Problem 17 - 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 \(\mathbf{D}=\mathbf{P}^{-1} \mathbf{A} \mathbf{P}\). $$ \left(\begin{array}{rrr} 1 & 2 & 0 \\ 2 & -1 & 0 \\ 0 & 0 & 1 \end{array}\right) $$

Step-by-Step Solution

Answer

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

Step 1: Verify if matrix is diagonalizable

To determine if the matrix \( \mathbf{A} \) is diagonalizable, we need to find its eigenvalues and check if the algebraic multiplicity of each eigenvalue is equal to its geometric multiplicity. First, find the characteristic polynomial of \( \mathbf{A} \) by calculating the determinant of \( \mathbf{A} - \lambda \mathbf{I} \), where \( \lambda \) is an eigenvalue and \( \mathbf{I} \) is the identity matrix.\[\text{Characteristic polynomial: } \det\begin{pmatrix} 1-\lambda & 2 & 0 \ 2 & -1-\lambda & 0 \ 0 & 0 & 1-\lambda \end{pmatrix} = (1-\lambda)((1-\lambda)(-1-\lambda) - 4) = (1-\lambda)(\lambda^2 - 1)\]

Step 2: Solve for eigenvalues

Set the characteristic polynomial to zero to solve for the eigenvalues:\[(1 - \lambda)(\lambda^2 - 1) = 0\]This implies:\[\lambda_1 = 1, \quad \lambda_2 = 1, \quad \lambda_3 = -1\]Thus, the eigenvalues are \( \lambda = 1 \) with multiplicity 2 and \( \lambda = -1 \).

Step 3: Find the eigenvectors

To determine the eigenvectors, solve \( (\mathbf{A} - \lambda \mathbf{I})\mathbf{v} = \mathbf{0} \) for each eigenvalue.For \( \lambda = 1 \):\[\begin{pmatrix} 0 & 2 & 0 \ 2 & -2 & 0 \ 0 & 0 & 0 \end{pmatrix} \begin{pmatrix} x \ y \ z \end{pmatrix} = \begin{pmatrix} 0 \ 0 \ 0 \end{pmatrix}\]This results in the eigenvector form \( y = 0 \) and \( x = z \), or \( \begin{pmatrix} 1 \ 0 \ 1 \end{pmatrix} \).For \( \lambda = -1 \):\[\begin{pmatrix} 2 & 2 & 0 \ 2 & 0 & 0 \ 0 & 0 & 2 \end{pmatrix} \begin{pmatrix} x \ y \ z \end{pmatrix} = \begin{pmatrix} 0 \ 0 \ 0 \end{pmatrix}\]This results in the eigenvector \( \begin{pmatrix} -1 \ 1 \ 0 \end{pmatrix} \).

Step 4: Construct the matrix P

Construct the matrix \( \mathbf{P} \) using the independent eigenvectors found:\[\mathbf{P} = \begin{pmatrix} 1 & 0 & -1 \ 0 & 1 & 1 \ 1 & 0 & 0 \end{pmatrix}\]

Step 5: Construct the diagonal matrix D

The diagonal matrix \( \mathbf{D} \) is constructed from the eigenvalues, listed in the order corresponding to the eigenvectors in \( \mathbf{P} \):\[\mathbf{D} = \begin{pmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & -1 \end{pmatrix}\]

Step 6: Verify the diagonalization

Confirm that \( \mathbf{D} = \mathbf{P}^{-1} \mathbf{A} \mathbf{P} \) by computing \( \mathbf{P}^{-1} \) and ensuring that the equation holds. Calculate \( \mathbf{P}^{-1} \):\[\mathbf{P}^{-1} = \begin{pmatrix} 0 & 1 & 0 \ 1 & 0 & 1 \ -1 & 1 & 1 \end{pmatrix}\]Check the multiplication:\[\mathbf{D} = \begin{pmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & -1 \end{pmatrix}\]

Key Concepts

Eigenvalues and Eigenvectors

Eigenvalues and eigenvectors are fundamental concepts in linear algebra that play a crucial role in matrix diagonalization. Eigenvalues represent the factors by which an eigenvector is stretched or compressed along its direction during a linear transformation. When we multiply a matrix by its eigenvector, the resulting vector is simply a scaled version of the eigenvector, and the scalar is the eigenvalue.

To find eigenvalues, you need to solve the \( \lambda \) for the characteristic equation \( \text{det}(\mathbf{A} - \lambda\mathbf{I}) = 0 \), where \( \mathbf{A} \) is your matrix and \( \mathbf{I} \) is the identity matrix of the same dimension.

Once eigenvalues are determined, eigenvectors can be found by solving \( (\mathbf{A} - \lambda \mathbf{I})\mathbf{v} = \mathbf{0} \). This means finding non-zero vectors \( \mathbf{v} \) where the transformation caused by \( \mathbf{A} \) only stretches or shrinks \( \mathbf{v} \) and does not alter its direction.

Characteristic Polynomial

The characteristic polynomial is a simplified expression used to find the eigenvalues of a matrix. It is calculated by taking the determinant of \( \mathbf{A} - \lambda \mathbf{I} \), effectively creating an equation that represents the matrix in its characteristic form.

For example, given a matrix \( \mathbf{A} \), you construct \( \mathbf{A} - \lambda \mathbf{I} \), replace each diagonal entry of \( \mathbf{A} \) with \( \lambda \), and compute the determinant. Solving the resulting polynomial equation provides the eigenvalues of the matrix. These eigenvalues are the roots of the characteristic polynomial.

In our specific example, the characteristic polynomial is \( (1 - \lambda)((1 - \lambda)(-1 - \lambda) - 4) = (1 - \lambda)(\lambda^2 - 1) \). Its roots \( \lambda_1 = 1 \) (with multiplicity 2) and \( \lambda_2 = -1 \) are calculated to find the eigenvalues.

Algebraic and Geometric Multiplicity

Algebraic and geometric multiplicities help us understand the diagonalizability of a matrix by providing information about its eigenvalues and eigenvectors.

- **Algebraic Multiplicity** refers to the number of times an eigenvalue appears as a root of the characteristic polynomial. For example, if the polynomial is \( (\lambda - 1)^2 \) for \( \lambda = 1 \), then \( \lambda = 1 \) has an algebraic multiplicity of 2.
- **Geometric Multiplicity** is the number of independent eigenvectors associated with an eigenvalue, essentially indicating the dimension of the eigenspace for that eigenvalue. To check this, you solve the equation \( (\mathbf{A} - \lambda \mathbf{I})\mathbf{v} = \mathbf{0} \) for its eigenvectors.

For a matrix to be diagonalizable, the algebraic multiplicity of each eigenvalue must equal its geometric multiplicity. In our example:
- The eigenvalue \( \lambda = 1 \) with algebraic multiplicity 2 has a corresponding pair of independent eigenvectors, fulfilling the condition for diagonalizability.
- The eigenvalue \( \lambda = -1 \) corresponds to its single eigenvector, also maintaining balance between its multiplicities, indicating the matrix is indeed diagonalizable.