# Matrices commuting with a completely reducible representation

By Schur's lemma, a matrix commuting with an irreducible representation (of a group over complex numbers, say) is a multiple of identity. What about a direct sum of irreducible representations (a.k.a. completely reducible or semisimple representation)? The first thought might be that each irreducible subspace is invariant, but that is false. Consider a representation that maps all group elements to $$I$$ on $$\mathbb{R}^n$$. Any 1D subspace is irreducible, but it is not necessarily invariant because any matrix commutes with $$I$$.

My guess is that one has to collect all isomorphic subrepresentations into a single subspace in what is called isotypic decomposition, and then isotypic subspaces are invariant under a commuting matrix. Is that correct? Even then, what can a commuting matrix be on each isotypic subspace? Apparently, there are no restrictions when the type is trivial, but what about sums of isomorphic non-trivial irreducible representations. Must commuting matrices have some special form?

If there is a reference where this is laid out I'd appreciate that too. I am interested in algebras of matrices commuting with some 'natural' representations of finite groups (permutational ones, for example), which often have repeated irreducible components.

Basically, isotypic components are invariant under them, and their restrictions to each are spanned by intertwiners between the component's irreducible subspaces (under some choice of basis). For the trivial component, those are the matrices $$E_{ij}$$ with a single entry $$1$$ in $$ij$$ positon, and the rest $$0$$-s. And in general, the restrictions form an algebra isomorphic to the matrix algebra of the dimension equal to the component's multiplicity, the number of irreducible subspaces in it. From this, Savage then constructs a natural basis in the commutant (Theorem 1.2.10), and expresses it explicitly for permutation representations (Lemma 1.4.12).