I'm reading through the proof in Artin's "Algebra" of Schur's lemma (second statement): if $T:V \to V$ is a $G$-invariant linear operator with respect to $\rho$ an irreducible representation, then $T = \lambda I$ for some $\lambda \in \mathbb{C}$.
The proof is simple, we let $\lambda$ be an eigenvalue of $T$ and see that $S = T - \lambda I$ is both $G$-invariant and non-invertible, and therefore zero. Thus $T = \lambda I$.
My question is in regards to $\lambda$. Since $T = \lambda I$ one can see that $T$ has in fact only one eigenvalue, with multiplicity $\text{dim}V$. So when we say
"let $\lambda$ be an eigenvalue of $T$..."
we could have said,
"let $\lambda$ be the eigenvalue of $T$...".
If only we knew that $\lambda$ was the only option. Or what? How could one find out that there is only one eigenvalue for $T$ without using this proof?
Thank you