Lyapunov stability question from Arnold's trivium

V.I. Arnold put the following question in his Mathematical Trivium:

Can an asymptotically stable equilibrium position become unstable in the Lyapunov sense under linearization?

It puzzled me for a while, since my experience doesn't include such a pathological case - but I'm still not sure about this one.

For the system $dx/dt=y-x^3$, $dy/dt=-y^3$, the origin is an asymptotically stable equilibrium. (It is even globally attracting, as can be seen by sketching the phase portrait with the help of the nullclines $y=x^3$ and $y=0$.)

The linearized system at the origin is $dx/dt=y$, $dy/dt=0$, which is not Lyapunov stable: the solution starting at $(x_0,y_0)=(0,\epsilon)$ is $(x(t),y(t)) = (\epsilon t, \epsilon)$, which goes to infinity if $\epsilon \neq 0$.

The matrix corresponding to the linearized system is $\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$, which has zero as a double eigenvalue. If both eigenvalues have nonzero real part, then the linearized system determines the stability of the original system. If the eigenvalues are $\pm c i$ (for some real $c \neq 0$), then the linearized system is a neutral center (hence Lyapunov stable). So the phenomenon in question can only occur if at least one eigenvalue is zero. (I haven't really thought about what can or cannot happen in the case with one zero eigenvalue and one real nonzero eigenvalue.)