How to determine $Hom (M,M)$ for an irreducible $R$-module $M$?

Question

More exactly, I'm considering $R$ to be a finite dimensional $\mathbb C$-algebra. For any $R$-module $M$, the $\mathbb C$ vector space $Hom_R(M,M)$ contains scalar multiplication and hence contains the whole of $\mathbb C$. What I want to show that is that if $M$ is irreducible as an $R$-module, then $Hom_R(M,M)=\mathbb C$. I've tried considering any such endomorphism as a map of $\mathbb C$-vector spaces. What next? I think I'll have to look at the eigenspace of any such map, but I don't know hoe to go from there. Can someone give a complete solution?

Answer 1

First note that an irreducible module is always cyclic, hence finite-dimensional if your underlying algebra is finite-dimensional. Now the trick, as you already guessed, is to consider a non-zero eigenspace of your endomorphism and show that it is in fact a submodule.