I know that a nilpotent operator with index of nilpotency equal to dimension of the vectorspace, then, has Jordan Canonical form. Does any nilpotent operator whose index of nilpotency is less than dimension of the vectorspace ,has Jordan Canonical form ?

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    $\begingroup$ Yes, the characteristic polynomial of every nilpotent operator splits over the field (it is infact equal to $(-1)^{n} t^n$ where $n=\dim V$), and this is true regardless of what the index of nilpotency is. Therefore, it has a Jordan canonical form $\endgroup$ – peek-a-boo Sep 6 at 15:40

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