Upon studying the Jordan normal form I came across the problem of determining the Jordan normal form of powers of a single base matrix and in that context I was wondering what happens to the eigenspaces. I am not yet entirely sure if the statement below is correct which is why I tried to prove it myself and would now like to know if the below proof is actually correct. Also, since I have not written many proofs yet as of now, please let me know if I have done any notational mistakes or if anything could be stated in a more elegant way.
Show that given any square-matrix $A\in \mathbb{K}^{n\times n}, \mathbb{K}$ is a field, $k \in \mathbb{N}$ the following statement holds true: $E_{\mu}(A^k) = \oplus_i E_{\lambda_i}(A)$ where $\lambda_i$ is an eigenvalue of $A$ such that ${\lambda_i}^k = \mu$.
"$\supseteq$": Let $\lambda_i$ be an eigenvalue of $A$ such that ${\lambda_i}^k = \mu$ and $v \in E_{\lambda_i}(A)$. Then: $Av = \lambda_iv$. $A^kv = A^{k-1}Av = \lambda_iA^{k-1}v = \ldots = {\lambda_i}^kv \implies v \in E_{{\lambda_i}^k}(A^k) = E_{\mu}(A^k)$.
"$\subseteq$": Assume $E_\mu(A^k) \supset \oplus_iE_{\lambda_i}(A)$. Then $\exists v \in E_\mu(A^k) \setminus \oplus_iE_{\lambda_i}(A)$. $A^kv = {\mu}v \implies A^kv^k = A^kvv^{k-1} = {\mu}vv^{k-1} = {\mu}v^k$ therefore $v^k$ is an eigenvector of $A^k$ and $\mu$. Note that $Av \neq \lambda_iv$ for any eigenvalue $\lambda_i$ of $A$. Thus: $(Av)^k \neq (\lambda_iv)^k \Leftrightarrow A^kv^k \neq \lambda_i^kv^k = \mu v^k$ which means that $v^k$ cannot be an eigenvector of $A^k$ and $\mu$ which is a contradiction. Such a vector $v$ can therefore not exist, implying that $E_\mu(A^k) \subseteq \oplus_iE_{\lambda_i}(A)$.