Suppose we have a linear matrix space $S\subset M_{n\times n}$, any $M\in S$ is a nilpotent matrix,

that is $M^n=0$.

Then for any finite subset of $S$, says $A=${$M_1,...,M_k$}, one can define the following map

$E(X)=\sum_{j=1}^k M_j X M_j^{+}$, where $M^{+}$ denotes the complex conjugate of $M$.

Does there always exist some integer $d$ such that

$E^d(X)=0$ holds for any $X$, where $E^d=E\circ E^{d-1}$.


The claim is not true. Suppose that $n= 3$ and take $A=\lbrace E_{1,2}+E_{2,3}, E_{3,2}-E_{2,1}\rbrace$, and let $S$ be the span of $A$. Note that all matrices from $S$ are nilpotent. However, if we compute $E(X)$, $E^2(X)$ and $E^3(X)$, we see that the entries from $E(X)$ and $E^3(X)$ are equal up to a scalar $1$ or $2$. In fact, the coefficients $x_{21},x_{12}, x_{23},x_{32}$ of $X=(x_{ij})$ are fixed, i.e., the same for $E(X)$ and $E^3(X)$. For example, $E(X)=E^3(X)$ for $$ X=\begin{pmatrix} 0 & 1 & 0 \cr 1 & 0 & 1\cr 0 & 1 & 0 \end{pmatrix}. $$ So there is no $d$ with $E^d(X)=0$ for all $X$, if I am not mistaken. The point is that not all subspaces $S$ of $M_{n,n}$ consisiting of nilpotent elements are conjugated to some linear space of stricly trianguar matrices.

Edit: $A=E_{1,2}$ and $B=E_{2,1}$ is not a counterexample, as pointed out by Abhinav Kumar.

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  • $\begingroup$ There's a problem with your counterexample: namely that the span of A doesn't just consist of nilpotent matrices ($E_{1,2} + E_{2,1}$ squares to the identity, or to a projection matrix if the dimension is larger than $2$). $\endgroup$ – Abhinav Kumar May 29 '13 at 15:14
  • $\begingroup$ Abhinav, I have edited my answer, because you were right. $\endgroup$ – Dietrich Burde May 29 '13 at 15:32
  • $\begingroup$ It looks correct now. I would suggest appending to your answer rather than editing it completely, so this comment thread makes sense ... $\endgroup$ – Abhinav Kumar May 29 '13 at 15:38
  • $\begingroup$ Yes, I apologize. I deleted the answer, because it was wrong. Then I wrote a new answer and used undelete. I did not know that the comments would come back. $\endgroup$ – Dietrich Burde May 29 '13 at 15:47
  • $\begingroup$ No worries! MO has its quirks. $\endgroup$ – Abhinav Kumar May 29 '13 at 16:08

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