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I have matrices $A,B$ of dimension $n$ with real coefficients which satisfy the following: $A^2-B^2=c(AB-BA)$ where $c$ is a real number. If $c\neq0$ , prove that $(AB-BA)^n = 0$.

So far, I've been able to show that $AB-BA$ is singular. Can someone help?

Asix
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  • Show $\text{trace}((AB-BA)^k)=0$ for $k\geq 1$ inductively. And use https://math.stackexchange.com/questions/191879/nilpotent-matrices-over-field-of-characteristic-zero – clark Jan 03 '18 at 20:37
  • @clark, but how do you prove that inductively? – jgon Jan 03 '18 at 20:38

1 Answers1

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As you mentioned in your another question, this is just a variant of that question.

Let $X=A-B$ and $Y=A+B$. Then \begin{align} XY&=(A-B)(A+B)=(c+1)(AB-BA),\\ YX&=(A+B)(A-B)=(c-1)(AB-BA). \end{align} Hence $XY-YX = 2(AB-BA)$ and $XY+YX = 2c(AB-BA)$, i.e. $XY+YX=c(XY-YX)$ for some $c\ne0$. So, the result from the aforementioned question shows that $(XY-YX)^n=0$. Hence $(AB-BA)^n=0$.

user1551
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