This question is a follow up on the question about marginal stability of LTI system: \begin{equation} \mathbf{x}[k] = \mathbf{A}\mathbf{x}[k-1] \end{equation} I am interested in the algorithm:
- $k := 0$
- $\mathbf{x}[0] := \mathbf{x}_0$
- $k := k + 1$
- $\mathbf{x}[k] := \mathbf{A}\mathbf{x}[k-1]$
- if $\lVert\mathbf{x}[k+1] - \mathbf{x}[k+1]\rVert > \varepsilon$ go to step 3
where $\varepsilon$ is a small value, e.g. $\varepsilon = 10^{-16}$. I am assuming that the matrix $\mathbf{A}$ has eigenvalues on the unit circle. This means that the system can be marginally stable at best, and not asymptotically stable. I am interested what are the conditions for above algorithm to terminate.
The following examples show that the marginal stability is not enough. An example of marginally stable matrix for which above algorithm doesn't converge is: \begin{equation} \mathbf{A} = \begin{bmatrix} 0 & -1\\ 1 & 0 \end{bmatrix} \end{equation} The above matrix has two complex eigenvalues $\lambda_{1,2} = \pm j$. The algorithm above will have a bounded oscillation and never finish with the iteration in the case of the presented matrix.
I don't think that complex eigenvalues are the problem since marginally stable matrix: \begin{equation} \mathbf{A} = \begin{bmatrix} -1 & 0\\ 0 & -1 \end{bmatrix} \end{equation} will also cause the above algorithm to oscillate with values $-x_0$ and $x_0$.
The algorithm will converge with an identity matrix $\mathbf{A} = \mathbf{I}$.
Are the eigenvalues of $1$ only ones allowed on the unit circle for the above algorithm to terminate? Is there some definition of stability that says that the above system is stable, but doesn't oscillate? Any answer and/or reference request would be helpful.