An endomorphism $f$ of a vector space is nilpotent; show that $\mathit{Id} - f$ is invertible

Question

An endomorphism $f$ of a vector space is nilpotent; show that $\mathit{Id} - f$ is invertible and give how is it written.

I’ve found this method:

$\mathit{Id} - f^n = \mathit{Id} = (\mathit{Id} - f)(Id + f + \dots + f^{n-1})$

but I'm looking for something more intuitive.

Answer 1

That's a pretty good intuition.

Since $f$ is nilpotent, there is an $n$ such that $f^n=0$, hence $\ \mathrm{Id}+f+f^2+\dots+f^{n-1}\ $ will be the inverse of $\ \mathrm{Id}-f\ $ by your formula.