Is a MIMO system with a pole $s=0$ stable? Is a MIMO (multi-input-multi-output) system $G(s)$ with a pole $s=0$ stable? Or what will happen if the MIMO system has a $s=0$ pole?
 A: Stability of MIMO LTI system depends on what kind of definition of stability you are using. For example BIBO (bounded input bounded output) stability is not satisfied if the pole at zero is controllable, because for example a constant input will then be integrated to a ramp.
Asymptotic/exponential stability is also not satisfied, because that requires that all poles have real parts which are strictly negative.
However Lyapunov stability is satisfied iff all poles have a negative real part or have a geometric multiplicity of one when they lie on the imaginary axis. For systems in state space form this comes down to that when writing the system matrix in Jordan form the Jordan blocks associated with eigenvalues on the imaginary axis have at most size one.
So if a MIMO LTI system has one pole at zero and the rest with a strictly negative real part then that will be at least Lyapunov stable. This is because the geometric multiplicity of that pole at zero is then always equal to one.
It can also be noted that a system can be BIBO stable but not Lyapunov stable if it has a uncontrollable unstable pole.
A: @Kwin van der Veen
I agree that controllability of unstable poles can be checked if you test a system for BIBO stabilitiy: If an unstable pole is uncontrollable, it cannot be excited by the input. But what about observability?
Example 1
Consider an LTI with a stable pole $\Re\{s_1\}<0$ and an unstable pole $\Re\{s_2\}>0$ that is observable but not controllable:


*

*It is clear that our input can not excite the unstable pole, as it is uncontrollable

*We can find an initial condition $x_0\neq 0$ that will let the unstable pole contribute to (at least) one output $y(t)$, because we assumed it was observable

*Now let us set the input to zero (clearly bounded), use the initial condition $x_0$ and consider the output $y(t)$. It will tend to infinity even though our input was bounded and the unstable pole was uncontrollable - a contradiction?


Example system:
$$\begin{align}
A_1&=\begin{pmatrix}
  -1&0\\
  0&1
\end{pmatrix},&
B_1&=\begin{pmatrix}
  1\\
  0
\end{pmatrix},&
C_1&=\begin{pmatrix}
  1&0\\
  1&1
\end{pmatrix},&
D_1&=\underline{0},&&&
x_0&=\begin{pmatrix}
  0\\
  1
\end{pmatrix}\\\\
y(t)&:= y_2(t)
\end{align}
$$
Example 2
Let us explore the dual case - an LTI with a stable pole $\Re\{s_1\}<0$ and an unstable pole $\Re\{s_2\}>0$ that is controllable but not observable:


*

*It is clear that our input can excite the unstable pole - it is controllable now!

*However, we cannot see the effect our input might have on the unstable pole, regardless of the output we may choose: We assumed the unstable pole was unobservable! That even includes any initial condition that might contribute to the unstable pole.

*If we set the input to any bounded function and choose any output $y(t)$, the unstable pole will have no effect on our ouput - we can only observe the stable part of the LTI! Our output will be bounded for any bounded input and any initial condition, so the LTI should be BIBO stable by definition. However, it has a controllable unstable pole - another contradiction?


Example system:
$$\begin{align}
A_2&=\begin{pmatrix}
  -1&0\\
  0&1
\end{pmatrix},&
B_2&=\begin{pmatrix}
  1\\
  1
\end{pmatrix},&
C_2&=\begin{pmatrix}
  1&0
\end{pmatrix},&
D_2&=\underline{0},&&&
x_0&=\begin{pmatrix}
  0\\
  1
\end{pmatrix}
\end{align}
$$
Conclusion
I'd say: The necessary and sufficient conditions for BIBO-stability of continous time LTIs in state space form without poles on the imaginary axis are that no (part of any) unstable pole may be observable at any output $y(t)$.
If one takes (possibly multiple) poles on the imaginary axis into account, one even has to consider which parts of the Jordan-Blocks of these poles are observable to get to necessary and sufficient contitions!
Sadly, I have no references that precisely state conditions for BIBO-stability in terms of poles/zeros of the underlying state space system instead of the $L_1$-norm of its implulse response.
Rem.: I know these example systems are purely academic - I only used them to get to grips with the different notions of stability, controllablity and oberservability and their necessary/sufficient conditions. Sadly, our control systems course did not cover these corner cases - I only learned about them in a numeric course for mathematicians...
Rem.: Often I found that people mix up the "poles of the system" with the poles they found in a particular transfer function of that system - when in general, a transfer function yields only a subset or all of the system poles.
