# Marginal stability of discrete linear time-invariant system

I have a question about marginal stability of a system: $$$$\mathbf{x}[k] = \mathbf{A}\mathbf{x}[k-1]$$$$ I would adapt the definition of marginal stability from this question to the above discrete system. The system is marginally stable if the signal $$\mathbf{x}[k]$$ is bounded, i.e.: $$$$\lim_{k\rightarrow\infty} \mathbf{x}[k] < M < \infty$$$$ 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: $$$$\mathbf{A} = \begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix}$$$$ In this case the matrix has two same eigenvalues with value $$1$$, and matrix is marginally stable. What confuses me is that you can have the following matrix: $$$$\mathbf{A} = \begin{bmatrix} 2 & -1\\ 1 & 0 \end{bmatrix}$$$$ 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?

• Isn't it because that $\mathbf{A}$ has a nontrivial Jordan block? – LeechLattice Oct 25 '18 at 3:24
• I simply didn't know this before, this is the answer to my question. Thank you! – Slaven Glumac Oct 25 '18 at 8:32

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 $$1$$.