# Regarding Schur's lemma, that $T = \lambda I$, the uniqueness of $\lambda$.

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

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We don't know that $\lambda$ is the only eigenvalue until we complete the proof, and then we do. The statement that $T$ has only one eigenvalue is already fairly close to Schur's lemma so I don't see the problem here. – Qiaochu Yuan May 1 '12 at 16:14
@QiaochuYuan right, I am just wondering if there is a way to see this before the proof... – user19447 May 1 '12 at 16:27
The proof is so short I can't imagine a reasonable way of proving that $T$ has only one eigenvalue which doesn't end up proving Schur's lemma. – Qiaochu Yuan May 1 '12 at 16:28
alright. That seems fair! Thank you for the help. – user19447 May 1 '12 at 16:28

Okay, so I lied slightly. There's a fairly natural way to see that $T$ has only one eigenvalue which doesn't prove Schur's lemma, but only because it doesn't use the full strength of the hypothesis: we only assume that $V$ is indecomposable (is not the direct sum of two non-trivial submodules) rather than irreducible.
Generally, let $R$ be a $k$-algebra where $k$ is an algebraically closed field and let $M$ be an indecomposable left module over $R$. (To get a group representation from this setup, let $R$ be a group algebra $k[G]$). Then $M$ is in particular a $k$-vector space. If $T : M \to M$ is a module endomorphism, then we know (e.g. by the theory of Jordan normal form) that $M$ breaks up into a direct sum of the generalized eigenspaces of $T$, and since every operator in $R$ commutes with $T$, this exhibits $M$ as a direct sum of submodules. By assumption $M$ is indecomposable, so there must only be one such nontrivial submodule, hence $T$ can have only one eigenvalue.
Since every irreducible module is indecomposable, the conclusion follows. Note that by Maschke's theorem, every indecomposable module is also irreducible when considering finite groups $G$ over a field of characteristic not dividing $|G|$, but in general there are indecomposable modules which are not irreducible. For example, $\mathbb{Z}$ acts on $\mathbb{C}^2$ by $$n \mapsto \left[ \begin{array}{cc} 1 & n \\\ 0 & 1 \end{array} \right]$$
and this representation is indecomposable but not irreducible (exercise). $1$ acts by a matrix which is an intertwining operator that has one eigenvalue but is not scalar.