LTI with all zero eigenvalues

Question

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?

Answer 1

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.