generalized eigenspace is the same as the eigenspace of the semi-simple part.

Question

Let $f: V\rightarrow V$ be an endomorphism of complex vector spaces and $ f=f_S+f_N$ it's decomposition into semi-simple and nilpotent parts. Then apparently we have that the generalized eigenspace $V^{\lambda}(f):=\{v\in V|(f-\lambda\cdot id_V)^kv=0 \text{ for some } k\in \mathbb{N}\}$ is the same as the eigenspace $V_{\lambda}(f_S)$ of the semisimple part. I was able to proof $V_{\lambda}(f_S)\subset V^{\lambda}(f)$ but for the converse i am unsure. what I tried was the following: As $f_S$ and $f_N$ commute, we may write $$(f_N+f_S-\lambda\cdot id_V)^n=\sum_{k=0}^nf_N^k (f_S-\lambda \cdot id_V)^{n-k}$$ Now applying this to some $v$ in the generalized eigenspaces with the respective $n$, I have to conclude that $v$ is in the eigenspace of $f_S$. Can anyone help me?

Answer 1

If $v \in V^{\lambda}(f)$ and write it as $v = \sum_{\mu} u_{\mu}$ where $u_{\mu}\in V_{\mu}\left(f_S\right)$.

\begin{align} 0 = \left(f - \lambda \mathbf{id}\right)^{n} v &= \sum_{\mu} \left(f - \lambda \mathbf{id}\right)^n u_{\mu}\\ &= \sum_{\mu}\sum_{k=0}^{n} f_N^k\left(f_S- \lambda\mathbf{id}\right)^{n-k}u_{\mu}\\ &= \sum_{\mu} \sum_{k=0}^{n} \left(\mu - \lambda\right)^{n-k}f_N^ku_{\mu} \end{align}

Since $f_N^ku_{\mu}\in V_{\mu}\left(f_S\right)$ then for all $\mu$,

$$\sum_{k=0}^{n} \left(\mu -\lambda\right)^{n-k} f_N^k u_{\mu} = 0$$

For $\mu \neq \lambda$, let $p = \min \left\{k\ge 0\, :\, f_N^ku_{\mu} = 0\right\}$ (it exists since $f_N$ is nilpotent). If $p \ge 1$, then $f_N^{p-1}u_{\mu} \neq 0$, multiply the previous equation by $f_N^{p-1}$ and you will have:

$$\left(\mu - \lambda\right)^{n}f^{p-1}_{N} u_{\mu} = 0$$ so $f_N^{p-1}u_{\mu} = 0$. This proves that $p = 0$ and $u_{\mu} = f_N^pu_{\mu} = 0$. Finally, $v = u_{\lambda} \in V_{\lambda}(f_S)$.