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We have $A \in M_n(C) $. prove $A$ is nil-potent iff for some matrix $B$ we have $AB-BA=A$.

i found this question in a math competition and i can prove the opposite side of this question but i have no idea for another side. any help will be appreciated.

VFV
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    Can't you construct a $B$ pretty easily if you assume $A$ is made up of blocks like $\begin{bmatrix} 0 & 1 \ & 0 & 1 \ &&\ddots&\ddots \ &&&\ddots &1 \ & & & & 0\end{bmatrix}$? – Ted Shifrin Nov 10 '17 at 18:19
  • A proof of one of the implications is provided here. – José Carlos Santos Nov 10 '17 at 18:21
  • $P^{-1}APP^{-1}BP-P^{-1}BPP^{-1}AP= P^{-1}AP^{-1}$ by using this idea from your comment and by $L(X)=AX-XA$ that is a linear transformation and calculating $L(e_{ij})=A_i - A'_j$ and using this fact in i'th column and row we have atmost one $1$. i think it's pretty trivial.$A_i and A'_j is i'th column and j'th row. am i right Ted Shifrin? – VFV Nov 10 '17 at 18:45
  • by the way thank you for help,Ted Shifrin. – VFV Nov 10 '17 at 18:48

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