<s>[INST] Stability equilibrium point of non linear system with nonpositive eigenvalues. We have the following non linear system:\begin{equation}
\dot{X}=f(X),
\end{equation}
and lets suppose that $f(X_0)=0$. We also have that $\lambda_1,\dots,\lambda_n$ are the eigenvales of $D_f(X_0)$ and there are $Re(\lambda_1),\dots,Re(\lambda_k)=0$ for some $k<n$ and $Re(\lambda_{k+1}),\dots,Re(\lambda_n)<0$ .
What can we say about the stability of the system? Can we say is stable because it is not unstable?
 [/INST] Here is a very simple example, a system on $\Bbb R^2$ which has a single equilibrium, a center, at $(0,0)$; the system is however unstable.
We let 
$\dot x = y + x^3, \tag 1$
$\dot y = -x + y^3; \tag 2$
then
$\dot x = \dot y = 0 \Longrightarrow y = -x^3, \; x = y^3, \tag 3$
whence
$x = y^3 = (-x^3)^3 = -x^9, \tag 4$
$x^9 + x = 0, \tag 5$
$x(x^8 + 1) = 0; \tag 6$
only real solution is evidently
$x = 0, \tag 7$
since $x^8 + 1 > 0$ for all real $x$; thus, the only equilibrium of (1)-(2) is 
$(x, y) = (x, -x^3) = (0, 0); \tag 8$
the Jacobian matrix of (1)-(2) is
$J(x, y) = \begin{bmatrix} 3x^2 & 1 \\ -1 & 3y^2 \end{bmatrix}, \tag 9$
whence
$J(0, 0) = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}; \tag{10}$
the characteristic polynomial of $J(0,0)$ is
$\det(J(0,0) - \lambda I) = \det \left (\begin{bmatrix} -\lambda & 1 \\ -1 & -\lambda \end{bmatrix} \right ) = \lambda^2 + 1; \tag{11}$
thus the eigenvalues of $J(0, 0)$ are $\pm i$; $(0,0)$ is a center; but with
$r^2 = x^2 + y^2, \tag{12}$
we have
$\dfrac{d}{dt}(r^2) = 2r\dot r = 2x \dot x + 2y\dot y$
$= 2xy + 2x^4 - 2yx + 2y^4 = 2(x^4 + y^4) > 0, \forall (x, y) \ne (0, 0); \tag{13}$
thus $r^2$, and hence $r$, is strictly monotonically increasing for all $r \ne 0$; the equilibrium point $(0,0)$, albeit a center, is unstable.
</s>