# Schur's lemma

This fact is related to: linear representation theory
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## Statement

Here, $G$ is a group and $k$ is a field. In the abstract formulation, $V_1,V_2$ are vector spaces over $k$.

Statement no. Condition Conclusion in abstract formulation for vector spaces: $\rho_1: G \to GL(V_1), \rho_2: G \to GL(V_2)$ are linear representations of $G$ over a field $k$. $f:V_1 \to V_2$ is a homomorphism from $\rho_1$ to $\rho_2$. Conclusion in concrete matrix formulation: $\rho_1:G \to GL(m,k)$, $\rho_2:G \to GL(n,k)$ are representations. $F \in \operatorname{Mat}_{n \times m}$ is a homomorphism of representations
1 $\rho_1$ is irreducible $f$ is either the zero map or is injective. $F$ is either the zero matrix or has full column rank, and $n \ge m$.
2 $\rho_2$ is irreducible $f$ is either the zero map or is surjective $F$ is either the zero matrix or has full row rank, and $m \ge n$.
3 $\rho_1, \rho_2$ are both irreducible. $f$ is either the zero map or is bijective and hence an isomorphism of representations. $F$ is either the zero matrix or is invertible, and $m = n$.
4 $\rho_1 = \rho_2$, it is irreducible, and $k$ is an algebraically closed field $f$ is a scalar multiplication map. $F$ is a scalar matrix and $m = n$.

## Proof

### Verbal proof

1. For this, we use the fact that the kernel of any homomorphism of representations is an invariant subspace. If $\rho_1$ is irreducible, the kernel is either the whole of $V_1$ (in which case we have the zero map) or the zero subspace (in which case we have an injective map).
2. For this, we use the fact that the image of any homomorphism of representations is an invariant subspace. If $\rho_2$ is irreducible, the image is either the whole of $V_2$ (in which case we have a surjective map) or the zero subspace (in which case we have the zero map).
3. This follows from the previous two parts.
4. Any homomorphism from a representation on that field to itself can be represented by a linear operator from the vector space to itself. This linear operator must have an eigenvalue, because the field is algebraically closed. Subtracting that eigenvalue times the identity matrix, we now get a linear operator that has zero as an eigenvalue, which is also a homomorphism of representations.

But part (3) tells us that any homomorphism of representations is either the zero map or an isomorphism. An isomorphism cannot have zero as an eigenvalue (since it is an invertible linear operator) and hence the new linear operator we get must be the zero map.

Thus the original linear operator must have been a scalar times the identity.