I cannot find any categorical definition of an eigenvalue, so I ask this question. Let $\mathbb{k}$ a be a field and $\mathcal{C}$ be a $\mathbb{k}$-linear abelian category. Let $f: X \rightarrow X \in \mathrm{End}\_\mathcal{C}(X)$. To me, it makes sense to call $\lambda \in \mathbb{k}$ an eigenvalue of $f$ if $\ker(f - \lambda 1\_X)$ is nonzero (and call this the corresponding eigenspace). By considering pullbacks one can show that these kernels do not "intersect" either for different $\lambda$.


If this indeed is the accepted definition, what are some reasonable set of conditions so that any such $f$ always has an eigenvalue (for instance, algebraic closedness of $\mathbb{k}$ will probably be necessary and some finiteness assumption)?


The greater context for such a question is from trying to prove categorical Schur's lemma for a tensor category, where any endomorphism of a simple object is a scalar multiple of the identity. And a similar statement about an endomorphism of an indecomposable being of the form $\lambda 1\_X + n$, where $n$ is nilpotent.



Existence of eigenvalues in a k-linear abelian category
Schur's lemma has the same proof in a $k$-linear abelian category $C$ as usual: if $T : M \to M$ is a nonzero endomorphism of a simple object, by simplicity it must have trivial kernel and cokernel, so is an isomorphism. Hence $\text{End}(M)$ is a division algebra over $k$. If furthermore $k$ is algebraically closed and $\text{End}(M)$ is finite-dimensional (e.g. if $C$ has finite-dimensional homsets) then $\text{End}(M) = k$.


Similarly if $k$ is algebraically closed and $\text{End}(M)$ is finite-dimensional then every endomorphism $T : M \to M$ has at least one eigenvalue (if $M$ is nonzero), because the natural map


$$k[x] \ni f(x) \mapsto f(T) \in \text{End}(M)$$


has nontrivial kernel (generated by the minimal polynomial of $T$). Working a little more carefully to check that all the details still work as usual without elements: if $m(t) = \prod (t - \lambda\_i)^{m\_i}$ is the minimal polynomial of $T$, then $m(T) = 0$ implies that (if $M \neq 0$) at least one of the factors $(T - \lambda\_i)^{m\_i}$ is not a monomorphism, hence has nontrivial kernel.


As for the indecomposable case, with the same hypotheses as above $M$ is naturally a module over $k[x]/m(x) \cong \prod k[x]/(x - \lambda\_i)^{m\_i}$. The primitive idempotents of this product split $M$ into the direct sum of generalized eigenspaces of $T$ (this is a general feature of idempotent endomorphisms in abelian categories and also does not require elements), so if $M$ is indecomposable then $T$ has exactly one eigenvalue $\lambda$ and $T - \lambda$ is nilpotent as usual.


