Checking the stability of an equilibrium point I have the linearization of a non-linear system about an equilibrium point as follows
$$
\dot x = (-A+M)x,
$$
where $x\in\mathbb{R}^3$, $A$ is a positive definite matrix and $M$ has its eigenvalues on the imaginary axis, and only one is equal to zero. Both $A$ and $M$ have real entries.
I am wondering whether this system is stable or not. Or on the other hand to find a counter-example where the system is not stable. For $M$ being a skew-symmetric matrix, then the proof becomes trivial, but in general my $M$ is not skew-symmetric.
 A: I will try to answer my own question.
Let the following decomposition $M = U^{-1}SU$, where $S$ is a skew symmetric matrix, and without loss of generality 
$$
S = \begin{bmatrix} 0 & \omega & 0 \\ -\omega & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}.
$$
Basically we have a circular rotational movement in the new X-Y axes, with a fixed frequency $\omega\in\mathbb{R}$, and the direction of the rotation is given by the sign of $\omega$. Decompose as well $A = U^{-1}TU$, and let $\tilde x = Ux$, therefore
$$
\dot{\tilde x} = -T\tilde x + S\tilde x
$$
Apply another change of coordinates to $\tilde x$, where basically the new frame of coordinates is the same as in $\tilde x$ but rotating with angular velocity $\omega$ and the rotational axis is parallel to the $Z$ axis of $\tilde x$ . The change of coordinates is
$$
y(t) = R(\theta(t))\tilde x,
$$
where
$$
R(\theta(t))= \begin{bmatrix}cos(\theta(t)) & -sin(\theta(t)) & 0 \\ sin(\theta(t)) & cos(\theta(t)) & 0 \\ 0 & 0 & 1
 \end{bmatrix},
$$
and
$$
\dot\theta(t) = \omega.
$$
Then the dynamics of $y$ are
$$
 \dot y = -\tilde{T}R(\theta)\tilde x + R(\theta)S\tilde x + \dot{R(\theta)}\tilde x= \\
 = -\tilde{T}y + \omega \left(\begin{bmatrix}s(\theta) & c(\theta) & 0 \\ -c(\theta) & s(\theta) & 0 \\ 0 & 0 & 0\end{bmatrix} + \begin{bmatrix}-s(\theta) & -c(\theta) & 0 \\ c(\theta) & -s(\theta) & 0 \\ 0 & 0 & 0\end{bmatrix}\right)y = -\tilde{T}(\theta(t))y,
$$
where we have applied the decomposition $T = R^T\tilde TR$. Note that $||y|| = ||\tilde x||$, so if the norm $||y||$ converges to zero, the same for $\tilde x$ and of course the same for $x$.
If $\omega$ is sufficiently small, then convergence follows based on results of slowly time variant systems... but this is quite conservative result, I believe that the system is stable for all $\omega$.
