# Linear operators with zero eigenvalues increase the kernel

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?

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 $$0$$.
• 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. Aug 8, 2020 at 1:46
• 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 $1$. Aug 8, 2020 at 1:53
• 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. Aug 8, 2020 at 2:06
• $f(V)=f^2(V)$ so $f\vert_{f(V)}$ is invertible. Aug 8, 2020 at 2:16