What can be inferred about the eigenvalues of this phase portrait?

Question

The trajectories above the red lines are moving "upwards"; and the trajectories below the red line are moving "downwards"

Is it correct to say that one of the eigenvalues contain 0, and the underlying matrix is singular.

Answer 1

First version (to be discarded in fact) : If the vector field is made of unit speed vectors (as one can see on your figure).

$$\binom{\dot x}{\dot y} \ = \ \binom{1}{2} \ \text{for} \ y<x \ \ \ \ \text{and} \ \ \ \ \binom{\dot x}{\dot y} \ = \ \binom{-1}{-2}\ \text{for} \ y>x $$

giving

$$\begin{cases}\dot x=1 \ \text{sign}(x-y)\\ \dot y=2 \ \text{sign}(x-y)\end{cases}\tag{1}$$

which is a deadend (as I said, non linear).

Second version (the good one, so I think).

Consider the linear differential system, very similar to (1) in fact :

$$\begin{cases}\dot x&=&-x+y\\ \dot y&=&-2x+2y\end{cases} \ \ \iff \ \ \binom{\dot x}{\dot y} \ = \underbrace{\begin{pmatrix}-1&1\\-2&2\end{pmatrix}}_{M} \binom{x}{y}\tag{2} $$

Using a little Matlab program, I got the following picture, where the stars indicate starting points, chosen at random in $[-2,2] \times [-2,2]$. This result is in full compliance with your figure :

Explanation :

The eigenvalues of $M$ are $0$ (indeed !) and $1$, with associated eigenvectors $V_0=\binom{1}{1}$, $V_1=\binom{1}{2}$, respectively. The first one is a directing vector of the red "separating line", the second one gives the direction of all the paths.

Important remark : if the (random) initial point is very close to the separating line (this is the case for two of them, one close to the origin, the other one at bottom left), they are bound to have a short trajectory : they haven't not enough initial speed to go far away... although they have been given the same amount of time as the others.

Explicit solutions of (2) are given by $X_t=\exp(tM).X_0$, where $X_0$ accounts for initial conditions, i.e., (using a CAS) :

$$\begin{pmatrix}x_t\\y_t\end{pmatrix}=\begin{pmatrix}2 - e^t& e^t - 1\\ 2 - 2e^t& 2e^t - 1\end{pmatrix}\begin{pmatrix}x_0\\y_0\end{pmatrix}$$

One can verify in particular that when $(x_0,y_0)$ is situated on the red line, i.e., when $y_0=x0$, the trajectory doesn't expand...

Answer 2

Short answer: You can say that one eigenvalue is $0$ with eigenvector $(1~~1)^T$ and other eigenvalue is positive with $(1~~2)^T$ eigenvector.

To see that why this is the case consider the equation $\dot{x}=Ax$ and $A$ has eigenpairs $(\lambda_i, u_i)$ (assuming real and distinct eigenvalues for simplicity). So, $x(t)=e^{\lambda_i t}u_i$ is a solution, which means all solutions can be characterized as $$x(t) = c_1 e^{\lambda_1 t} u_1 + c_2 e^{\lambda_2 t} u_2$$ for a 2nd order system where $c_i$ depends on the initial condition. In particular if you select $c_2=0$, it means all solutions are along the vector $u_1$, which intuitively means if you start on $u_1$ then you stay on $u_1$, forever.

In your example we know that $\lambda_1=0$ and $u_1$ is the red line, because if it wasn't the case, other trajectories should be curved because of the $e^{\lambda_1 t}$ term. This means you have infinite equilibrium points (all the points on the red line). Since all equilibrium points are unstable, we can conclude that $\lambda_2 > 0$.