Marginal stability with non-simple poles on the imaginary axis It is known that a system marginally stable if and only if the real part of every pole in the system's transfer-function is non-positive, one or more poles have zero real part, and all poles with zero real part are simple roots (i.e. the poles on the imaginary axis are all distinct from one another).[Wikipedia].
My question is based on the definition, a system with state space representation (A,B,C), where 
$A = \begin{bmatrix} 0 & 0\\ 0 & 0 \end{bmatrix}$ that has a non simple root at origin, is considered unstable? why?
 A: Suppose that
$$\mathrm A = \mathrm O_2 \qquad \qquad \mathrm b = \begin{bmatrix} 0\\ 1\end{bmatrix} \qquad \qquad \mathrm c = \begin{bmatrix} 1 & 1\end{bmatrix}$$
and that the initial condition is $\mathrm x_0 := (x_{10}, x_{20})$. Hence, the states are
$$x_1 (t) = x_{10} \qquad \qquad \qquad x_2 (t) = x_{20} + \int_0^{t} u (\tau) \,\mathrm{d}\tau$$
and the output signal is
$$y (t) = (x_{10} + x_{20}) + \int_0^{t} u (\tau) \,\mathrm{d}\tau$$
Suppose that we start from zero initial conditions and that the input signal is constant, say, $u = 1$. Hence, the output signal is given by
$$y (t) = \int_0^{t} u (\tau) \,\mathrm{d}\tau = t$$
Thus, even though the LTI system is internally marginally stable (the zero eigenvalue has two $1 \times 1$ Jordan blocks, rather than one $2 \times 2$ Jordan block), it is not BIBO (bounded input, bounded output) stable, as a bounded input can produce an unbounded output.
A: Since not all the eigenvalues of $A$ are negative, then your system is not asymptotically stable. However, even having all the eigenvalues on the complex axis you system can be unstable in the sense BIBO (Bounded Input / Bounded Output) but marginally stable (if all the zero eigenvalues are simple). Check the answer from @Rodrigo for more details.
