Question

Using equations (13) and (14), find a fundamental set of solutions for the linear system \(\mathbf{y}^{\prime}=A \mathbf{y} .\) In each exercise, you are given an eigenvalue \(\lambda\), where \(\lambda\) has algebraic multiplicity 3 and geometric multiplicity 1 and an eigenvector \(\mathbf{v}_{1}\).\(A=\left[\begin{array}{lll}4 & 0 & 0 \\ 2 & 4 & 0 \\ 1 & 3 & 4\end{array}\right]\)

Short answer

Based on the given matrix: $$ A = \begin{bmatrix} 4 & 0 & 0 \\ 2 & 4 & 0 \\ 1 & 3 & 4 \end{bmatrix} $$ and the information about the eigenvalue \(\lambda = 4\) with algebraic multiplicity 3 and geometric multiplicity 1, we derived the fundamental set of solutions for the linear system \(\mathbf{y}^{\prime} = A \mathbf{y}\). The fundamental set of solutions is given by: $$ \mathbf{y}_1 = \begin{bmatrix} e^{4t} \\ 0 \\ 0 \end{bmatrix}, \quad \mathbf{y}_2 = \begin{bmatrix} te^{4t} \\ 0 \\ e^{4t} \end{bmatrix}, \quad \mathbf{y}_3 = \begin{bmatrix} \frac{1}{2}t^2 e^{4t} \\ te^{4t} \\ (1 - t)e^{4t} \end{bmatrix}. $$

Step-by-step solution

Compute the eigenvalue

We are given the eigenvalue \(\lambda\) has an algebraic multiplicity of 3 and a geometric multiplicity of 1. Since the matrix A is an upper triangular matrix, we can directly read the eigenvalue from the diagonal: $$ \lambda = 4 $$

Compute the eigenvectors

We are given the first eigenvector \(\mathbf{v}_1\). To find the second and third eigenvectors, we need to compute the generalized eigenvectors associated with the eigenvalue \(\lambda = 4\): Solve the equation: $$ (A - \lambda I) \mathbf{v}_2 = \mathbf{v}_1, $$ And: $$ (A - \lambda I) \mathbf{v}_3 = \mathbf{v}_2. $$ With the given matrix A and eigenvalue 4, we have: $$ (A - 4I) \mathbf{v}_2 = \textbf{v}_1 \Longrightarrow \left[\begin{array}{ccc}0 & 0 & 0 \\ 2 & 0 & 0 \\ 1 & 3 & 0 \end{array}\right] \mathbf{v}_2 = \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} $$ Let \(\mathbf{v}_2 = [a, b, c]^T\), solving the system we get a solution: $$ \mathbf{v}_2 = \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} $$ And for the third eigenvector, $$ (A - 4I) \mathbf{v}_3 = \mathbf{v}_2 \Longrightarrow \left[\begin{array}{ccc}0 & 0 & 0 \\ 2 & 0 & 0 \\ 1 & 3 & 0 \end{array}\right] \mathbf{v}_3 = \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} $$ Let \(\mathbf{v}_3 = [a, b, c]^T\), solving the system we get a solution: $$ \mathbf{v}_3 = \begin{bmatrix} 0 \\ 1 \\ -1 \end{bmatrix} $$

Compute the fundamental set of solutions

Now that we have the eigenvalue \(\lambda\) and eigenvectors \(\mathbf{v}_1\), \(\mathbf{v}_2\), and \(\mathbf{v}_3\), we can construct the fundamental set of solutions using the equation \(\mathbf{y} = \mathbf{v} e^{\lambda t}\). Our three solutions will be in the form: $$ \mathbf{y}_1 = \mathbf{v}_1 e^{\lambda t}, \quad \mathbf{y}_2 = \mathbf{v}_2 e^{\lambda t} + t \mathbf{v}_1 e^{\lambda t}, \quad \mathbf{y}_3 = \mathbf{v}_3 e^{\lambda t} + t \mathbf{v}_2 e^{\lambda t} + \frac{1}{2} t^2 \mathbf{v}_1 e^{\lambda t}. $$ Substitute the eigenvalue (4) and eigenvectors \(\mathbf{v}_1\), \(\mathbf{v}_2\), and \(\mathbf{v}_3\): $$ \mathbf{y}_1 = \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} e^{4t}, \quad \mathbf{y}_2 = \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} e^{4t} + t\begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} e^{4t}, \quad \mathbf{y}_3 = \begin{bmatrix} 0 \\ 1 \\ -1 \end{bmatrix} e^{4t} + t\begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} e^{4t} + \frac{1}{2}t^2\begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} e^{4t}. $$ Finally, we have the fundamental set of solutions: $$ \mathbf{y}_1 = \begin{bmatrix} e^{4t} \\ 0 \\ 0 \end{bmatrix}, \quad \mathbf{y}_2 = \begin{bmatrix} te^{4t} \\ 0 \\ e^{4t} \end{bmatrix}, \quad \mathbf{y}_3 = \begin{bmatrix} \frac{1}{2}t^2 e^{4t} \\ te^{4t} \\ (1 - t)e^{4t} \end{bmatrix}. $$

