# Existence of eigenvalues in a k-linear abelian category

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.

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.