To determine the phase portrait of the matrix $\begin{bmatrix} 1 & 0.5 \\ 0.5 & 1 \end{bmatrix}$, we need to analyze the eigenvalues and eigenvectors of the matrix. These will help us understand the dynamics of the system represented by the matrix.

Step 1: Eigenvalues

The eigenvalues $\lambda$ are found by solving the characteristic equation: $\det(A - \lambda I) = 0$ Where $A = \begin{bmatrix} 1 & 0.5 \\ 0.5 & 1 \end{bmatrix}$.

$\det\left(\begin{bmatrix} 1 & 0.5 \\ 0.5 & 1 \end{bmatrix} - \lambda \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\right) = 0$

$\det\begin{bmatrix} 1-\lambda & 0.5 \\ 0.5 & 1-\lambda \end{bmatrix} = (1-\lambda)^2 - (0.5)^2 = 0$

$(1-\lambda)^2 - 0.25 = 0 \implies (1-\lambda + 0.5)(1-\lambda - 0.5) = 0$

$\lambda = 1.5 \quad \text{and} \quad \lambda = 0.5$

Step 2: Eigenvectors

For each eigenvalue, solve $(A - \lambda I)v = 0$:

For $\lambda = 1.5$:

$\begin{bmatrix} 1-1.5 & 0.5 \\ 0.5 & 1-1.5 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = \begin{bmatrix} -0.5 & 0.5 \\ 0.5 & -0.5 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = 0$

$-0.5x_1 + 0.5x_2 = 0 \quad \implies \quad x_1 = x_2$

Eigenvector: $v_1 = \begin{bmatrix} 1 \\ 1 \end{bmatrix}$.

For $\lambda = 0.5$:

$\begin{bmatrix} 1-0.5 & 0.5 \\ 0.5 & 1-0.5 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = \begin{bmatrix} 0.5 & 0.5 \\ 0.5 & 0.5 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = 0$

$0.5x_1 + 0.5x_2 = 0 \quad \implies \quad x_1 = -x_2$

Eigenvector: $v_2 = \begin{bmatrix} 1 \\ -1 \end{bmatrix}$.

Step 3: Phase Portrait Description

• The eigenvalues are real and distinct ($\lambda_1 = 1.5 > 0$, $\lambda_2 = 0.5 > 0$), indicating that the origin is an unstable node.
• The eigenvectors $\begin{bmatrix} 1 \\ 1 \end{bmatrix}$ and $\begin{bmatrix} 1 \\ -1 \end{bmatrix}$ define the directions of the trajectories.
• The trajectories move away from the origin along these eigenvector directions, with growth rates determined by the corresponding eigenvalues.

Step 4: Visualizing the Phase Portrait

In the phase portrait:

1. Trajectories emerge radially outward from the origin.
2. Along $\begin{bmatrix} 1 \\ 1 \end{bmatrix}$ (associated with $\lambda_1 = 1.5$), trajectories grow faster.
3. Along $\begin{bmatrix} 1 \\ -1 \end{bmatrix}$ (associated with $\lambda_2 = 0.5$), trajectories grow slower.

Would you like a plotted phase portrait for better visualization?

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5 Related Questions:

1. How do eigenvalues affect the stability of a system?
2. What are the differences between nodes, saddles, and spirals in phase portraits?
3. How can you interpret phase portraits for systems with complex eigenvalues?
4. How do you calculate the direction of eigenvectors geometrically?
5. Can eigenvalues with the same sign result in other types of equilibrium points?

Tip: Always check eigenvalues to determine the nature of the critical point before attempting to sketch the phase portrait.