Key concepts

eigenvalues and eigenvectors

Eigenvalues and eigenvectors play a crucial role in understanding linear transformations and systems of differential equations. An eigenvalue is a scalar, often represented by \( \lambda \), associated with a linear transformation that stretches or compresses the eigenvector but does not change its direction. If \( \mathbf{v} \) is an eigenvector of a matrix \( A \) associated with eigenvalue \( \lambda \), it satisfies the equation: \[ A \mathbf{v} = \lambda \mathbf{v} \] In our linear system, the matrix \( A \) is an upper triangular matrix: \[ A = \begin{bmatrix} 4 & 0 & 0 \ 2 & 4 & 0 \ 1 & 3 & 4 \end{bmatrix} \] We can instantly identify the eigenvalue from the diagonal elements, which is \( \lambda = 4 \). This value indicates no rotation and consistent stretching along the direction of its eigenvectors. Understanding these concepts delves deep into the structure and behavior of matrix transformations.

generalized eigenvectors

Generalized eigenvectors extend the concept of eigenvectors to handle cases where a matrix is not diagonalizable. This happens when a matrix does not have a complete set of linearly independent eigenvectors. To find generalized eigenvectors, we solve different equations iteratively. When \( A - \lambda I \) is raised to a power equal to its algebraic multiplicity (here it is 3), it provides information about chains of eigenvectors, which include generalized eigenvectors. The process involves:

• Starting with an eigenvector, \( \mathbf{v}_1 \), as given.
• Finding another vector \( \mathbf{v}_2 \) such that \( (A - \lambda I) \mathbf{v}_2 = \mathbf{v}_1 \).
• Identifying the next vector \( \mathbf{v}_3 \) with the equation \( (A - \lambda I) \mathbf{v}_3 = \mathbf{v}_2 \).

These steps create a sequence that helps in forming a generalized set of solutions. In our exercise:

• \( \mathbf{v}_1 = \begin{bmatrix} 1 \ 0 \ 0 \end{bmatrix} \)
• \( \mathbf{v}_2 = \begin{bmatrix} 0 \ 0 \ 1 \end{bmatrix} \)
• \( \mathbf{v}_3 = \begin{bmatrix} 0 \ 1 \ -1 \end{bmatrix} \)

These vectors together form a basis that captures the solution's essence when the matrix \( A \) does not provide enough eigenvectors on its own.

fundamental set of solutions

The fundamental set of solutions provides the building blocks for understanding a linear differential equation system's general solution. These solutions are derived using eigenvalues and eigenvectors. In the context of systems of differential equations, a fundamental set of solutions offers linearly independent solutions that span the solution space. For our linear system \( \mathbf{y}^{\prime}=A \mathbf{y} \), using eigenvalues and the vector chains we computed, the fundamental set is:

• \( \mathbf{y}_1 = \mathbf{v}_1 e^{\lambda t} \)
• \( \mathbf{y}_2 = \mathbf{v}_2 e^{\lambda t} + t \mathbf{v}_1 e^{\lambda t} \)
• \( \mathbf{y}_3 = \mathbf{v}_3 e^{\lambda t} + t \mathbf{v}_2 e^{\lambda t} + \frac{1}{2} t^2 \mathbf{v}_1 e^{\lambda t} \)

Here:

• \( \lambda = 4 \)
• \( \mathbf{v}_1 = \begin{bmatrix} 1 \ 0 \ 0 \end{bmatrix} \)
• \( \mathbf{v}_2 = \begin{bmatrix} 0 \ 0 \ 1 \end{bmatrix} \)
• \( \mathbf{v}_3 = \begin{bmatrix} 0 \ 1 \ -1 \end{bmatrix} \)

The time-dependent terms \( t \) and \( \frac{1}{2} t^2 \) illustrate the system's non-normal modes that arise from the generalized eigenvectors. These solutions allow us to express any solution to the system as a linear combination of these three formulas.

algebraic and geometric multiplicity

The concepts of algebraic and geometric multiplicity are important for understanding the properties of eigenvalues and their corresponding eigenvectors. Algebraic multiplicity of an eigenvalue refers to the number of times it appears as a root of the characteristic polynomial of a matrix. Geometric multiplicity, on the other hand, is the number of linearly independent eigenvectors associated with that eigenvalue. For this problem, the eigenvalue \( \lambda = 4 \) has an algebraic multiplicity of 3, indicating it is a repeated root three times in the characteristic polynomial. Yet, its geometric multiplicity is only 1, meaning there is a lack of distinct eigenvectors; only one independent eigenvector is directly associated with it initially. This discrepancy is what necessitates the introduction of generalized eigenvectors to complete the solution space. It reflects the need for additional, linearly independent, generalized eigenvectors to supplement the eigenvector deficit, providing a richer structure necessary to describe transformations by \( A \). This highlights how the algebraic and geometric multiplicity of eigenvalues helps predict whether the matrix can be diagonalized over the real numbers or not.