Problem 16 - 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 \(\mathbf{D}=\mathbf{P}^{-1} \mathbf{A P}\). $$ \left(\begin{array}{lll} 1 & 1 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{array}\right) $$

Step-by-Step Solution

Answer

The matrix \( \mathbf{A} \) is diagonalizable with \( \mathbf{P} = \begin{pmatrix} 1 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} \) and \( \mathbf{D} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{pmatrix} \).

Step 1: Determine Eigenvalues

Find the eigenvalues of the matrix \( \mathbf{A} \) by solving the characteristic equation \( \det(\mathbf{A} - \lambda \mathbf{I}) = 0 \). For our matrix, \( \mathbf{I} \) is the identity matrix and \( \lambda \) is the eigenvalue. \[\det\left(\begin{array}{ccc} 1 - \lambda & 1 & 0 \ 0 & 2 - \lambda & 0 \ 0 & 0 & 3 - \lambda \end{array}\right) = 0\]The determinant can be simplified as:\[(1-\lambda)((2-\lambda)(3-\lambda)) = 0\]The eigenvalues are \( \lambda = 1, 2, 3 \).

Step 2: Check Algebraic and Geometric Multiplicity

A matrix is diagonalizable if, for each eigenvalue, its algebraic multiplicity equals its geometric multiplicity. Here, we have three distinct eigenvalues \( \lambda = 1, 2, 3 \), each repeated once (i.e., algebraic multiplicity is 1 for each). Therefore, we only need to find a linearly independent set of eigenvectors corresponding to each eigenvalue.

Step 3: Find Eigenvectors for Each Eigenvalue

For each eigenvalue \( \lambda_i \), solve \( (\mathbf{A} - \lambda_i \mathbf{I}) \mathbf{x} = \mathbf{0} \) to find the eigenvectors.**Eigenvalue \( \lambda = 1 \):**Solve \((\mathbf{A} - \mathbf{I}) \mathbf{x} = \mathbf{0}\), which results in:\[\begin{pmatrix} 0 & 1 & 0 \ 0 & 1 & 0 \ 0 & 0 & 2 \end{pmatrix} \mathbf{x} = \mathbf{0}\]Eigenvector: \( \mathbf{x}_1 = \begin{pmatrix} 1 \ 0 \ 0 \end{pmatrix} \).**Eigenvalue \( \lambda = 2 \):**\[\begin{pmatrix} -1 & 1 & 0 \ 0 & 0 & 0 \ 0 & 0 & 1 \end{pmatrix} \mathbf{x} = \mathbf{0}\]Eigenvector: \( \mathbf{x}_2 = \begin{pmatrix} 1 \ 1 \ 0 \end{pmatrix} \).**Eigenvalue \( \lambda = 3 \):**\[\begin{pmatrix} -2 & 1 & 0 \ 0 & -1 & 0 \ 0 & 0 & 0 \end{pmatrix} \mathbf{x} = \mathbf{0}\]Eigenvector: \( \mathbf{x}_3 = \begin{pmatrix} 0 \ 0 \ 1 \end{pmatrix} \).

Step 4: Construct Matrix P

Compose matrix \( \mathbf{P} \) using the eigenvectors as columns. So for our case:\[\mathbf{P} = \begin{pmatrix} 1 & 1 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{pmatrix}\]

Step 5: Construct Diagonal Matrix D

Matrix \( \mathbf{D} \) is simply the diagonal matrix composed of the eigenvalues in the same order as their corresponding eigenvectors in \( \mathbf{P} \):\[\mathbf{D} = \begin{pmatrix} 1 & 0 & 0 \ 0 & 2 & 0 \ 0 & 0 & 3 \end{pmatrix}\]

