When are all trajectories of $x_{k+1}=Ax_k+b$ bounded?

Question

Suppose $x_0\in \mathbb{R^d}$, $b\in \mathbb{R^d}$, $A$ is a symmetric $d \times d$ matrix, and I'm looking at the following recurrence

$$x_{k+1}=Ax_k+b$$

How do I determine whether all trajectories are bounded? And how do I bound them?

Answer 1

I don't know the complete answer, but I can try to explain some cases and indicate why a complete answer is complicated.

Since $A$ is symmetric, it is conjugate to a diagonal matrix $D$, meaning there exists an invertible matrix $U$ and real numbers $\lambda_1,\lambda_2,\ldots,\lambda_k$ such that $$D = U^{-1}AU = \begin{pmatrix} \lambda_1 & 0 & ... & 0 \\ 0 & \lambda_2 & ... & 0 \\ \vdots & \vdots & \ldots & \vdots \\ 0 & 0 & \ldots & \lambda_n \\ \end{pmatrix} $$ Thus, the whole dynamical system $f(x) = Ax+b$ is conjugate to $$f'(x) = D (x) + b', \qquad b' = U^{-1} b U $$ and $f$ has bounded orbits if and only if $f'$ has bounded orbits.

Now there are cases to consider:

• If there exists $i = 1,\ldots,n$ such that $|\lambda_i| > 1$ then $f'$, hence $f$, has an unbounded orbit, in which the magnitude of points in that orbit grows exponentially, with the base of the exponential being $|\lambda_i|$.
• If for all $i=1,\ldots,n$ we have $|\lambda_i| < 1$ then $f'$, hence $f$, has bounded orbits. In fact each orbit converges exponentially to fixed point, with the base of the exponential being bounded above by the largest value of $|\lambda_i|$.

Notice that in these two cases the nature of $b$ is irrelevant. The exact details of why it is irrelevant come out in the details of the proofs of the above two cases, but the point is that one can in fact prove that $f'$ has a fixed point, and using that one can prove that $f'$ is conjugate to $f''(x) = D(x)$ using a conjugation which is simply a translation that moves the fixed point to the origin. It is then quite simple to find a proof of the above statements for $f''(x)$, by thinking about eigenvectors.

The remaining case is where $|\lambda_i| \le 1$ for all $i$, and $|\lambda_i|=1$ for some $i$. This is a tricky case and harder to analyze, and it depends on delicate interactions between $A$ and $b$.