Question

Let \(A\) be a real \((2 \times 2)\) matrix. Assume that \(A\) has eigenvalues \(\lambda_{1}\) and \(\lambda_{2}\), and consider the linear homogeneous system \(\mathbf{y}^{\prime}=A \mathbf{y}\). (a) Prove that if \(\lambda_{1}\) and \(\lambda_{2}\) are both nonzero, then \(\mathbf{y}_{e}=\mathbf{0}\) is an isolated equilibrium point. (b) Suppose that eigenvalue \(\lambda_{1} \neq 0\) but that \(\lambda_{2}=0\) with corresponding eigenvector \(\left[\begin{array}{r}\beta \\\ -\alpha\end{array}\right]\). Show that all points on the phase-plane line \(\alpha x+\beta y=0\) are equilibrium points. (In this case, \(\mathbf{y}_{e}=\mathbf{0}\) is not an isolated equilibrium point.)

Short answer

In summary, when both given eigenvalues are nonzero, the only equilibrium point is the origin, making it an isolated equilibrium point. However, when one of the eigenvalues is zero, all the points on the phase-plane line corresponding to the eigenvector associated with the zero eigenvalue are equilibrium points, making the origin not isolated.

Step-by-step solution

Analyze the given system with eigenvalues

Given a \((2 \times 2)\) matrix \(A\) with eigenvalues \(\lambda_1\) and \(\lambda_2\), we consider the linear homogeneous system \(\mathbf{y'} = A\mathbf{y}\). We know that if \(\lambda_1\) and \(\lambda_2\) are eigenvalues of matrix \(A\), then there exist nonzero vectors \(\mathbf{v_1}\) and \(\mathbf{v_2}\) such that \(A\mathbf{v_1} = \lambda_1\mathbf{v_1}\) and \(A\mathbf{v_2} = \lambda_2\mathbf{v_2}\). We are first tasked to show that if both \(\lambda_1\) and \(\lambda_2\) are nonzero, then the origin is an isolated equilibrium point.

Finding equilibrium points

An equilibrium point for a linear system is a point where the rate of change of the variables is equal to zero, i.e., \(\mathbf{y'} = \mathbf{0}\). We are looking for points \(\mathbf{y} = \begin{bmatrix} x \\ y \end{bmatrix}\) such that \(A\mathbf{y} = \mathbf{0}\). Since \(A\mathbf{y} = \lambda_1\mathbf{v_1}+\lambda_2\mathbf{v_2}\), and both \(\lambda_1\) and \(\lambda_2\) are nonzero, to have \(A\mathbf{y} = \mathbf{0}\), it must be the case that both \(\mathbf{v_1}\) and \(\mathbf{v_2}\) are also zero vectors. This implies that the only possible equilibrium point is the origin, i.e., \(\mathbf{y}_e = \begin{bmatrix} 0 \\ 0 \end{bmatrix}\). Thus, we have proven part (a) of the exercise.

Analyze the system with one eigenvalue being zero

For part (b), we are given that \(\lambda_1 \neq 0\) and \(\lambda_2 = 0\). Corresponding to eigenvalue \(\lambda_2\), there is a nonzero eigenvector \(\mathbf{v}=\begin{bmatrix}\beta \\ -\alpha\end{bmatrix}\). We now need to show that all points on the phase-plane line \(\alpha x+\beta y=0\) are equilibrium points.

Showing equilibrium points on the phase-plane line

We know that the eigenvector corresponding to the zero eigenvalue, \(\mathbf{v}=\begin{bmatrix}\beta \\ -\alpha\end{bmatrix}\), satisfies the equation \(A\mathbf{v} = \lambda_2\mathbf{v} = \mathbf{0}\). This indicates that all scalar multiples of \(\mathbf{v}\) also satisfy the equation \(A\mathbf{y} = \mathbf{0}\) since the scalar multiples of eigenvectors are also eigenvectors corresponding to the same eigenvalue. So, for any scalar \(k\), the vector \(\mathbf{y} = k\begin{bmatrix}\beta \\ -\alpha\end{bmatrix}\) is an equilibrium point. Now, any point on the phase-plane line \(\alpha x + \beta y = 0\) can be represented as a scalar multiple of \(\begin{bmatrix}\beta \\ -\alpha\end{bmatrix}\) for some scalar \(k\). Therefore, all points on this phase-plane line satisfy the equation \(A\mathbf{y} = \mathbf{0}\), making them equilibrium points. In this case, since there are infinitely many points on the phase-plane line, the origin is not an isolated equilibrium point.

Key concepts

Eigenvalues and Eigenvectors

Understanding the significance of eigenvalues and eigenvectors is foundational when examining differential equations, particularly in the context of linear homogeneous systems. When we have a matrix A, eigenvalues are special scalars denoted as λ that provide important information about the system's behavior. For each eigenvalue, there is a corresponding eigenvector, a non-zero vector denoted as v, which satisfies the equation Av = λv.

These vectors are crucial because they maintain their direction after being transformed by the matrix. In the context of our differential equation \( \mathbf{y'} = A\mathbf{y} \), the eigenvectors aligned with the eigenvalues dictate the geometry of the solutions. If both eigenvalues are non-zero, as shown in the exercise, this has significant implications on the system's equilibrium points, and particularly the stability of these equilibria. It often results in all solutions spiraling towards or away from the equilibrium point, which in our case is the origin point in the phase-plane.

Phase-Plane Analysis

Phase-plane analysis is a visual tool for studying the behavior of solutions to two-dimensional systems of differential equations. It represents trajectories or paths of the system in the state space, which usually consists of two variables x and y.

In our exercise, the phase-plane aids in identifying equilibrium points—points where the system doesn't change over time. When both eigenvalues are non-zero, the origin is the only equilibrium point, and it's isolated; no other equilibria exist in the immediate vicinity, making the point a unique state of balance. However, with one eigenvalue being zero, we face a line of equilibrium points on the phase-plane, which are all scalar multiples of the corresponding eigenvector. This suggests that there is an entire line in the phase-plane where the system is at rest, indicating a fundamentally different dynamic behavior than the isolated equilibrium scenario.

Linear Homogeneous Systems

Linear homogeneous systems can be demystified by focusing on their name. 'Linear' implies that all terms are to the first power, and coefficients are constant or functions of the independent variable only. 'Homogeneous' indicates that these systems have zero on the right side of the equations when written in standard form.

Our example of the system \( \mathbf{y'} = A\mathbf{y} \) fits this definition. In such systems, solutions are straightforwardly connected to eigenvalues and eigenvectors. The behavior of the system is dictated by the nature of the eigenvalues: whether they are real or complex, positive, negative, or zero. An equilibrium solution, such as the origin in our case, is stable or unstable based on the eigenvalues' sign and magnitude.

The exercise showcases how a single zero eigenvalue changes the landscape: the system no longer has an isolated equilibrium point and instead, possesses a continuum of equilibrium states along a line in the phase-plane. This subtlety is essential for students to grasp, as it impacts the interpretation of the system's long-term behavior and demonstrates the profound impact eigenvalues have in the realm of linear homogeneous systems.