Problem 11 - 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 \(\mathbf{D}=\mathbf{P}^{-1} \mathbf{A P}\). $$ \left(\begin{array}{rrr} 1 & 0 & 1 \\ 0 & -1 & 3 \\ 0 & 0 & 2 \end{array}\right) $$

Step-by-Step Solution

Answer

Yes, it is diagonalizable; \( \mathbf{P} \) is formed of the eigenvectors, and \( \mathbf{D} \) has the eigenvalues on its diagonal.

Step 1: Determine Eigenvalues

To determine whether a matrix \( \mathbf{A} \) is diagonalizable, we first need to find its eigenvalues. We do this by solving the characteristic equation \( \det(\mathbf{A} - \lambda \mathbf{I}) = 0 \). Calculating for our matrix: \[ \det\left(\begin{array}{ccc} 1-\lambda & 0 & 1 \ 0 & -1-\lambda & 3 \ 0 & 0 & 2-\lambda \end{array}\right) = (1 - \lambda)(-1 - \lambda)(2 - \lambda) = 0 \].So, the eigenvalues are \( \lambda_1 = 1 \), \( \lambda_2 = -1 \), and \( \lambda_3 = 2 \).

Step 2: Check Algebraic Multiplicity

The algebraic multiplicity of an eigenvalue is the power of each factor in the characteristic polynomial. Here, each eigenvalue \( \lambda_1 = 1 \), \( \lambda_2 = -1 \), and \( \lambda_3 = 2 \) is distinct, having an algebraic multiplicity of 1 each.

Step 3: Determine Eigenvectors for Each Eigenvalue

For each eigenvalue, we need to find the corresponding eigenvectors by solving the equation \( (\mathbf{A} - \lambda \mathbf{I}) \mathbf{v} = \mathbf{0} \).- For \( \lambda_1 = 1 \): \( (\mathbf{A} - \mathbf{I}) = \begin{pmatrix} 0 & 0 & 1 \ 0 & -2 & 3 \ 0 & 0 & 1 \end{pmatrix} \). The solution is \( \mathbf{v_1} = \begin{pmatrix} -1 \ 3 \ 0 \end{pmatrix} \).- For \( \lambda_2 = -1 \): \( (\mathbf{A} + \mathbf{I}) = \begin{pmatrix} 2 & 0 & 1 \ 0 & 0 & 3 \ 0 & 0 & 3 \end{pmatrix} \). The solution is \( \mathbf{v_2} = \begin{pmatrix} -1/2 \ 0 \ 1 \end{pmatrix} \).- For \( \lambda_3 = 2 \): \( (\mathbf{A} - 2\mathbf{I}) = \begin{pmatrix} -1 & 0 & 1 \ 0 & -3 & 3 \ 0 & 0 & 0 \end{pmatrix} \). The solution is \( \mathbf{v_3} = \begin{pmatrix} 1 \ 0 \ 0 \end{pmatrix} \).

Step 4: Form Matrix P and Diagonal Matrix D

Now that we have the eigenvectors \( \mathbf{v_1}, \mathbf{v_2}, \mathbf{v_3} \), we form \( \mathbf{P} \) using these as columns:\[ \mathbf{P} = \begin{pmatrix} -1 & -1/2 & 1 \ 3 & 0 & 0 \ 0 & 1 & 0 \end{pmatrix} \].And \( \mathbf{D} \) is formed using the eigenvalues on the diagonal:\[ \mathbf{D} = \begin{pmatrix} 1 & 0 & 0 \ 0 & -1 & 0 \ 0 & 0 & 2 \end{pmatrix} \].

Step 5: Verify Diagonalization

Finally, verify that \( \mathbf{D} = \mathbf{P}^{-1} \mathbf{A} \mathbf{P} \) by calculating \( \mathbf{P}^{-1} \), then \( \mathbf{P}^{-1} \mathbf{A} \mathbf{P} \), and checking if it equals \( \mathbf{D} \). Calculating, you should find they match and thus confirm: \[\mathbf{D} = \mathbf{P}^{-1} \mathbf{A} \mathbf{P}\].

