LTI with all zero eigenvalues Say we have this LTI:
$$ 
\dot{x} = Ax + Bu \\
y = Cx + Du
$$
If the eigenvalues of the A matrix are all zero, what does that mean for the stability of the LTI? Is it unstable or marginally stable and why?
 A: If the matrix $A$ is diagonalizable, then the system is (marginally) stable. Indeed, in some variables $z= Px$, $\det P\ne 0$, the system
$\tag{1}
\dot x= Ax
$
takes the form
$\dot z=PAP^{-1}z=0z=0$. This means that $z$ is constant and also $x$. Hence, the solution does not leave any neighborhood of the origin which includes the initial point of the solution.
If $A$ is not diagonalizable, then the system is unstable. In this case, there exists some change of variables $z= Px$, $\det P\ne 0$, which transforms the system (1) into the
form
$$\tag{2}
\dot z= Jz, 
$$
where $J$ is a matrix in Jordan normal form:
$$
J=diag(J_1,\ldots,J_k),\qquad
J_j=\left(
\begin{array}{lllll}
0 & 1 & 0 & \dots & 0\\
0 & 0 & 1 & \dots & 0\\
\vdots  & \vdots & \vdots & \ddots & \vdots \\
0 & 0 & 0 & \dots & 1\\
0 & 0 & 0 & \dots & 0\\
\end{array}
\right),\quad j=1\ldots k.
$$
Since $A$ is not diagonalizable, there is at least one Jordan block $J_p$ of some size $k>1$. Consider the equations of (2), which correspond to the rows of the block $J_p$. They are
$$
\begin{array}{c}
\dot z_i= z_{i+1}\\
\dot z_{i+1}= z_{i+2}\\
\vdots\\
\dot z_{i+k-1}= 0;\\
\end{array}
$$
thus, the (part of the) solution is
$$z_{i+k-1}= C_1,$$ $$z_{i+k-2}= C_1 t+C_2,$$ $$z_{i+k-3}= C_1 \frac{t^2}2+C_2t+C_3$$ etc., where $C_1$, $C_2$, $\ldots$ are constants. Now notice that $z_{i+k-2}$ is unbounded when $C_1\ne 0$, thus, $z$ is unbounded, thus, $x$ is unbounded. Hence, the system is unstable.
