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Let for the linear operator $A:E\to E$, $A^k=0$, then $A^n=0$.

Here $n=dim E$ and $n <k.$

I tried to find the characteristic oolynomial that I think it is the clue, however did nit succeed.

Any suggestion?

Majid
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    If $A^k=0$, the minimal polynomial of $A$ is equal to some $X^m$ and it shares the same roots as the characteristic polynomial, so the characteristic polynomial is $X^n$. – Gabriel Romon May 22 '19 at 19:59
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    Do you know the Cayley-Hamilton theorem? – SdV May 22 '19 at 20:00
  • This could help: https://math.stackexchange.com/questions/108422/how-to-show-that-the-nth-power-of-a-n-times-n-nilpotent-matrix-equals-to-zero. – Minus One-Twelfth May 22 '19 at 20:02
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    I know Cayley -Hamilton. Wrt the first comment the roots are zero and so it is solved! yes? – Majid May 22 '19 at 20:04

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