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$\DeclareMathOperator{\Conj}{Conj} \DeclareMathOperator{\GL}{GL} \DeclareMathOperator{\id}{id} \newcommand\Z{\mathbb{Z}}$

For an $n \times n$ integer matrix $A \in \GL_n(\Z)$, let $\Conj(A)$ be the set of $n \times n$ integer matrices that are $\GL_n(\Z)$-conjugate to $A$. Also, let $A \oplus \id_m$ be the $(n+m) \times (n+m)$ integer matrix obtained by adjoining an $m \times m$ identity matrix to the bottom right hand corner of $A$.

Question: There is a natural map $\Conj(A) \rightarrow \Conj(A \oplus \id_m)$. Are there conditions on $A$ that ensure this is always a surjection/injection/bijection? Also, is it the case that for sufficiently large $m$ we always have that the map $\Conj(A \oplus \id_m) \rightarrow \Conj(A \oplus \id_{m+1})$ is a bijection?

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    $\begingroup$ The natural map $X \rightarrow X \oplus \mathrm{id}_m$ is always an injection. It is a bijection if and only if $A=\mathrm{id}_n$ since $\mathrm{id}_m \oplus A$ belongs to $\mathrm{Conj}(A \oplus \mathrm{id}_m)$ by perturbation. $\endgroup$
    – Zerox
    Commented Mar 6 at 19:25
  • $\begingroup$ Shoot, I stated this incorrectly, and as you point out it is obviously false. I'd delete it, but there is no button for me to do that. $\endgroup$
    – Ingrid
    Commented Mar 6 at 19:36
  • $\begingroup$ Related question: If $A \oplus \operatorname{id}_m$ is conjugate to $B \oplus \operatorname{id}_m$, does it then follow that $A$ is conjugate to $B$? $\endgroup$
    – darij grinberg
    Commented Mar 6 at 23:09
  • $\begingroup$ @darijgrinberg: Versions of that were what I really meant to ask about, but I totally screwed it up when writing the question. I'll probably ask a corrected question in a few days, but I don't want to spam MO with constant variants on the same thing. $\endgroup$
    – Ingrid
    Commented Mar 6 at 23:14
  • $\begingroup$ Actually, I can answer this one too now: The $2\times 2$-matrix $A := \begin{pmatrix} 8 & 2 \\ 0 & 1 \end{pmatrix} \in \mathbb Z^{2\times 2}$ is not similar to its transpose $A^T$ (this was claimed in math.stackexchange.com/questions/1732276 and can be easily checked solving the relevant system by hand), but the $3\times 3$-matrix $A \oplus \operatorname{id}_1$ is similar to $A^T \oplus \operatorname{id}_1$ via the invertible conjugating matrix $M = \begin{pmatrix} 1&-4&2 \\ -4&14&-7 \\ 2&-7&3 \end{pmatrix} \in \operatorname{GL}_3\left(\mathbb Z\right)$ (just ... $\endgroup$
    – darij grinberg
    Commented Mar 6 at 23:23

1 Answer 1

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${\rm Conj}(A)\to{\rm Conj}(A\oplus I_m)$ is not surjective, because $$I_m\oplus A\in{\rm Conj}(A\oplus I_m) \quad\hbox{but}\quad\not\in R({\rm Conj}(A)).$$

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