# Phase space/Differential equation

You have two solutions of an differential equation with the same eigenvalues. How can you see, that the phase space are lines:

$$y_1=c_1e^{\lambda t} \begin{pmatrix} 1 \\ 0 \end{pmatrix}$$

$$y_2=c_2e^{\lambda t} \begin{pmatrix} 0 \\ 1 \end{pmatrix}$$

• Exactly what solutions do you have? There are two rather different possibilities when you have a repeated eigenvalue. – amd Feb 16 at 23:17
• The eigenvalues are greater than 0. – SvenMath Feb 16 at 23:21
• That’s not the most interesting condition. A defective eigenvalue generates a very different phase portrait than if both its algebraic and geometric multiplicities are 2. – amd Feb 16 at 23:24
• Sorry for that. The algebraic and geometric multiplicities are 2. – SvenMath Feb 16 at 23:33
• $\dot y_1=\lambda_1 y_1$ and $\dot y_2=\lambda_2 y_2$ are lines, no? – Cosmas Zachos Aug 2 at 22:06