Generalised eigenspace of inverse $\newcommand{\id}{\operatorname{id}}$Let $V$ be a finite-dimensional vector space and $\Phi \in \operatorname{End}(V)$ be invertible. Let $\lambda \in \mathbb{F}, \lambda \neq 0$.

$\ker (\Phi - \lambda\id)^{\dim V} = \ker (\Phi^{-1} - \lambda^{-1}\id)^{\dim V}$

How to prove this? My attempt:
Let $n = \dim V$. Suppose that $\Phi^n v \in \ker(\Phi - \lambda\id)^n$. Expand
$$(\Phi^{-1} - \lambda^{-1}\id)^n = \sum_{k=0}^n \binom{n}{k} \Phi^{-k} \lambda^{k - n}, \qquad(\Phi - \lambda\id)^n = \sum_{k=0}^n \binom{n}{k} \Phi^{n - k} \lambda^{k}
$$
Apply:
$$(\Phi^{-1} - \lambda^{-1}\id)^n\Phi^n v = \sum_{k=0}^n \binom{n}{k} \Phi^{n-k} \lambda^{k - n} v = \lambda^{-n} \sum_{k=0}^n \binom{n}{k} \Phi^{n-k} \lambda^{k} v = \lambda^{-n} (\Phi - \lambda\id)^n v
$$

You forgot a few signs.
$$(A-\lambda I)^n = \sum_{k=0}^n \binom{n}{k} (-\lambda)^k A^{n-k}$$
$$(A^{-1} - \lambda^{-1} I)^n = \sum_{k=0}^n \binom{n}{k} (-\lambda)^{k-n} A^{-k}$$
But actually the expansion is unnecessary. You can show the following directly.
$$(A^{-1} - \lambda^{-1} I)^n v = (-\lambda A)^{-n} (A-\lambda I)^n v.$$
Since $(-\lambda A)^{-n}$ is invertible, the two kernels are the same.
