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A phase portrait provides a visual representation of the behavior of a dynamical system in the phase plane. By plotting the trajectories of the system, we can observe how the system evolves over time. In the context of our linear differential system, each point in the plane represents a state of the system, described by the variables \(x\) and \(y\). Drawing trajectories for various initial points helps us understand how these states change with time.

For the given system, the phase portrait includes trajectories converging towards the critical point \((0,0)\). This behavior indicates that the system is stable around this point. Since both eigenvalues of the system are negative, the phase portrait confirms an asymptotically stable node, meaning that over time all paths lead towards the critical point. This type of stability is often seen visually as all paths point to \((0,0)\) without circling or diverging away.

The direction field, also known as a vector field, complements the phase portrait by showing the direction of the system's trajectory at any given point in the phase plane. Each vector indicates how the system evolves when it starts from that point.

In our case, plotting the direction field involves drawing short arrows starting from several points, with each arrow pointing in the direction given by the system of differential equations. For this system:

• Each arrow for \(\frac{dx}{dt} = -2x\) points to the left, as \(x\) decreases.
• Each arrow for \(\frac{dy}{dt} = -y\) points downwards, as \(y\) decreases.

The direction field collectively visualizes how the system's variables \(x\) and \(y\) change over time and how they converge to the critical point \((0,0)\). This confirms the asymptotically stable node we see in the phase portrait.

Linear differential systems involve equations where the variables and their derivatives appear linearly. In our system, equations like \(\frac{dx}{dt} = -2x\) and \(\frac{dy}{dt} = -y\) demonstrate this linearity.

Such systems can often be written in matrix form: \(\frac{d\mathbf{x}}{dt} = A\mathbf{x}\), where \(A\) is a matrix consisting of the coefficients of the variables. The example given is: \[A = \begin{bmatrix} -2 & 0 \ 0 & -1 \end{bmatrix}\]
This matrix helps in analyzing the system using tools such as eigenvalues and eigenvectors, providing insights into the system's overall behavior and stability. Linear systems, like the one in the exercise, are fundamental to understanding more complex, non-linear systems.

Eigenvalues are key when analyzing the stability of a linear differential system. Calculating eigenvalues involves solving the characteristic equation \(\det(A - \lambda I) = 0\), where \(A\) is the system’s coefficient matrix and \(I\) is the identity matrix.

In our exercise, the matrix \(A\) is \[\begin{bmatrix} -2 & 0 \ 0 & -1 \end{bmatrix}\]
The characteristic equation then becomes \((-2 - \lambda)(-1 - \lambda) = 0\), giving eigenvalues \(\lambda_1 = -2\) and \(\lambda_2 = -1\).

• Negative eigenvalues: Indicate stability, as solutions decay over time moving towards the critical point.
• Zero eigenvalues: Represent a neutral stability, contributing neither to growth nor decay.
• Positive eigenvalues: Indicate instability, as solutions grow without bounds.

Thus, the negative eigenvalues in our system confirm that the critical point \((0,0)\) is asymptotically stable. This helps us predict the system's long-term behavior based on initial conditions.