Vector recurrences (asymptotic property)

Question

Fix $m\in \mathbb{N}.$ For each $n\in \mathbb{N},$ let $A_n\in \mathbb{M}_{m}(\mathbb{C}),$ $X_n\in \mathbb{C}^m,$ and $B_n\in \mathbb{C}^m.$ Suppose that $$X_{n+1}=A_n X_n+B_n,$$ $$\lim_{n\rightarrow \infty} A_n=A_0,$$ $$|\det(A_0)|>1,$$ the moduli of all entries of $A_0$ are greater than 1, and $$\limsup_{n\rightarrow \infty}\|B_n\|^{1/n}<1,$$ where $\|\cdot\|$ is the Euclidean norm.

Is it possible that $$\limsup_{n\rightarrow \infty}\|X_n\|^{1/n}=1?$$ I have the feeling that the answer should be "No." But I do not know how to prove it. For $m=1,$ it is not difficult to show that the answer is no. Could somebody suggest how to prove or disprove it? Any references (books or papers) are very welcome? Thank you.

Answer 1

$\newcommand\1{\mathbf1}\newcommand\R{\mathbb R}$Yes, this is of course possible. E.g., this will be so if for all $n$ the matrix $A_n$ is the diagonal matrix with $1,2\dots,2$ on the diagonal, $B_n=0$, and $X_n=[1,0,\dots,0]^\top$.

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The OP has changed the question, thereby invalidating the above answer. After the change, the answer is still the same: yes, this is possible. E.g., let $m=2$ and for all $n$ let $A_n:=\begin{bmatrix}3&2\\ 2&3\end{bmatrix}=:A_0$ and $B_n:=0$.

Then all your conditions on $A_n$ and $B_n$ hold. However, $1$ is an eigenvalue of $A_n=A_0$ for all $n$. It remains to let $X_1$ be an eigenvector corresponding to the eigenvalue $1$ of $A_n$; say, we can let $X_1:=[-1,1]^\top$.

More generally, for any $m\ge2$ and for all $n$ we can let $A_n=A_0$ have (say) $3$ everywhere on the diagonal and $2$ everywhere off the diagonal, so that the eigenvalues of $A_n=A_0$ are $1$ (of multiplicity $m-1$) and $1+2m$ (so that $\det A_0=1+2m>1$). Indeed, the vector $\1:=[1,\dots,1]^\top\in\R^m$ is an eigenvector of $A_n=A_0$ corresponding to the eigenvalue $1+2m$, and all nonzero vectors in $\R^m$ orthogonal to $\1$ are eigenvectors of $A_n=A_0$ corresponding to the eigenvalue $1$.