Let $A$ and $B$ be two fixed similar matrices, and $X$ and $Y$ be matrices such that $A=XBX^{-1}$ and $A=YBY^{-1}$. Clearly we cannot conclude that $X=Y$ since for instance $X$ could be a scalar multiple of $Y$. Indeed, we know that if $X$ and $Y$ are any two matrices such that $Y^{-1}X$ commutes with $B$, then $$ YBY^{-1} = YB(Y^{-1}X)X^{-1} = Y(Y^{-1}X)BX^{-1} =XBX^{-1}.$$

Question: Is this the worst that can happen? That strikes me as unlikely— I see no obvious reason that the set of "similarity matrices for $(A,B)$" couldn't be even larger. Assuming this is true, is there a nice characterization of this set? (If not, have its properties been studied by folks before?)

Motivation: I have a module homomorphism $T:\Bbb Z^n\to \Bbb Z^n$ which is naturally expressed in a specialized basis $(f_i)$, and I also computed its action with respect to the standard basis $(e_i)$. I extended it to a linear map $\Bbb Q^n\to\Bbb Q^n$ and computed its rational canonical form $R$.

In the abstracted notation above, the $A$ that I care about is $[T]_{(f_i)}$ and the $B$ that I care about is $R$. Using a particular (naive) algorithm, I can compute a particular choice of conjugating matrix to get $A$ to be in rational canonical form; that is the $X$ that I care about. The $Y$ that I care about is what this algorithm spits out if I first do a change of basis to $[T]_{(e_i)}$.

In doing so, I observed a phenomenon that, because I was wrapped up in some problem-specific details, I found very counterintuitive at the time: there is a prime number $p$ which shows up in the denominators of the entries in $Y$ but does not appear in any denominators of the entries in $X$. Considering the objection from the first paragraph, I am less mystified. But this got me to wonder if anything useful can be recovered from this general situation.

  • $\begingroup$ I just saw math.stackexchange.com/questions/275746/…; this looks like the same question. Feel free to delete this for decluttering purposes, or keep it for searchability purposes; I don't care to dig through meta to find out what the trend of the moment is. $\endgroup$ May 11 '19 at 17:40
  • $\begingroup$ Answer: no, this isn't the worst which can happen. We sometimes have more freedom to chose from such conjugating matrices (examples for $3\times 3$-matrices already show this). Are you working in $M_n(\Bbb Q)$, or in $M_n(\Bbb Z)$? A trivial example is $A=B=I$. Then you can choose any invertible $X$. $\endgroup$ May 11 '19 at 18:10

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