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Question Find matrices $A,B$: such that there exists matrices $X,Y$: $AX-BY=[A,B]$ and $XA-YB=[A,B]$, here $X,Y$ are also matrices, [A,B]=AB-BA - commutator.

"Find" means "say something worth" any help is welcome. In particular it is not clear for me:

0) Are there some nice families of solutions except $[A,B]=aA +bB$?

1) What is dimension of variety $A,B$?

2) Is it reduced? If not how many components this variety has?

... and further in future one might be asking algebraic-geometric questions about is the corresponding ideal radical ? normality, singularities...


Remarks. For fixed $A,B$ we have linear system of equations on $X,Y$. however this system is always degenerate. For generic matrices $A,B$ it would not have solutions.

See also these questions:

$\det(A \otimes B - B \otimes A) = 0$ why? Why $rk(M) = n^2-n$ ? Why x and -x in Spec(M) ?

http://mathoverflow.net/questions/97036/relaxing-commutativity-for-c1-c2-find-q1-q2-1-c1-c2q1c2-q2c1-2-q1-q20/

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up vote 4 down vote accepted

If $A$ and $B$ are invertible, the following condition is necessary. If $A$ and $B$ are invertible and $A B^{-1}$ has distinct eigenvalues, it is necessary and sufficient:

For all integers $n$, we have $$Tr{\Large (}[A,B] (AB^{-1})^k {\Large )} = Tr{\Large (}[A,B] (B^{-1} A)^k {\Large )}. \quad (\ast)$$

Proof of necessity: I'll write out the case of $k$ positive; $k$ negative is similar. The left hand side of $(\ast)$ is $$Tr((AX-BY) (A B^{-1})^k) = Tr (X (AB^{-1})^k A - Y (AB^{-1})^k B) = Tr (XAB^{-1} A B^{-1} \cdots A - Y A B^{-1} \cdots A)$$ where we have used the cyclic symmetry of trace. Similarly, the right hand side of $(\ast)$ is $$Tr((XA-YB) (B^{-1} A)^k) = Tr(X A (B^{-1} A)^k - Y B (B^{-1} A)^k) = Tr (XAB^{-1} A B^{-1} \cdots A - Y A B^{-1} \cdots A).$$ The two expressions are identical. $\square$

Proof of sufficiency: Let's study the more general question of, given matrices $P$ and $Q$, when there exist $X$ and $Y$ with $AX-BY=P$ and $XA-YB=Q$. Let $V \subseteq \mathrm{Mat}_n^2$ be the set of pairs $(P,Q)$ which can be written in this way.

Clearly, $V$ is a linear space, so it is defined by some linear relations. Every linear function on $\mathrm{Mat}_n^2$ is of the form $(P,Q) \mapsto Tr(PT-QU)$ for some matrices $T$ and $U$. Let's figure out for which $(T,U)$ the function $Tr(PT-QU)$ vanishes on $V$.

This occurs if and only if $$Tr((AX-BY) T -(XA-YB)U) = Tr (X (TA-AU)) - Tr (Y (TB-BU))=0.$$ for all $X$, $Y$. This occurs if and only if $TA=AU$ and $TB=BU$.

So we deduce the more general, but less explicit, condition:

Your equation is solvable if and only if we have $Tr([A,B] T) = Tr([A,B] U)$ for all $T$ and $U$ such that $TA=AU$ and $TB=BU$.

Now suppose $A$ and $B$ are invertible. Then $T A B^{-1} = A U B^{-1} = A B^{-1} T$. So $T$ commutes with $A B^{-1}$ and $U$ can be computed from $T$ as $U=A^{-1} T A$. With the additional hypothesis that $A B^{-1}$ has distinct eigenvalues, the space of matrices that commute with $A B^{-1}$ is spanned by the powers of $A B^{-1}$. $\square$.

Some comments on dimension: Note that the sequence $(A B^{-1})^k$ satisfies a linear recurrence of length $n$, by the Cayley-Hamilton theorem. So condition $(\ast)$ for any $n$ consecutive values implies $(\ast)$ for all $k$. Also, condition $(\ast)$ is trivial for $k=0$. So condition $(\ast)$ cuts out a space of codimension $\leq n-1$ in the open set of matrices for which $A$ and $B$ are invertible and $A B^{-1}$ has distinct eigenvalues. The space of solutions to your equation thus has dimension $\geq 2 n^2 - (n-1)$.

I didn't check carefully, but I think that the condition $[A,B] \in \mathrm{Span}(A,B)$ has dimension $n^2 + n+1$, so much smaller.

A general comment: The set of $(A,B)$ such that this equation is solvable will be constructable, by Chevalley's theorem. But I see no reason to expect that it is closed. Without a result saying that the image is closed, I don't see a way to equip the image with the structure of an algebraic variety. Without this, questions about singularities and more subtle invariants don't make sense.

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Thank you very much !!! That is great and unexpected, I did not thought this line of thought... Some replies 1) it seems to me dim of [A,B]=a A +bB is n^2+n the same as [A,B]=0 2) (A,B) - only contructible - seems yes - probably I was too hurrying up - what I actually keep in mind is variety (A,B,X,Y) with additional equation [X,Y]=0 "relaxed commuting variety". May be it is the same good as usual "commuting variety". –  Alexander Chervov Jun 9 '12 at 20:01
    
See math.stackexchange.com/questions/146615/… for my arguments for n^2+n for [A,B]=aA+bB –  Alexander Chervov Jun 9 '12 at 20:27
    
Note that my solution (like your question above) does not impose that $X$ and $Y$ commute. That extra nonlinear condition sounds hard to deal with. Also, you're right about $n^2+n$; I was working too fast. –  David Speyer Jun 9 '12 at 22:00
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