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In recent days, I learned a linear algebra problem from one of my friends. It can be stated as follows.

Given four matrices $A,B,C,D$, find three matrices $E,G,F$, not simultaneously zero, such that the following conditions (1), (2), (3) are satisfied: $$ \begin{align*} (1) &\quad AE=EA, \cr (2) &\quad BG=GB, \cr (3) &\quad AF-FB=ED-CG. \end{align*} $$ The question is whether such $E,G,F$ always exist.

Also it is obvious that we can obtain $E,G$ by (1) and (2) easily. However the hard die is to satisfy condition (3). I just know when $A$ and $B$ have different spectra, we can obtain $F$ in a unique way.

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A solution is E=G=F=0. – Did Jan 3 '11 at 10:43
$E$ is usually the identity... – Wadim Zudilin Jan 3 '11 at 11:37
@Wadim: ...then why (1)? Anyway, my point was that the OP could wish to state the question more precisely. And a bit of context would not hurt either. – Did Jan 3 '11 at 13:13
@yaoxiao: it is considered impolite to edit your question in a way that makes existing answers unintelligible. I have restored the question. – Qiaochu Yuan Jan 6 '11 at 16:44
I have flagged for the moderators to lock the question. @yaoxiao: What you are doing now is spam. – Andrés E. Caicedo Jan 6 '11 at 16:53
up vote 9 down vote accepted

Let $E=x I_n$, $G=y I_n$, then 1-2 are satisfied and the 3rd is a system of $n^2$ linear homogeneous equations with total number of variables equals to $n^2 + 2$, thus there are simultaneously non-zero solutions. of course, one can do better estimates on the dimension of the solution.

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