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I have matrices $X,Y$ of dimension $n$ with real coefficients which satisfy the following: $XY+YX=c(YX-XY)$ where $c$ is a real number. If $c\neq0$ , prove that $(YX-XY)^n = 0$.

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

Asix
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1 Answers1

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By the given condition, $(c+1)XY=(c-1)YX$. Since $c\ne0$, either $XY=kYX$ or $YX=kXY$ for some $|k|\ne1$. However, as $XY$ and $YX$ have identical spectra, they must be nilpotent. Hence $(YX-XY)^n=(1-k)^n(YX)^n=0$ when $XY=kYX$ and the analogous holds when $YX=kXY$.

user1551
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  • Why must $XY$ and $YX$ be nilpotent? I don't get your argument. – Gabriel Romon Jan 04 '18 at 09:15
  • @GabrielRomon On one hand their spectral radii are identical: $\rho(XY)=\rho(YX)$ (this is a general truth, even for rectangular matrices $X$ and $Y$); on the other hand, since $XY=kYX$, $\rho(XY)=|k|\rho(YX)$. As $|k|\ne1$, we must have $\rho(XY)=\rho(YX)=0$, i.e. $XY$ and $YX$ are nilpotent. – user1551 Jan 04 '18 at 09:31