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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.
user1551
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  • I made some edits, please check them! – Daniel Donnelly Nov 10 '13 at 16:20
  • 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 – math110 Nov 10 '13 at 16:23
  • $V:C^{n}\to C^n$ – math110 Nov 10 '13 at 16:24
  • @nanchangjian how do you have that $trace(ZX_0)=0$ it seems nontrivial. Could you state proof? – Daniel Donnelly Nov 10 '13 at 17:43
  • sorry, it;s such $trance(ZX_{0})=0$ –  Nov 11 '13 at 07:07
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    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. – GA316 Nov 11 '13 at 07:41
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    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. – Tomasz Lenarcik Nov 19 '13 at 09:30
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    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 $Z\in M$? – Marc van Leeuwen Nov 25 '13 at 15:48

1 Answers1

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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 $Z \in M$.

Wei Wang
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  • 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. – user1551 May 03 '14 at 09:14