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Let $V$ be a $n$-dimensional vector space. In a proof that any linear operator $f$ on $V$ with only zero eigenvalues is nilpotent, the following reasoning is made:

For any linear operator $f$, $\ker f\subseteq \ker f^2$. However, since $f$ only has zero eigenvalues, $\ker{f}\neq \ker{f^2}$ unless $\ker{f}=V$. Therefore the following chain $$ \ker f\subset \ker f^2 \subset \ker f^3 \subset \ldots $$ implies $\ker f^k=V$ for some $k\leq n$, so that $f$ is nilpotent.

I do not see why for any $f$ with only zero eigenvalues $\ker{f}\neq \ker{f^2}$ unless $\ker{f}=V$. Is it so clear?

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One way: Suppose $\ker f=\ker f^2$. Apply rank-nullity to get $f(V)=f^2(V)$, and so if $f(V)\neq 0$ we would get a nonzero eigenvalue for $f\vert_{f(V)}$. But each eigenvalue of $f\vert_{f(V)}$ must be an eigenvalue of $f$, so $f(V)$ must be $0$.

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  • $\begingroup$ thanks for your answer! Do you mean that if $f(V)=f^2(V)$ and $f(V)\neq0$, $f|_{f(V)}$ would have 1 as an eigenvalue? I do not see this. $\endgroup$ Aug 8, 2020 at 1:46
  • $\begingroup$ Not necessarily 1 at that point. e.g., $f=\begin{pmatrix}0\\&2\end{pmatrix}$ has $\ker f=\ker f^2=\operatorname{span}\{e_1\}$, $f(\mathbb{R}^2)=f^2(\mathbb{R}^2)=\operatorname{span}\{e_2\}$ and doesn't have eigenvalue $1$. $\endgroup$ Aug 8, 2020 at 1:53
  • $\begingroup$ ok, but then why an eigenvalue of $f|_{f(V)}$ should be nonzero if $f(V)\neq0$? I mean, $f|_{f(V)}$ is an endomorphism on $f(V)\neq\{0\}$ which may have zero or nonzero eigenvalues. I think I'm missing some part of the argument. $\endgroup$ Aug 8, 2020 at 2:06
  • $\begingroup$ $f(V)=f^2(V)$ so $f\vert_{f(V)}$ is invertible. $\endgroup$ Aug 8, 2020 at 2:16
  • $\begingroup$ Ah, ok, you are applying rank-nullity again. $\endgroup$ Aug 8, 2020 at 2:22

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