Reducible modules and representations

Question

I've started to study groups representations theory with the book of the same title, by Larry Dornhoff (part A). In the introduction chapter, there's an observation regarding matrices that I'm not able to understand (see picture below). I'll appreciate if anyone can explain to me how to interpret those lines, and what is the reason this happen. Thanks in advance

Answer 1

Since we are asked to consider an $RG$-module, any module subspace must be closed with respect to multiplication induced by the $RG$-linear map.

There is another way to speak of this without reference to the group ring: We can consider the natural $R$-module structure, and also $\cdot_{G}: G \times M \to M$, which is "multiplication" by $G$ that behaves well with respect to our other operations. This is essentially saying that we have an $R$-module with one additional operation, and that any submodule is closed with respect to all three operations ($+,\cdot,\cdot_{G})$.

Thankfully, we can cast the main point in yet a different language: a module can be viewed as an abelian group equipped with a ring homomorphism $\phi: R \to \mathrm{End(M})$, into the ring of endomorphisms. Similarly, we can view our group representation as an $R$-module equipped with a homomorphism of $G$ into $GL(V)$, if $V$ is free.

Hence, we can consider linear transformations induced by the action of $G$, denoted $T_g$ for each $g \in G$ (indeed every element in the image of this homomorphism can be viewed as an invertible linear transformation.)

Hence, an $RG$-Submodule is the same thing as asking a subspace of just an $R$-module to be invariant under $T_g$ for each $g \in G$. So, letting $U \subset M$ be a regular ole' submodule, we just need that $T_g(U) \subset U$ for every $g \in G$.

Once we have an invariant subspace, since $V$ was free, we can specify a basis, $u_1,...,u_m$ and rewrite our linear transformations with respect to the basis. However, since $T_g(u_1),...T_g(u_m) \subseteq U$ we know that the matrix representation for $T_g$ is going to have a giant zero under the first $m$ rows, since nothing gets sent "out" of the space. Extending the basis to all of $V$, we get the rest of the matrix, which can in general be a little messier.

If We consider the complement $U^{\perp}$ for an invariant subspace $U$, and it is also invariant, we get a giant $0$ in the top right of our matrix, since the image of any $v \in U^{\perp}$ has a zero coefficient for all the basis elements in $U$. In this case, we say that $T_g$ is a completely reducible representation.

Answer 2

What that is saying is that when a representation is reducible and completely reducible, we may think of those subspaces as invariant under the group action. Hence in the first case we know that the matrix must be of the block form as shown since if the bottom left matrix was not zero, then in that basis we would have the group element not holding the subspace invariant. It is similar for the other matrix.