Tom Leinster's eventual kernel + eventual image decomposition of a linear map over an arbitrary field? Let $\mathsf{C}$ be a category. Then we have a subcategory $\mathrm{End}(\mathsf{C}) \hookrightarrow \mathsf{C}^\to$ of the arrow category which consists of endomorphisms of objects of $\mathsf{C}$. Equivalently, this subcategory can be described as the functor category $\mathsf{C}^{\mathbf{B}\mathbb{N}}$ where $\mathbf{B}\mathbb{N}$ is the natural numbers viewed as a one-object monoid. In the case where $\mathsf{C}=K\mathsf{Mod}$ for $K$ a commutative ring, then there is another equivalent description of this endomorphism category as $K[x]\mathsf{Mod}$. 
Working with an endomorphism $f: V \to V$ in $K\mathsf{Mod}$, we say that the eventual kernel of $f$ is
$$\mathrm{Ker}^\infty (f) = \bigcup_{i} \mathrm{Ker}(f^i)$$
and the eventual image of $f$ is
$$\mathrm{Im}^\infty (f) = \bigcap_{i} \mathrm{Im}(f^i).$$
In Tom Leinster's blog post, he states that given an endomorphism $f: V \to V$ of a finite-dimensional vector space over an algebraically closed field $\mathbb{K}$, we can decompose this endomorphism as $f=f|_{\mathrm{Ker}^\infty (f)} \oplus f|_{\mathrm{Im}^\infty (f)}$ where 
$$f|_{\mathrm{Ker}^\infty (f)}: \mathrm{Ker}^\infty (f) \to \mathrm{Ker}^\infty (f)$$ 
is nilpotent and 
$$f|_{\mathrm{Im}^\infty (f)}: \mathrm{Im}^\infty (f) \to \mathrm{Im}^\infty (f)$$ 
is invertible.
I am just wondering if this decomposition can still be carried out for a field which is not algebraically closed? I have tried to see if Tom Leinster has any papers containing the proof, to see which part depends on the algebraic closure, but I haven't been able to find anything other than the blog post.
 A: Yes, everything is fine over an arbitrary field. The assumption of algebraic closure is just so you can deduce this from the existence of Jordan normal form: the eventual kernel corresponds to the generalized eigenvectors with eigenvalue $0$ and the eventual image corresponds to the other ones. 
But over an arbitrary field $k$ you can just use generalized Jordan normal form or rational canonical form. Here is one fairly clean way to state the argument. An endomorphism $f : V \to V$ of a finite-dimensional vector space $k$ has a minimal polynomial $P(x) \in k[x]$ such that $P(f) = 0$. Write the irreducible factorization of $P$ over $k$ as
$$P(x) = \prod_i P_i(x)^{m_i}$$
where the $P_i$ are irreducible. Then $V$ naturally becomes a module over
$$k[x]/\prod_i P_i(x)^{m_i} \cong \prod_i k[x]/P_i(x)^{m_i}$$
where $x$ acts by $T$ and the isomorphism uses the Chinese remainder theorem. Now, it's a general fact that modules $M$ over a finite direct product $\prod_i R_i$ of rings canonically decompose as a finite direct product $\prod_i M_i$ where the action of $R$ on $M_i$ factors through the projection to $R_i$, and using this we get a canonical decomposition 
$$V \cong \bigoplus_i V_i$$
of $V$ into $T$-invariant subspaces where the minimal polynomial of $T$ on $V_i$ is $P_i(x)^{m_i}$. If $k$ is algebraically closed these are the generalized eigenspaces of $T$. Geometrically, $V$ is a quasicoherent sheaf over $\text{Spec } k[x]/P(x)$, which is disconnected, and we are looking at the restrictions to each connected component. 
In particular, if $P_i(x) = x$ for some $i$ then $V_i$ is the eventual kernel of $T$, and the direct sum of $V_j, j \neq i$ is the eventual image of $T$. 
