Proof of a simplified version of Schur's Lemma

I have given the Schur’s Lemma in following version:

Let $R:G rightarrow text{U}(mathcal{H})$ be an irreducible representation of $G$ on $mathcal{H}$. If $A in text{L}(mathcal{H})$ satisfies

$$A R(g) = R(g) A quad forall g in G$$

then $A = c I$ for some $c in mathbb{C}$.

Here $mathcal{H}$ states a finite dimensional Hilbert space and $text{U}(mathcal{H})$ denotes the subspace of unitary operators. $R$ describes a homomorphism of a group $G$ on $text{U}(mathcal{H})$.

The proof I have given shows that it is sufficient to only prove this for hermitian $A$. So far so good. It continues with introducing a eigenvector $left| psi rightrangle$ of $A$ so that the eigenspace of this operator is given by $text{Eig}_lambda(A) = {left| psi rightrangle: Aleft| psi rightrangle = lambda left| psi rightrangle}$. Then it states that $R(g) left| psi rightrangle in text{Eig}_lambda(A)$, because of $AR(g) left| psi rightrangle = R(g) A left| psi rightrangle = lambda R(g) left| psi rightrangle$. From that it is concluded that $text{Eig}_lambda(A)$ is an invariant subspace and because of $R$ being irreducible $text{Eig}_lambda(A)=mathcal{H}$ follows.

My problem with this proof is now that I don’t have a clue why you can conclude that $text{Eig}_lambda(A)$ is an invariant subspace of $R$ only because of the statement $A R(g)left| psi rightrangle= lambda R(g) left| psi rightrangle$.