Throughout is a finite group and is a representation of

We defined an irreducible representation in here. The definition can also be given in terms of matrices.

**Theorem 1**. Let Then is reducible if and only if there exist a -basis of and an integer such that for all

where here means an zero matrix. The matrices and depend on

*Proof*. Well, by definition, is reducible if and only if is not simple as a – module. So is reducible if and only if has a non-zero -submodule Since is a semisimple ring (see the corollary in this post), is a semisimple -module. Thus for some -submodule of Now, choose a -basis for and extend it to a basis for Note that because Let Since is a -module, we have for all So is a -linear combinations It is now clear that the matrix looks like the matrix given in the theorem. Conversely, if for some basis of the matrix looks like the one given in the theorem, then we let It is clear from the form of that for all and hence is a -submodule of Thus cannot be irreducible.

**Theorem 2**. There exists a basis of and square matrices such that for every

*Proof*. Since is a semisimple -module, for some integer and some simple -submodules of In fact, from the theory of semisimple modules, it is clear that are all simple -submodules of Now, choose a – basis for each and let Let Obviously is a basis for and is the matrix given in the theorem.