I'm trying to solve the following problem:
"Look at the image of trajectories of a linear, time-invariant system with the form: $\frac{d\textbf x}{dt}=\textbf {Ax}:$
Determine possible eigenvectors $\textbf p_1$ and $\textbf p_2$. What can you say about the corresponding eigenvalues? "
So from the picture I can see that one eigenvalue has to be $<0$, because they are attracted to one point. For the eigenvectors I would take $\textbf p_1= \begin{pmatrix} 0 \\ 1 \end{pmatrix}$ and $\textbf p_2= \begin{pmatrix} 1 \\ 0 \end{pmatrix}$. Is this correct so far? How can I determine the eigenvalues?
Thank you very much!
First, notice that the point $x_e = \begin{bmatrix}0.5 & 0.5\end{bmatrix}^T$ is a fixed point, hence it must be in the kernel space of $A$ since $A x_e = 0$. Now, obviously $A$ is not a full rank matrix, also since $x_e$ is attracting, $A$ also has a negative real eigenvalue (because the orbit is not a spiral). Therefore $A$ has eigenvalues $\lambda_1 = 0$ and $\lambda_2 < 0$.
$x_e$ is already an eigenvector for the eigenvalue $\lambda_1 = 0$. The other eigenvector is $\begin{bmatrix}1 & 0\end{bmatrix}^T$, because the orbit is in that direction and $x_e$ has no effect since its eigenvalue is $0$. Hence, the phase portrait consists of parallel lines to $x_1$ axes where they goes to a point on $x_2 = x_1$ line.
To see this explicitly, the solution of the system is
$$x(t) = \begin{bmatrix}x_1(t) \\ x_2(t)\end{bmatrix} = c_1 \begin{bmatrix}1 \\ 1\end{bmatrix} e^{\lambda_1 t} + c_2 \begin{bmatrix}1 \\ 0\end{bmatrix} e^{\lambda_2 t}$$
where $c_1$ and $c_2$ is determined by initial values. Let's assume $x(0) = \begin{bmatrix}a & b\end{bmatrix}^T$ where $a, b$ are real numbers. Then, $c_1 = b$ and $c_2 = a - b$. Hence,
$$\lim_{t \to \infty} x(t) = \begin{bmatrix}b \\ b\end{bmatrix}$$
since $\lambda_2 < 0$, as we expected.