Question

In each exercise, consider the linear system \(\mathbf{y}^{\prime}=A \mathbf{y}\). Since \(A\) is a constant invertible \((2 \times 2)\) matrix, \(\mathbf{y}=\mathbf{0}\) is the unique (isolated) equilibrium point. (a) Determine the eigenvalues of the coefficient matrix \(A\). (b) Use Table \(6.2\) to classify the type and stability characteristics of the equilibrium point at the phase-plane origin. If the equilibrium point is a node, designate it as either a proper node or an improper node. $$ \mathbf{y}^{\prime}=\left[\begin{array}{rr} -2 & 1 \\ -1 & -2 \end{array}\right] \mathbf{y} $$

Short answer

Answer: The eigenvalues of the given matrix A are -3 and -1. The equilibrium point at the phase-plane origin is a stable proper node for the linear system.

Step-by-step solution

(a) Find the eigenvalues of A

To find the eigenvalues of the given matrix \(A=\begin{bmatrix}-2 & 1 \\ -1 & -2 \end{bmatrix}\), we begin by solving the characteristic equation $$\text{det}(A-\lambda I)=\text{det}\left(\begin{bmatrix}-2-\lambda & 1 \\ -1 & -2-\lambda \end{bmatrix}\right)=0.$$ The determinant is calculated as: $$(-2-\lambda)((-2)-\lambda)-(1)(-1)=\lambda^2+4\lambda+3.$$ Now, we can factor the characteristic polynomial: $$\lambda^2+4\lambda+3=(\lambda+3)(\lambda+1).$$ Setting the factors to 0, we find the eigenvalues: $$\lambda_1=-3 \text{ and } \lambda_2=-1.$$

(b) Classify the equilibrium point

Using Table 6.2 or the general understanding of linear system stability, we can classify the equilibrium point by examining the eigenvalues \(\lambda_1=-3\) and \(\lambda_2=-1\). Since both eigenvalues are real and have the same sign (negative), the origin \((0,0)\) is a stable node. To check if it is a proper or improper node, we observe that the matrix has distinct eigenvalues (\(\lambda_1\neq\lambda_2\)), which means the origin is a proper node for this linear system. In summary, the equilibrium point at the phase-plane origin is a stable proper node for the given linear system.

Key concepts

Eigenvalues

Eigenvalues are a fundamental concept in linear algebra and play a crucial role in analyzing dynamic systems. Imagine you have a matrix, as in our given problem, \( A = \begin{bmatrix} -2 & 1 \ -1 & -2 \end{bmatrix} \), which describes a linear transformation. Eigenvalues are special numbers associated with a matrix which give us insights into the system's dynamics.Essentially, when you multiply a matrix by a vector, the vector's direction doesn't change when it's an eigenvector, just its magnitude changes by the eigenvalue factor. The eigenvalues are found by solving the characteristic equation, which is derived from the determinant of the matrix \( A - \lambda I = 0 \), where \( I \) is the identity matrix and \( \lambda \) represents the eigenvalues.In our example:

• We find the determinant: \( (-2-\lambda)((-2)-\lambda)-1 \times -1 = \lambda^2+4\lambda+3 \).
• We factor the polynomial: \( (\lambda+3)(\lambda+1) = 0 \).
• The solutions \( \lambda_1 = -3 \) and \( \lambda_2 = -1 \) are our eigenvalues.

These eigenvalues help us examine the behavior of systems at equilibrium points, a concept we discuss in the next section.

Equilibrium Point

An equilibrium point in a dynamic system is where the system doesn't change over time. For the given linear system, \( \mathbf{y}^{\prime}=A\mathbf{y} \), the origin \( \mathbf{y} = \mathbf{0} \) is the equilibrium point.At this point, if you start your system, it will "stay put," meaning there are no forces or flows moving the system away from there. This is particularly relevant in the context of linear systems where the equilibrium point is isolated; in simpler terms, it's the only spot where this happens without external changes.The uniqueness of this equilibrium point is due to the matrix \( A \) being invertible, ensuring no other constant \( \mathbf{y} \) can satisfy the equation without reverting back to zero. Understanding the role of equilibrium points allows us to determine how systems behave in response to small disturbances. This leads us to perform stability analysis, detailed below.

Stability Analysis

Stability analysis is crucial to understanding how a system behaves in response to disturbances, especially near its equilibrium point. It's like checking if a ball placed slightly off-center at the bottom of a bowl will roll back to the center or continue away.In the solution given, we performed a stability analysis by examining the eigenvalues of matrix \( A \). Since both eigenvalues\( \lambda_1 = -3 \) and \( \lambda_2 = -1 \) are real and negative, we determined that the system is stable.The rules of thumb for stability analysis involve:

• Negative eigenvalues: System is stable. Small disturbances will return to the equilibrium point.
• Positive eigenvalues: System is unstable. Even small disturbances can lead to large deviations.
• Mixed signs: Indicates a saddle point; the system diverges in certain directions and converges in others.

Furthermore, distinct eigenvalues indicate that the equilibrium point is a proper node, meaning the solution paths smoothly converge to the equilibrium point as straight lines. Stability analysis allows us to predict long-term behavior of systems, which is valuable in fields ranging from engineering to economics.