Let $\mathbb{F}$ be a non-Archimedean local field. Let $\{T_a\}_{a=1}^\infty$ be an arbitrary sequence of linear operators $\mathbb{F}^n\to\mathbb{F}^n$. After a choice of subsequence, is it possible to construct sequences of vectors $\{e^i_a\}\subset \mathbb{F}^n$, $i=1,\dots, n$ such that the following properties are satisfied:

(1) for any $a$ the vectors $e^1_a,\dots, e^n_a$ form a basis of $\mathbb{F}^n$ and when $a\to \infty$ they converge to another basis of $\mathbb{F}^n$;

(2) the vectors $T_a(e^1_a),\dots, T_a(e^n_a)$ also form a basis of $\mathbb{F}^n$ and when $a\to \infty$ appropriate multiples of these vectors converge to another basis of $\mathbb{F}^n$?

Remark. In the Archimedean case the answer is positive. Indeed fix a Euclidean (resp. Hermitian) metric. $T_a$ admits an orthonormal basis $e^1_a,\dots, e^n_a$ such that $T_a(e^1_a),\dots, T_a(e^n_a)$ are pairwise orthogonal vectors. After a choice of subsequence all the required properties will be satisfied.

Let $\mathbb{F}$ be a non-Archimedean local field. Let $\{T_a\}_{a=1}^\infty$ be a sequence of linear operators $\mathbb{F}^n\to\mathbb{F}^n$ of rank $n$. After a choice of subsequence, is it possible to construct sequences of vectors $\{e^i_a\}\subset \mathbb{F}^n$, $i=1,\dots, n,$ such that the following properties are satisfied:

(1) for any $a$ the vectors $e^1_a,\dots, e^n_a$ form a basis of $\mathbb{F}^n$ and when $a\to \infty$ they converge to another basis of $\mathbb{F}^n$;

(2) the vectors $T_a(e^1_a),\dots, T_a(e^n_a)$ also form a basis of $\mathbb{F}^n$ and when $a\to \infty$ appropriate multiples of these vectors converge to another basis of $\mathbb{F}^n$?

Remark. In the Archimedean case the answer is positive. Indeed fix a Euclidean (resp. Hermitian) metric. $T_a$ admits an orthonormal basis $e^1_a,\dots, e^n_a$ such that $T_a(e^1_a),\dots, T_a(e^n_a)$ are pairwise orthogonal vectors. After a choice of subsequence all the required properties will be satisfied.