According to PlanetMath Schur's lemma is this:
Let $G$ be a finite group and let $V$ and $W$ be irreducible $G$-modules. Then, every $G$-module homomorphism $\,f: V \to W$ is either invertible or the trivial zero map.
Let $V$ be a finite-dimensional, irreducible $G$-module taken over an algebraically closed field. Then, every $G$-module homomorphism $\,f: V \to V$ is equal to a scalar multiplication.
Since the ground field is algebraically closed, the linear transformation $f: V\to V$ has an eigenvalue; call it $\lambda$. By definition, $\,f - \lambda 1$ is not invertible, and hence equal to zero by Schur's lemma. In other words, $\,f = \lambda$, a scalar.
In the proof above, why can Schur's lemma be applied to $\,f - \lambda 1$? Is it assumed that $\,f - \lambda 1$ is also a $G$-module homomorphism?