Step 6: Verify the Result

Check that \( \mathbf{AP} = \mathbf{PD} \). If true, the matrix is correctly diagonalized. Calculate:\[\mathbf{AP} = \begin{pmatrix} 1 & 1 & 0 \ 0 & 2 & 0 \ 0 & 0 & 3 \end{pmatrix} \begin{pmatrix} 1 & 1 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 2 & 0 \ 0 & 2 & 0 \ 0 & 0 & 3 \end{pmatrix}\]Since \( \mathbf{PD} \) also equals:\[\begin{pmatrix} 1 & 1 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 & 0 \ 0 & 2 & 0 \ 0 & 0 & 3 \end{pmatrix} = \begin{pmatrix} 1 & 2 & 0 \ 0 & 2 & 0 \ 0 & 0 & 3 \end{pmatrix}\]Both equals verify the result.

Key Concepts

Eigenvalues

Eigenvalues are special numbers associated with a matrix that give us valuable information about the matrix's properties. To determine these, we solve the characteristic equation, which involves finding the roots of the polynomial derived from the matrix. In the context of matrix diagonalization, eigenvalues are essential. They populate the diagonal of the diagonal matrix

• An eigenvalue, denoted by \( \lambda \), satisfies \( \det(A - \lambda I) = 0 \), where \( A \) is our matrix and \( I \) is the identity matrix.
• For the example matrix, the diagonalization process gives us eigenvalues \( 1, 2, \) and \( 3 \).

Eigenvalues define the scaling factor that alters the direction of an eigenvector but not its span. Without detecting eigenvalues first, we cannot progress to finding eigenvectors and thus confirming if a matrix is diagonalizable.

Eigenvectors

Eigenvectors are vectors that, when multiplied by a matrix, yield an output that's a scalar multiple of themselves. These vectors point in a direction that is invariant when transformed by the matrix. The importance of eigenvectors in diagonalization is significant as they form the basis for the columns of matrix \( P \), which diagonalizes matrix \( A \).

• Eigenvectors are found by solving \( (A - \lambda I)x = 0 \) for each eigenvalue \( \lambda \).
• In our matrix example, the eigenvectors for \( \lambda = 1, 2, \text{ and } 3 \) are \( \begin{pmatrix} 1 \ 0 \ 0 \end{pmatrix}, \begin{pmatrix} 1 \ 1 \ 0 \end{pmatrix}, \begin{pmatrix} 0 \ 0 \ 1 \end{pmatrix}\).

Eigenvectors must be linearly independent to successfully construct the matrix \( P \). This ensures that the matrix is truly diagonalizable in conjunction with its eigenvalues.

Characteristic Equation

The characteristic equation is central to finding the eigenvalues of a matrix. It is derived from the equation \( \det(A - \lambda I) = 0 \) and results in a polynomial equation in terms of \( \lambda \). Solving this polynomial provides the potential eigenvalues.

• This equation is obtained by subtracting \( \lambda I \) from \( A \) and setting its determinant equal to zero.
• For our matrix, the characteristic polynomial leads to \((\lambda - 1)(\lambda - 2)(\lambda - 3)\).

By finding the roots of this equation, you obtain the matrix's eigenvalues. This methodology allows us to understand the structure and behavior of a matrix in relation to its eigenvectors and consequently its diagonal form.

Diagonal Matrix

A Diagonal Matrix is a special kind of matrix where the entries outside the main diagonal are all zero. A matrix is considered diagonalizable if it can be expressed as \( P^{-1}AP = D \) where \( D \) is a diagonal matrix containing the eigenvalues of \( A \).

• The diagonal matrix \( D \) has eigenvalues as its diagonal entries.
• For our problem, \( D = \begin{pmatrix} 1 & 0 & 0 \ 0 & 2 & 0 \ 0 & 0 & 3 \end{pmatrix} \), containing the eigenvalues \( 1, 2, \text{ and } 3 \).

Diagonal matrices possess simple properties that simplify many types of matrix calculations, especially when dealing with powers and exponential functions. Recognizing matrix diagonalization can drastically ease analyzing complex systems in advanced algebra.