After linearizing a system of differential equations (non-linear, on two variables), the Jacobian matrix at the equilibrium $(0,1)$ is as follows:
$$ J_{(0,1)}= \begin{bmatrix} 0 & 0 \\ 1 & 2 \\ \end{bmatrix} $$
As you can see, the eigenvalues are $0$ and $2$. I know that this linear system has a line of equilibriums, but what does this say about the original (non-linear) system? The Hartman-Grobman theorem (or linearization theorem) is inconclusive in this case, since it does not consider the case where the determinant of the Jacobian is $0$, but there must be something we can say about the system.
Edit: decided to add the non-linear system below.
$$x'=x(x-2)(y-1), y'=y(y^2-x^2+x-1)$$
The equilibrium $(0,1)$ is unstable, since the line $x=0$ is invariant, and on this line the system reduces to single ODE $\dot y=y(y^2-1) = y(y+1)(y-1)$, which has an unstable equilibrium at $y=1$. (That is, perturbations of the equilibrium $(0,1)$ in the $y$ direction will cause the system to move away from $(0,1)$ along the $y$ axis.)