# Given a chain of commuting matrices over the complex numbers, can we build one over the real numbers?

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.)

• 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.
– YCor
Jul 3, 2020 at 21:17
• 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). Jul 3, 2020 at 21:49
• 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. Jul 3, 2020 at 22:04