How many fixed points does a linear dynamical system have? I am watching a video on Koopman theory, which is about linear embeddings for nonlinear dynamical systems. The professor giving talk is showing a system described by:
$$
\frac{dx}{dt} = f(x, y, t).
$$
This system is the represented as by a linear system
$$
\frac{dx}{dt} \approx \bf{A}x.
$$
But the professor said that while the original system may have multiple fixed points, a linear system only has a single fixed point at the zero vector. By fixed point I assume the Professor is referring to the point at which $\dot{x} = 0$, as we would see in any nonlinear ODE textbook like Strogatz. 
I was trying to confirm that linear systems only have a single fixed point. If the matrix $A$ is differentiable, then the vectors in the nullspace of the Jacobian of $A$ would also be fixed points, right? Perhaps the video will go on to talk about some constraints on the matrix given a Koopman type problem, but that point about linear systems and only a single fixed point made me wonder.
 A: First of all, some clarification:
If you perform a linearisation, $A$ is the Jacobian of $f$.
It does not have a meaningful Jacobian on its own.
Fixed points occur if and only if $\frac{\mathrm{d}x}{\mathrm{d}t}=0$.
Thus, for the linearised system, we have a trivial fixed point for $x=0$, which exists in every case and which is the “single fixed point at the zero vector” your professor is referring to.
There may be non-trivial fixed points, i.e., $x≠0$ such that $\mathbf{A}x=0$.
These are per definition eigenvectors of $\mathbf{A}$ with eigenvalue zero.
However, the probability of this happening for a random $\mathbf{A}$ is 0.
They are therefore not of great concern for techniques based on linearisation, which after all are only approximative anyway. For any linearisation with zero eigenvalue there are plenty of slightly different ones without it, which are almost as good as approximations of the dynamics.
Moreover If you look at all linearisations along a trajectory, you usually only need to go a small step into the future to lose a zero eigenvector.
