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Suppose we have two $n\times n$ matrices $A$ and $B$ with entries in $\mathbb{R}$, and two non-scalar matrices $X$ and $Y$ with entries in $\mathbb{C}$, such that $AX=XA$, $XY=YX$, and $BY=YB$.

Is it necessarily the case that there exist non-scalar matrices $X'$ and $Y'$ with entries in $\mathbb{R}$ such that $AX'=X'A$, $X'Y'=Y'X'$, and $BY'=Y'B$?

(Here "non-scalar" just means that the matrices aren't scalar multiples of the identity matrix.)

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    $\begingroup$ Changing $X$ into $X-(Tr(X)/n)I_n$ and similarly for $Y$, this can be restated as follows: define $V$ as the set of pairs $(X,Y)$ with the given condition and with trace zero. If $V$ contains $(X,Y)$ both nonzero, does it contain a real point with the same condition? I have no idea whether this can be useful, but at worst it's a harmless comment. $\endgroup$
    – YCor
    Commented Jul 3, 2020 at 21:17
  • $\begingroup$ Just a comment (hence in the right place): the obvious thing to do is to put $X' = X + \overline X$ and $Y' = Y + \overline Y$, but that might be scalar, so use $X' = i(X - \overline X)$ and/or $Y' = i(Y - \overline Y)$ if necessary; but then of course it's no longer obvious that $X'$ and $Y'$ commute. If $X$ is semisimple, then the centraliser of either choice for $X'$ is both larger than that of $X$, and closed under conjugation, hence contains either choice for $Y'$; and similarly for $Y$; but I don't see how to reduce to this case (since $X_\text s$ or $Y_\text s$ could be scalar). $\endgroup$
    – LSpice
    Commented Jul 3, 2020 at 21:49
  • $\begingroup$ In fact, this shows that you can reduce to the case where $X$ and $Y$ are both nilpotent, which means, since they commute, they can be simultaneously conjugated to strictly upper triangular elements; but I think that this conjugation cannot also be assumed to keep $A$ and $B$ real. $\endgroup$
    – LSpice
    Commented Jul 3, 2020 at 22:04

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Update: it turns out the answer is no! See for instance Example 4.14 here.

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