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Is every $n \times n$ matrix with entries in $\mathbb{Z}_p$ (or even $\mathbb{Z}$) conjugate to its transpose via a matrix in $GL_n(\mathbb{Z}_p)$?

On the one hand, I know the analogous fact is false for matrices over $\mathbb{Z}$ with counterexamples constructed via the Latimer-MacDuffee theorem (but the only counterexamples I've seen break down in $\mathbb{Z}_p$ for all $p$.)

On the other hand I know there are various conditions under which $GL_n(\mathbb{Q}_p)$-conjugacy implies $GL_n(\mathbb{Z}_p)$-conjugacy, as well as lifting criteria from $GL_n(\mathbb{F}_p)$, so it seems possible that it could hold here.

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    $\begingroup$ The word "any" is sometimes confusing to non-native English speakers, since it can mean either "all" (e.g., "any positive number has a real square root") or "some" (e.g., "does this equation have any solutions?"). I think it is better to write "each" instead of "any" at the start of your question. $\endgroup$
    – KConrad
    Commented Dec 15, 2018 at 3:07
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    $\begingroup$ Have you seen the second paragraph of arxiv.org/pdf/1212.6157.pdf? $\endgroup$
    – KConrad
    Commented Dec 15, 2018 at 3:09
  • $\begingroup$ For $n = 2$, this reduces to the question of when a matrix can be expressed as the product of two symmetric matrices, one nonsingular. More generally, if we have $A = BC$ with $B,C$ symmetric and $B$ nonsingular, then $A^t = CB$ is obviously similar to $A$. I wouldn't be surprised to see (though I don't have a full proof) that your problem can be reduced to that. $\endgroup$
    – user44191
    Commented Dec 15, 2018 at 4:10
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    $\begingroup$ I'd be interested to see your example over $\Bbb{Z}$. $\endgroup$
    – abx
    Commented Dec 15, 2018 at 10:51
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    $\begingroup$ @MartinBrandenburg see kconrad.math.uconn.edu/blurbs/gradnumthy/matrixconj.pdf $\endgroup$
    – KConrad
    Commented Oct 21, 2023 at 6:22

2 Answers 2

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No for $n\geq 3$.

If $A\in M_n(\mathbf Z_p)$ were similar to $A^T\in M_n(\mathbf Z_p)$, then going modulo $p^2$, its image in $M_n(\mathbf Z/p^2\mathbf Z)$ would be similar to the image of its transpose.

However, Pooja Singla, Steven Spallone and I have shown in Similarity of matrices over local rings of length two that this fails for $n\geq 3$ (Theorem 7.9). For $n=3$ and $4$ we provide a classification and exact counts for matrices where this fails (see Table 9.4/ Table 8 in the arXiv version).

The answer is yes for $n=2$. This follows from Theorem 2.2 of Similarity classes of 3x3 matrices over a local principal ideal ring . I thank @KConrad for nudging me to think this through.

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    $\begingroup$ Your $M_n(\mathbf Z/p^\mathbf Z)$ should probably be $M_n(\mathbf Z/p^2)$. And what about $n = 2$? (I'm glad to see your reply, since I had cited the arXiv version of it in a comment earlier.) $\endgroup$
    – KConrad
    Commented Dec 15, 2018 at 5:11
  • $\begingroup$ Thanks, fixed the typo, and answered the 2x2 case. $\endgroup$
    – Amritanshu Prasad
    Commented Dec 15, 2018 at 5:57
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No, the matrix $$ \begin{pmatrix}0 & 1 & 0\\ 0 & 0 & p\\ 0 & 0 & 0 \end{pmatrix}\in\mathrm{M}_{3}(\mathbb{Z}_{p}) $$

is not similar to its transpose. This has been known for some time. McDonald, Linear Algebra over Commutative Rings (1984) atttributes this to M. Hochster in Exercise V.D.17, p. 424. More generally, it is noted there that this counter-example works for any commutative ring $R$ that is not von Neumann regular when instead of $p$ one uses an element $x\in R$ such that $x\not\in (x^2)$.

The "reason" why there is no counter-example for $2\times 2$ matrices over $\mathbb{Z}_p$ is that any such matrix can be written as $aI+p^iX$, where $a$ is a scalar and $X$ is a regular matrix, that is, $X$ is similar to a companion matrix. The point is that it is known that companion matrices over commutative rings are similar to their transpose (I think this is due to Gustafson).

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