Isomorphism between Hom-space and $k$

Let $$\mathrm{vect}$$ be the category of vector spaces over $$k$$ and let $$V \in \mathrm{vect}$$. Can one say $$\mathrm{Hom}(V, V)$$ (the hom-space in the category $$\mathrm{vect}$$) is isomorphic to $$k$$?

• Certainly not for all $V$... – Arnaud D. Feb 25 at 11:37

Say $$V$$ is finite dimensional so $$V\cong \oplus_{i=1}^n k_i$$ where $$k_i=k$$. Then
$$\mathrm{Hom}(V,V)\cong \mathrm{Hom}(\oplus_{i=1}^n k_i, \oplus_{j=1}^n k_j)=\oplus_{j=1}^n \mathrm{Hom}(\oplus_{i=1}^n k_i,k_j)=$$ $$\oplus_{i=1}^n \oplus_{j=1}^n \mathrm{Hom}(k_i,k_j) \cong \oplus_{i=1}^n \oplus_{j=1}^n\mathrm{Hom}(k,k)\cong \oplus_{i=1}^n \oplus_{j=1}^n k,$$
which is not $$k$$ in general, but finitely many copies of it. In the infinite dimensional case this is even further from being true. Your claim is of course true when $$\dim(V)=1$$.
• Thanks for you answer! In the answer to my question math.stackexchange.com/questions/3118727/…, the person answering uses $\mathrm{Hom}_{\mathcal{M}}(M, M) = k$ where $\mathcal{M}= vect$ the module category of vector spaces. Do you have an idea why it holds there? I remember my professor saying something similar, so I suppose that the claim itself is correct. – P. Schulze Feb 25 at 11:49
• I have in my case a module category $\mathcal{M}$ over a category $\mathcal{C}$ and $M$ is irreducible. Does irreducible mean that $M$ has dimension 1? – P. Schulze Feb 25 at 11:50