# 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 $$\left| \psi \right\rangle$$ of $$A$$ so that the eigenspace of this operator is given by $$\text{Eig}_\lambda(A) = \{\left| \psi \right\rangle: A\left| \psi \right\rangle = \lambda \left| \psi \right\rangle\}$$. Then it states that $$R(g) \left| \psi \right\rangle \in \text{Eig}_\lambda(A)$$, because of $$AR(g) \left| \psi \right\rangle = R(g) A \left| \psi \right\rangle = \lambda R(g) \left| \psi \right\rangle$$. 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 $$A R(g)\left| \psi \right\rangle= \lambda R(g) \left| \psi \right\rangle$$.

When you prove that $$A R(g)\left| \psi \right\rangle= \lambda R(g) \left| \psi \right\rangle$$ for $$\left| \psi \right\rangle$$ a $$\lambda$$-eigenvalue of $$A$$, you're proving that $$R(g)\left| \psi \right\rangle$$ is a $$\lambda$$-eigenvalue of $$A$$ for every $$g\in G$$. This directly implies that $$R(g)\left| \psi \right\rangle \in \mathrm{Eig}_\lambda(A)$$ $$\forall g\in G$$, $$\forall \left| \psi \right\rangle \in \mathrm{Eig}_\lambda(A)$$, which is what $$R$$-invariance means for $$\mathrm{Eig}_\lambda(A)$$.