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Let $V=M_n(\mathbb C)$ and $A\subseteq B$ are subspaces of $V$. Let also $$ M=\{X\in V:\ \forall Y\in B,\ XY-YX\in A\}. $$ Suppose $X_0\in M$ enjoys the property that $\operatorname{tr}(ZX_0)=0$ for any $Z\in M$. Show that $X_0$ is nilpotent.


My attempt: I know that $$\operatorname{tr}(XY-YX)=0$$ but I cannot continue from there. Thank you for any help. The problem appears on a Chinese bulletin board. A soluton can be found in section 4.3 of James E. Humphreys, Introduction to Lie Algebras and Representation Theory, but I want to know if this problem can be solved by other methods. Thank you.

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  • $\begingroup$ I made some edits, please check them! $\endgroup$ Nov 10, 2013 at 16:20
  • $\begingroup$ I have see this problem in china BBS,But I also not see solution, this problem is this 设V是C^n到自己的线性变换组成的空间,B是V的子空间,A是B的子空间。设M={X∈V:任Y∈B,XY-YX∈A}.设X_0∈M使得任Z∈M,有tr(ZX_0)=0.求证,X_0是幂零变换 $\endgroup$
    – math110
    Nov 10, 2013 at 16:23
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    $\begingroup$ Its a nice problem. See this book : Introduction to Lie Algebras and Representation Theory James E.Humphreys. section 4.3. proof is given nicely here. $\endgroup$
    – GA316
    Nov 11, 2013 at 7:41
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    $\begingroup$ Something is missing on your assumptions. If $A=B=0$ then $M$ is just equal to $V$. Take any $X\in V$ and $Z=0$, then clearly $\text{trace}(ZX)=0$, but $X$ can be arbitrary. $\endgroup$ Nov 19, 2013 at 9:30
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    $\begingroup$ Something is clearly wrong with the statement. Nothing prevents taking $Z=0$, since $0\in M$ always, and for this $Z$ you get no information on $X_0$. Do you mean $\operatorname{tr}(ZX_0)=0$ for all $Z\in M$? $\endgroup$ Nov 25, 2013 at 15:48

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I wanted to add a comment but did not have sufficient reputation points. So I put my comment here which extends that of Marc.

As $A$ contains $0$, the constant diagonal matrices $k I$ ($I$ is the identity matrix) are in $M$. So, if $tr(ZX_{0}) = 0$ for all $Z \in M$ then $tr(X_{0}) = 0$. But there are invertible matrix with zero trace.

So the statement is wrong, either $tr(ZX_{0}) = 0$ for all $Z \in M$ or $tr(ZX_{0}) = 0$ for some $Z \in M$.

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  • $\begingroup$ Fix a traceless invertible matrix $X_0$. That $tr(ZX_0)=0$ for $Z=kI\in M$ doesn't mean $tr(ZX_0)=0$ for all $Z\in M$, so I don't see how your argument serves as a counterexample to the problem statement. $\endgroup$
    – user1551
    May 3, 2014 at 9:14

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