Remark

Suppose there exists an integer $n\geqslant 0$ such that Ker($u^n$) $=$ Ker($u^{n+1}$) (resp. Im($u^n$) $=$ Im($u^{n+1}$)); one then has Ker($u^m$) $=$ Ker($u^n$) (resp. Im($u^m$) $=$ Im($u^n$)) for every integer $m\geqslant n$.

Suppose $u$ of index zero. For every integer $n\geqslant 0$, one has

$n$ ind($u$) $=$ ind($u^n$) $=$ codim$_E$(Im($u^n$))$-$ dim(Ker($u^n$)).

The conditions Ker($u^n$) $=$ Ker($u^{n+1}$) and Im($u^n$) $=$ Im($u^{n+1}$) are therefore equivalent. When they are satisfied, the vector space E is the direct sum of Ker($u^n$) and of Im($u^n$) (A, VIII, p. 25, prop. 2), the space Ker($u^n$) is finite-dimensional, and $u$ induces, by passing to the subspace, an automorphism of Im($u^n$).