a matrix could potentially be diagonalizable.

Eigenvectors

Eigenvectors are vectors that, when transformed by a matrix, only scale the vector rather than changing its direction. In other words, for a matrix \( \mathbf{A} \) and its eigenvalue \( \lambda \), an eigenvector \( \mathbf{v} \) satisfies the equation \( (\mathbf{A} - \lambda \mathbf{I})\mathbf{v} = \mathbf{0} \).
In our exercise, we attempted to find eigenvectors corresponding to the eigenvalue \( \lambda = 1 \). Upon solving \( (\mathbf{A} - \mathbf{I})\mathbf{v} = \mathbf{0} \), we found that there are no non-trivial solutions as it simplified to trivial equations \( 2y = 0 \) and \(-\frac{1}{2}x = 0 \). This indicates that their geometric multiplicity does not match the algebraic multiplicity, leading to the conclusion that not enough eigenvectors exist to diagonalize the matrix under given criteria.

Characteristic Equation

The characteristic equation of a matrix is a polynomial equation that is fundamental in finding the eigenvalues of a matrix. This is defined by \( \det(\mathbf{A} - \lambda \mathbf{I}) = 0 \), where each solution \( \lambda \) is an eigenvalue of the matrix \( \mathbf{A} \). The characteristic equation forms the basis for exploring deeper properties of matrices including stability, diagonalization, and more.
From the exercise, forming the characteristic equation was crucial in determining if the matrix \( \mathbf{A} \) is diagonalizable. By solving \( \det(\mathbf{A} - \lambda \mathbf{I}) = 0 \), we derived \( (1 - \lambda)^2 = 0 \), giving the eigenvalue \( \lambda = 1 \). The equations ultimately dictated the search for eigenvectors and the analysis of both algebraic and geometric multiplicities.