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 \(\mathbf{v}_{1}\).\(A=\left[\begin{array}{lll}2 & 1 & 0 \\ 0 & 2 & 1 \\ 0 & 0 & 2\end{array}\right]\)

Short answer

Answer: The fundamental set of solutions is given by \(\mathbf{y}(t)=c_1e^{\lambda t}\begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}+c_2\left(te^{\lambda t}\begin{bmatrix} 0 \\ \frac{1}{2(2-\lambda)} \\ 0 \end{bmatrix}\right)+c_3\left(t^2e^{\lambda t}\begin{bmatrix} 0 \\ 0 \\ \frac{1}{6(2-\lambda)^2} \end{bmatrix}\right)\), where \(c_1, c_2\), and \(c_3\) are arbitrary constants, and \(\lambda\) is the eigenvalue.

Step-by-step solution

Identify the given matrix and eigenvector

The given matrix \(A\) and eigenvector \(\mathbf{v}_{1}\) are: \(A=\left[\begin{array}{lll}2 & 1 & 0 \\\ 0 & 2 & 1 \\\ 0 & 0 & 2\end{array}\right]\) and \(\mathbf{v}_{1}=\begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}\).

Find the geometric multiplicity

The geometric multiplicity is given as 1, which implies that there is only one linearly independent eigenvector corresponding to the eigenvalue \(\lambda\).

Compute generalized eigenvectors

To find the generalized eigenvectors, considering the difference between \(A\) and \(\lambda I\): \((A-\lambda I)\), where \(I\) is the identity matrix. \(\left[\begin{array}{lll}2-\lambda & 1 & 0 \\\ 0 & 2-\lambda & 1 \\\ 0 & 0 & 2-\lambda\end{array}\right]\). Since algebraic multiplicity is 3, we need to find generalized eigenvectors up to the second order. To find the second-order generalized eigenvector \(\mathbf{v}_2\), solve the system of equations \((A-\lambda I)^2\mathbf{v}_2=\mathbf{v}_1\). \((A-\lambda I)^2=\begin{bmatrix}(2-\lambda) & 1 & 0 \\ 0 & (2-\lambda) & 1 \\ 0 & 0 & (2-\lambda)\end{bmatrix}\begin{bmatrix}(2-\lambda) & 1 & 0 \\ 0 & (2-\lambda) & 1 \\ 0 & 0 & (2-\lambda)\end{bmatrix}\) Simplifying the matrix product, we get: \((A-\lambda I)^2=\begin{bmatrix}(2-\lambda)^2 & 2(2-\lambda) & 1 \\ 0 & (2-\lambda)^2 & 2(2-\lambda) \\ 0 & 0 & (2-\lambda)^2\end{bmatrix}\) Now, solve the system of equations: \(\begin{bmatrix}(2-\lambda)^2 & 2(2-\lambda) & 1 \\ 0 & (2-\lambda)^2 & 2(2-\lambda) \\ 0 & 0 & (2-\lambda)^2\end{bmatrix}\mathbf{v}_2=\begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}\) By solving this system, we have \(\mathbf{v}_2=\begin{bmatrix} 0 \\ \frac{1}{2(2-\lambda)} \\ 0 \end{bmatrix}\). To find \(\mathbf{v}_3\), we solve the system: \((A-\lambda I)^3\mathbf{v}_3=\mathbf{v}_2\). \((A-\lambda I)^3=\begin{bmatrix}(2-\lambda)^3 & 3(2-\lambda)^2 & 3(2-\lambda) \\ 0 & (2-\lambda)^3 & 3(2-\lambda)^2 \\ 0 & 0 & (2-\lambda)^3\end{bmatrix}\) And the system to solve is: \(\begin{bmatrix}(2-\lambda)^3 & 3(2-\lambda)^2 & 3(2-\lambda) \\ 0 & (2-\lambda)^3 & 3(2-\lambda)^2 \\ 0 & 0 & (2-\lambda)^3\end{bmatrix}\mathbf{v}_3=\begin{bmatrix} 0 \\ \frac{1}{2(2-\lambda)} \\ 0 \end{bmatrix}\) Solving this system, we get \(\mathbf{v}_3 = \begin{bmatrix} 0 \\ 0 \\ \frac{1}{6(2-\lambda)^2} \end{bmatrix}\).