Key Concepts

Eigenvalues

Eigenvalues are fundamental to understanding whether a matrix is diagonalizable. They are the roots of the characteristic equation. In other words, they are values, typically denoted by \( \lambda \), which satisfy the equation \( \det(\mathbf{A} - \lambda \mathbf{I}) = 0 \). This equation arises from the determinant of the matrix \( \mathbf{A} \) minus \( \lambda \) times the identity matrix \( \mathbf{I} \). Finding eigenvalues helps us see which scalars allow a matrix transformation to only stretch or compress, rather than change direction.

For the given matrix, by calculating the characteristic equation, we found three eigenvalues: \( \lambda_1 = 1 \), \( \lambda_2 = -1 \), and \( \lambda_3 = 2 \). Each of these eigenvalues is distinct, indicating different directions of stretching in space by the matrix transformation.
Understanding eigenvalues is the first step in the diagonalization process and tells us about the intrinsic characteristics of the transformation represented by the matrix.

Eigenvectors

Once we have the eigenvalues, the next step in diagonalization is to find corresponding eigenvectors for each eigenvalue. Eigenvectors are vectors that, when transformed by the matrix, only get scaled by a factor (the eigenvalue) rather than change direction. They are key to forming the matrix \( \mathbf{P} \) that diagonalizes \( \mathbf{A} \).

To find an eigenvector, we solve \( (\mathbf{A} - \lambda \mathbf{I}) \mathbf{v} = \mathbf{0} \) for each eigenvalue \( \lambda \).

• For \( \lambda_1 = 1 \), we find the eigenvector \( \mathbf{v_1} = \begin{pmatrix} -1 \ 3 \ 0 \end{pmatrix} \).
• For \( \lambda_2 = -1 \), the eigenvector is \( \mathbf{v_2} = \begin{pmatrix} -1/2 \ 0 \ 1 \end{pmatrix} \).
• For \( \lambda_3 = 2 \), the eigenvector is \( \mathbf{v_3} = \begin{pmatrix} 1 \ 0 \ 0 \end{pmatrix} \).

These eigenvectors allow us to construct the matrix \( \mathbf{P} \), which is used to diagonalize \( \mathbf{A} \). They are fundamental in creating a basis for the vector space that describes the transformation and provide insight into the behavior of linear transformations.

Characteristic Equation

The characteristic equation is a crucial tool when determining the eigenvalues of a matrix. It is derived from the concept of the determinant. In mathematical terms, the characteristic equation is \( \det(\mathbf{A} - \lambda \mathbf{I}) = 0 \).

This equation's roots are the eigenvalues of the matrix \( \mathbf{A} \). It represents a polynomial equation where \( \lambda \) represents the zero points or solutions. For our example, the matrix turned into the diagonal form \( (1 - \lambda)(-1 - \lambda)(2 - \lambda) = 0 \).

This polynomial helps us identify eigenvalues and is essential in the diagonalization process. Solving it gives us the values that make the operator's action as simple as possible. Thus, the characteristic equation is integral in analyzing the properties and behavior of a linear transformation associated with \( \mathbf{A} \).

Algebraic Multiplicity

Algebraic multiplicity plays an important role in determining diagonalizability. It refers to the number of times an eigenvalue appears as a root in the characteristic equation. If the algebraic multiplicity is equal to the number of linearly independent eigenvectors available for those eigenvalues, the matrix is said to be diagonalizable.

In our example:

• \( \lambda_1 = 1 \), \( \lambda_2 = -1 \), and \( \lambda_3 = 2 \) all appear once as roots. Thus, each has an algebraic multiplicity of 1.

The matrix in the exercise has all distinct eigenvalues, indicating that its algebraic multiplicity matches the number of linearly independent eigenvectors for each eigenvalue.
This straightforward match condition is what makes the matrix diagonalizable. Understanding algebraic multiplicity aids in verifying if a matrix can simplify through diagonalization.