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Mar 7 at 2:59 history became hot network question
Mar 6 at 23:58 comment added darij grinberg @Nate: I don't know; the counterexample was just a guess that turned out right.
Mar 6 at 23:46 comment added Nate Neat example @darijgrinberg! Do you know if $A$ and $A^T$ are always conjugate after some stabilization?
Mar 6 at 23:23 comment added darij grinberg ... check that $\det M = 1$ and $\left(A \oplus \operatorname{id}_1\right) M = M \left(A^T \oplus \operatorname{id}_1\right)$). See also the Example on page 15 of web.archive.org/web/20141004072809/http://hermite.cii.fc.ul.pt/… ("Direct sums") for a counterexample to a similar question.
Mar 6 at 23:23 comment added darij grinberg 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 ...
Mar 6 at 23:14 comment added Ingrid @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.
Mar 6 at 23:09 comment added darij grinberg 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 $B$?
Mar 6 at 19:37 vote accept Ingrid
Mar 6 at 19:36 comment added Ingrid 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.
Mar 6 at 19:29 answer added Denis Serre timeline score: 5
Mar 6 at 19:25 comment added Zerox 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.
S Mar 6 at 18:52 review First questions
Mar 6 at 19:42
S Mar 6 at 18:52 history asked Ingrid CC BY-SA 4.0