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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 $A R(g)\left| \psi \right\rangle= \lambda R(g) \left| \psi \right\rangle$.

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It's possible you're overthinking this -- the step is not too complicated.

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 $\mathrm{Eig}_\lambda(A)$.

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  • $\begingroup$ Thank you for your fast response. My thinking about this was much to complicated.. $\endgroup$ – enco909 Nov 6 '20 at 17:07

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