Eigenspaces in Jordan Chevalley Decomposition If $V$ be a finite dimensional vector space over an algebrically closed field $F$ and $x\in\operatorname{End}(V)$, then by Jordan Chevalley Decomposition we have $x=x_s+x_n$ where $x_s$ and $x_n$ are semisimple and nilpotent parts of $x$ respectively, $x_n$ and $x_s$ both being polynomials in $x$ without constant term. I want to show that if $a$ is an eigenvalue of $x$, then the generalized eigenspace of $x$ corresponding to $a$ is exactly the same as the eigenspace of $x_s$ corresponding to $a$. I have managed to show that the eigenspace of $x_s$ corresponding to $a$ is 
contained in the generalized eigenspace of $x$ corresponding to $a$. But I am unable to prove the converse.

You can see this from an explicit construction of $x_s$. Write the characteristic polynomial of $x$ as $\prod_{i=1}^k (T - \lambda_i)^{m_i}$ where $\lambda_i \neq \lambda_j$ for $i \neq j$. The generalized eigenspaces of $x$ are $V_i = \ker((x - \lambda_i \cdot \mathrm{id})^{m_i})$. Let $p \in F[T]$ be a polynomial satisfying the congruence relations
$$ p(T) \equiv \lambda_i \mod (T - \lambda_i)^{m_i} \,\,\, \forall 1 \leq i \leq k. $$
Then $x_s = p(x)$ and $x_n = x - p(x)$. Since $x_s$ is a polynomial in $x$ and $V_i$ is $x$-invariant, we see that $V_i$ is also $x_s$-invariant. The congruence relations show that $x_s|_{V_i}$ acts on $V_i$ as a scalar operator multiplying each vector by $\lambda_i$. Thus, each $V_i$ consists of regular eigenvectors of $x_s$ corresponding to the eigenvalue $\lambda_i$. Since $V = \bigoplus_{i=1}^k V_i$, we see that the generalized eigenspaces of $x_s$ are the $V_i$'s and are in fact regular eigenspaces.
If you are not familiar with this explicit construction, then this shows that $x_s$ (defined as $p(x)$) is semisimple. Each $V_i$ is also $x_n = x - p(x)$ invariant and the congruence relations show that $x_n|_{V_i}$ is nilpotent and thus $x_n$ is nilpotent. Since $x_s$ and $x_n$ are polynomials in $x$, they commute and so by the uniqueness of the Jordan-Chevalley decomposition, this shows that the $x_s$ described above is the $x_s$ guaranteed by the Jordan-Chevalley decomposition.
