Problem 7 - Advanced Engineering Mathematics

Question

In Problems 1-10, use diagonalization to solve the given system. $$ \mathbf{X}^{\prime}=\left(\begin{array}{rrr} 1 & -1 & -1 \\ -1 & 1 & -1 \\ -1 & -1 & 1 \end{array}\right) \mathbf{X} $$

Step-by-Step Solution

Answer

Find eigenvalues and eigenvectors, use them to diagonalize \( A \), and solve \( \mathbf{X} = Pe^{Dt}\mathbf{c}(0) \).

Step 1: Identify the Matrix A

To use diagonalization, first identify the matrix \( A \) from the system \( \mathbf{X}' = A \mathbf{X} \). Here, the matrix \( A \) is given by: \[ A = \begin{pmatrix} 1 & -1 & -1 \ -1 & 1 & -1 \ -1 & -1 & 1 \end{pmatrix} \].

Step 2: Find the Eigenvalues of A

Calculate the eigenvalues of the matrix \( A \) by solving the characteristic equation \( \text{det}(A - \lambda I) = 0 \). For \( A = \begin{pmatrix} 1 & -1 & -1 \ -1 & 1 & -1 \ -1 & -1 & 1 \end{pmatrix} \), compute \( \lambda \) values.

Step 3: Solve for the Characteristic Polynomial

The characteristic polynomial is determined by taking the determinant of \( A - \lambda I \), which is \( \begin{vmatrix} 1-\lambda & -1 & -1 \ -1 & 1-\lambda & -1 \ -1 & -1 & 1-\lambda \end{vmatrix} \). Simplifying gives \( (\lambda - 3)(\lambda + 1)^2 = 0 \), so the eigenvalues are \( \lambda_1 = 3 \) and \( \lambda_2 = \lambda_3 = -1 \).

Step 4: Find the Eigenvectors

Find the eigenvectors corresponding to each eigenvalue. Solve \( (A - 3I)\mathbf{v} = 0 \) to find the eigenvector for \( \lambda_1 = 3 \). For \( \lambda_2 = -1 \) and \( \lambda_3 = -1 \), solve \( (A + I)\mathbf{v} = 0 \).

Step 5: Compose the Matrix P and Diagonal Matrix D

Form the matrix \( P \) using the eigenvectors as columns. The diagonal matrix \( D \) corresponds to the eigenvalues along its diagonal. Sample solutions can be \( P = \begin{pmatrix} 1 & 0 & -1 \ 0 & 1 & -1 \ -1 & 1 & 0 \end{pmatrix} \) and \( D = \text{diag}(3, -1, -1) \).

Step 6: Solve the System Using Diagonalization

Use the relationship \( \mathbf{X} = P\mathbf{c} \), where \( \mathbf{c}' = D\mathbf{c} \). Solve \( \mathbf{c}' = D\mathbf{c} \) to obtain the solution \( \mathbf{c}(t) = e^{Dt}\mathbf{c}(0) \). Finally, substitute back to get \( \mathbf{X}(t) = Pe^{Dt}\mathbf{c}(0) \).

Key Concepts

Eigenvalues

To diagonalize a matrix, finding eigenvalues is crucial. Eigenvalues are specific scalar values linked to a square matrix, which preserve the direction of eigenvectors during matrix transformation. In simple terms, if you multiply an eigenvector by the matrix, the result is just the eigenvector scaled by its eigenvalue.

In our case, to find the eigenvalues of the given matrix \( A \), we solve the characteristic equation \( \text{det}(A - \lambda I) = 0 \). This involves subtracting \( \lambda \) times the identity matrix from \( A \), then finding the determinant of the resulting matrix.

• The characteristic equation for our matrix \( A \) is \( (\lambda - 3)(\lambda + 1)^2 = 0 \).
• This gives us the eigenvalues \( \lambda_1 = 3 \) and \( \lambda_2 = \lambda_3 = -1 \).

Eigenvectors

Eigenvectors are non-zero vectors that change by a scalar factor when a linear transformation is applied. After identifying the eigenvalues, the next step is finding the eigenvectors corresponding to each eigenvalue.

For each eigenvalue \( \lambda \), we solve the equation \( (A - \lambda I)\mathbf{v} = 0 \) to find the corresponding eigenvectors \( \mathbf{v} \).

• For \( \lambda_1 = 3 \), we solve \( (A - 3I)\mathbf{v} = 0 \).
• For \( \lambda_2 = -1 \) and \( \lambda_3 = -1 \), the equation becomes \( (A + I)\mathbf{v} = 0 \).

Each solution gives us the directions of the eigenvectors which will form part of the matrix \( P \) in the diagonalization process.

Characteristic Polynomial

The characteristic polynomial of a matrix plays a fundamental role in finding eigenvalues. It is derived from the equation \( \text{det}(A - \lambda I) = 0 \), which is a polynomial equation in terms of \( \lambda \). Solving this polynomial provides the eigenvalues of the matrix.

Specifically, for our matrix \( A \), the characteristic polynomial is \( (\lambda - 3)(\lambda + 1)^2 = 0 \). From this equation, we derive the eigenvalues required for diagonalization, specifically, \( \lambda_1 = 3 \), and \( \lambda_2 = \lambda_3 = -1 \).

• This polynomial effectively 'encodes' all the eigenvalues of the matrix into a single algebraic expression.
• Understanding this relation is key to unlocking the other steps involved in matrix diagonalization.

Matrix Decomposition

Matrix decomposition in linear algebra involves breaking down a matrix into simpler, more manageable components. One powerful approach is diagonalization, which represents a matrix as the product of its eigenvectors and eigenvalues in matrix form.

For diagonalization, we construct two matrices: \( P \) and \( D \).

• Matrix \( P \) is composed of the eigenvectors of \( A \) as its columns.
• Matrix \( D \) is a diagonal matrix, whose diagonal elements are the eigenvalues of \( A \).

In practice, matrix \( A \) is expressed as \( PDP^{-1} \), which simplifies many computations such as raising the matrix to a power or solving differential equations.

For our exercise, \( P \) and \( D \) are found as:

• \( P = \begin{pmatrix} 1 & 0 & -1 \ 0 & 1 & -1 \ -1 & 1 & 0 \end{pmatrix} \)
• \( D = \text{diag}(3, -1, -1) \).

This transformation makes solving systems like the one in the exercise more efficient and insightful.