Matrices with real entries such that $(I -(AB-BA))^{n}=0$ I was just trying out some problems, when i couldn't solve this question:


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*Does there exist $n \times n$ matrices $A$ and $B$ with real entries such that $$ \Bigl(I - (AB-BA)\Bigr)^{n}=0?$$


I really don't know how to proceed for this question. What i did was to find some examples, but that didn't quite work. Any ideas on how one goes about thinking on such problems and how to solve them would be of great help.
 A: First off, there is no point in having the exponent be equal to the dimension of the space. On the other hand one does not want that dimension to be zero, since there exists an example with $0\times 0$ matrices (one has $I=0$ there). So I'll exclude that, and prove

For $\def\N{\Bbb N}n\in\N_{>0}$ and $k\in\N$, and $F$ a field of characteristic zero, there are no $A,B\in M_n(F)$ with
  $$ (I-(AB-BA))^k=0. $$ 

Suppose such $A,B$ existed. In other words $M=I-AB+BA$ is nilpotent. This implies the characteristic polynomial of $M$ is $X^n$, and in particular the trace of $M$ (which is minus the coefficient of $X^{n-1}$ in that characteristic polynomial) is zero. But $\def\tr{\operatorname{tr}}\tr M=\tr I-\tr(AB)+\tr(BA)=\tr I=n\neq0$, contradiction.
A: Chandru1: Here is a hint to get you started: If the equation $(I-(AB-BA))^n=0$ holds, then for any $v$, if $w=(I-(AB-BA))^{n-1}v$, then $(I-(AB-BA))w=0$, or $(AB-BA)w=w$. If $w=0$ for all $v$, then $(I-(AB-BA))^{n-1}=0$. Proceeding inductively this way, you get that either $AB-BA=I$, or else 1 is an eigenvalue of $AB-BA$. 
Now, the first option is impossible, because the traces of $AB$ and $BA$ coincide, but the trace of $I$ is $n$ rather than 0.
