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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.

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    $\begingroup$ If I'm correct, all you need is a "Cartan" $KDK$ decomposition for $D$ the group of invertible diagonal matrices, for some compact subgroup $K$ of $\mathrm{GL}_n(\mathbb{F})$. In the non-archimedean case this is done in Bruhat-Tits (Publ IHES 1972), Section 4.4, if I'm correct. $\endgroup$
    – YCor
    Commented Dec 20, 2022 at 14:47

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$\def\FF{\mathbb{F}}\def\GL{\text{GL}}$This is basically the same thing YCor sketches in his comment: Let $R$ be the ring of integers of $\FF$. Let $K$ be the group of invertible matrices with entries in $R$ whose inverses are also in $R$, and let $T$ be the group of diagonal matrices.

The ring $R$ is a dvr so, by the Smith normal form theorem, each of your matrices $T_a$ can be factored as $U_a D_a V_a$ with $D_a \in T$ and $U_a$, $V_a \in K$.

The group $K$ is compact in the non-archimedean topology. Proof: $K$ can be written as $\{ (g,h) : gh = \text{Id}_n,\ g,h \in \text{Mat}_{n \times n}(R) \}$. This is a closed subspace of $\text{Mat}_{n \times n}(R)^2$, and $\text{Mat}_{n \times n}(R)^2 \cong R^{2n^2}$ is obviously compact. So we can extract a subsequence where the $U_a$ approach a limit $U$ and the $V_a$ approach a limit $V$.

Then we take $(e_a^1, e_a^2, \ldots, e_a^n)$ to be the columns of $V_a^{-1}$, just as in the archimedean case.

I learned the slogan "Smith normal form is non-archimedean singular value decomposition" from Kiran Kedlaya.

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    $\begingroup$ Nice! I'm wondering if there is an algorithm for finding the SVD that somehow resembles the Smith normal form algorithm (an "Euclidean algorithm" on the first row, followed by one on the first column, then again on the first row, etc., until the only nonzero entry in the first row and the first column becomes the diagonal entry; then moving on to the next row). A "geometric Euclidean algorithm", so to speak, using Givens rotations perhaps (but probably infinitely many, since the SVD is not rational). $\endgroup$
    – darij grinberg
    Commented Dec 20, 2022 at 15:48
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    $\begingroup$ Interesting! If I use a left rotation to zero out the bottom $n-1$ entries in the left column, then use a right rotation to zero out the right $n-1$ entries in the top row, then go back and do the left column again, then the top row again and so forth, then the upper left entry increases at each step, and it seems likely that the limit is the first singular value. No idea how fast it would converge, though. $\endgroup$
    – David E Speyer
    Commented Dec 21, 2022 at 4:09
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    $\begingroup$ I just thought about this a bit more, and I think a better idea would be a variant of the QR algorithm for eigenvalues. Let $A$ be our initial matrix and write $A = Q_1 R_1$. Then let $R_1 = R_2 Q_2$, $R_2 = Q_3 R_3$ etcetera. Here $Q_k$ is always orthogonal, and $R_k$ is either upper or lower triangular depending on the parity of $k$. Then $A$, $R_1$, $R_2$, $R_3$, etcetera all have the same singular values, and a little experimentation suggests that the off diagonal entries of the $R$'s go to zero. $\endgroup$
    – David E Speyer
    Commented Dec 28, 2022 at 15:47
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    $\begingroup$ @darijgrinberg It looks like Dan at math.SE also had this idea math.stackexchange.com/a/4309570/448 . $\endgroup$
    – David E Speyer
    Commented Dec 28, 2022 at 15:55
  • $\begingroup$ Ah, that's an interesting variant of QR iteration! I guess it's really pretty similar to what I suggested, except that we don't focus entirely on the first row/column before stepping to the next ones, but rather triangularize everything at once as well we can. $\endgroup$
    – darij grinberg
    Commented Dec 28, 2022 at 16:16

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