I have a question about marginal stability of a system: \begin{equation} \mathbf{x}[k] = \mathbf{A}\mathbf{x}[k-1] \end{equation} I would adapt the definition of marginal stability from this question to the above discrete system. The system is marginally stable if the signal $\mathbf{x}[k]$ is bounded, i.e.: \begin{equation} \lim_{k\rightarrow\infty} \mathbf{x}[k] < M < \infty \end{equation} I have trouble finding the correct book reference.

Most of the references I have found talk about asymptotic stability, and state that spectral radius of matrix should be $\rho(\mathbf{A}) < 1$. If at least one eigenvalue of matrix $\mathbf{A}$ is outside unit circle, the above system is unstable.

I have read in few references that multiple same eigenvalues result in the unstable matrix. However, I don't think this is the case for the unit matrix: \begin{equation} \mathbf{A} = \begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix} \end{equation} In this case the matrix has two same eigenvalues with value $1$, and matrix is marginally stable. What confuses me is that you can have the following matrix: \begin{equation} \mathbf{A} = \begin{bmatrix} 2 & -1\\ 1 & 0 \end{bmatrix} \end{equation} with same eigenvalues, but this system is unstable.

Can marginal stability be characterized by the location of eigenvalues? How can I determine whether the system above is stable by analyzing the matrix $\mathbf{A}$. If possible, could you provide a reference?

  • 2
    $\begingroup$ Isn't it because that $\mathbf{A}$ has a nontrivial Jordan block? $\endgroup$ Oct 25, 2018 at 3:24
  • $\begingroup$ I simply didn't know this before, this is the answer to my question. Thank you! $\endgroup$ Oct 25, 2018 at 8:32

1 Answer 1


The matrix $\mathbf{A}$ is similar to the matrix $\begin{bmatrix} 1 & 1\\ 0 & 1 \end{bmatrix}$, which is not marginally stable. Hence the original matrix is not stable.

In general, one can apply Jordan decomposition on $\mathbf{A}$, and it is marginally stable if and only if there are no eigenvalues larger than $1$ and there are no nontrivial jordan block with diagonal $1$.


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