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I was just trying out some problems, when i couldn't solve this question:

  • 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.

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up vote 11 down vote accepted

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.

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I understadn that $AB-BA=I$ does not hold but then i did not understand the point behind $1$ being an eigen value of $AB-BA$.. could you please explain that! – Praphulla Koushik Jan 8 '14 at 11:03
@Marc Thank you, but I prefer the phrasing as it was. – Andrés E. Caicedo Aug 31 '14 at 5:07
So near the end of your first sentence "then X, or Y" should be read as "then X, or in other words Y"? This really put me off when I was reading this. – Marc van Leeuwen Aug 31 '14 at 5:10
Yes, that's the meaning. – Andrés E. Caicedo Aug 31 '14 at 5:12
Still I don't get the punchline. Either you've got an eigenvector for $1$, or $AB-BA=I$ (in which case you've got plenty such eigenvectors as well). There's no real need to brandish the impossibility of the stronger second condition in order to conclude that the weaker first condition must hold. Where does one go from there? (Commutators can have $1$ as eigenvalue.) – Marc van Leeuwen Aug 31 '14 at 5:37

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.

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Nice and to the point. (Do you know a counterexample in characteristic different from $0$?) – Andrés E. Caicedo Aug 31 '14 at 5:30
Henning Makholm gave one here: – Marc van Leeuwen Aug 31 '14 at 5:41
Hehe. I even commented on that question. Bookmarked now, so I don't forget again. Thank you! – Andrés E. Caicedo Aug 31 '14 at 5:43

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