Write the fundamental set of solutions

Now, the fundamental set of solutions can be written using the eigenvector and generalized eigenvectors just found: \(\mathbf{y}(t)=c_1e^{\lambda t}\mathbf{v}_{1}+c_2\left(te^{\lambda t}\mathbf{v}_{2}\right)+c_3\left(t^2e^{\lambda t}\mathbf{v}_{3}\right)\) Substituting the eigenvector and generalized eigenvectors, we get: \(\mathbf{y}(t)=c_1e^{\lambda t}\begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}+c_2\left(te^{\lambda t}\begin{bmatrix} 0 \\ \frac{1}{2(2-\lambda)} \\ 0 \end{bmatrix}\right)+c_3\left(t^2e^{\lambda t}\begin{bmatrix} 0 \\ 0 \\ \frac{1}{6(2-\lambda)^2} \end{bmatrix}\right)\) This is the fundamental set of solutions for the linear system \(\mathbf{y}^{\prime}=A\mathbf{y}\).

Key concepts

Eigenvalues and Eigenvectors

When we deal with matrices, two of the most essential concepts we come across are eigenvalues and eigenvectors. These concepts are particularly important in systems where the matrix represents a transformation.

Eigenvalues, denoted by \( \lambda \), are scalars associated with a linear system of equations that provide important information about the system's behavior. When a non-zero vector \( \mathbf{v} \) is multiplied by the matrix \( A \) and the result is a scalar multiple of the same vector, \( \mathbf{v} \), then \( \mathbf{v} \) is an eigenvector of \( A \) and the scalar is its corresponding eigenvalue.

Mathematically, the relationship is described by the equation \( A\mathbf{v} = \lambda\mathbf{v} \). The eigenvector does not change direction in this transformation process, instead, it's simply scaled by the eigenvalue. Identifying the eigenvalues and eigenvectors of a matrix is a crucial step in analyzing linear transformations and can be particularly useful in solving differential equations and stability analysis.

Geometric Multiplicity

Geometric multiplicity is related to the number of linearly independent eigenvectors associated with a particular eigenvalue. It is a measure of the dimensions of the eigenspace corresponding to that eigenvalue.

In our scenario, the geometric multiplicity of 1 indicates that there's a one-dimensional eigenspace for the eigenvalue \( \lambda \). Essentially, this tells us that even though \( \lambda \) is repeated three times (has an algebraic multiplicity of 3), there is only one unique direction, defined by the eigenvector \( \mathbf{v}_{1} \), that remains unaffected by the transformation \( A \). All other vectors in the eigenspace can be expressed as scalar multiples of this single eigenvector.

This is an important property because it impacts the solution to the system of linear equations. When the geometric multiplicity is less than the algebraic multiplicity, it indicates that there will be a need for generalized eigenvectors to form a complete set of solutions.

Generalized Eigenvectors

Generalized eigenvectors come into play when the geometric multiplicity is less than the algebraic multiplicity. In such cases, the canonical set of eigenvectors is not sufficient to form a basis for the corresponding eigenspace.

We can find generalized eigenvectors by looking at higher powers of the matrix \( A - \lambda I \), where \( I \) is the identity matrix. Essentially, we solve \( (A - \lambda I)^k\mathbf{v} = \mathbf{0} \) for some integer \( k > 1 \), and \( \mathbf{v} \) in this context is a generalized eigenvector. In the provided exercise, we're dealing with second and third-order generalized eigenvectors.The process of calculating these vectors broaden our capability to solve for a full set of solutions to the system of equations when a repeated eigenvalue has fewer linearly independent eigenvectors than its algebraic multiplicity. This guarantees that we'll have enough independent solutions to span the solution space.

Linear Systems of Differential Equations

Linear systems of differential equations are a cornerstone in understanding dynamic systems in various fields such as engineering, physics, and economics. When expressed in matrix form as \( \mathbf{y}^{\text{\textquotesingle}} = A\mathbf{y} \), where \( A \) is a square matrix and \( \mathbf{y} \) is a vector of functions, the solutions to these systems can reveal the evolution of the system over time.

The fundamental set of solutions to such a system is a set of vectors that provides the general solution to the differential equations. This involves finding the eigenvectors and corresponding eigenvalues, and when necessary, generalized eigenvectors. Once these vectors are identified, they're used to construct solutions of the form \( e^{\lambda t}\mathbf{v} \) or its generalized forms such as \( t^ke^{\lambda t}\mathbf{v}_k \) for some non-negative integer \( k \) and generalized eigenvector \( \mathbf{v}_k \).In the context of our exercise, the solution demonstrates how these concepts are assembled to form a comprehensive solution that addresses the behavior of the system over time for all possible initial conditions. Students can use this approach to analytically solve any linear system of differential equations with constant coefficients.