Question regarding the definition of direct sum decomposition of a representation Please bear with me. I am trying to learn representation theory of finite groups from J.P. Serre's book by myself.
Here, the author has used the word 'representation' for the homomorphism $\rho : G\rightarrow GL(V)$, as well as the vector space $V$ interchangeably. But I am a little confused about the definition of direct sum of subrepresentations.
Please correct me if I have understood it right : A $subrepresentation$ of $V$ is a subspace $W\subseteq V$ such that $\rho_g (W) \subseteq W $ $\forall g\epsilon G$,and $V$ is said to be the $direct$ $ sum$ of two subrepresentations $W,W'$ if $V=W\oplus W'$ as '$a$ $vector$ $space$', where the direct sum is $internal$.
What I am confused about is what it means for a representation to be the direct sum of two subrepresentations, where by 'representation' we mean the 'homomorphism' $\rho$ itself, i.e. the exact formulation of $\rho$ in terms of $\rho_{|W}$ and $\rho_{|W'}$. I understand that if a proper basis is chosen, then the matrix of $\rho_g$ can be written as$$[\rho_g] = \begin{pmatrix} [\rho_{g|W}] & 0\\ 0&[\rho_{g|W'}]\end{pmatrix}$$
But I want a description of $\rho_g$ in terms of $\rho_{g|W}$ and $\rho_{g|W'}$ which does not depend on the choice of a basis, e.g. if $T_1: V_1 \rightarrow W_1, T_2:V_2 \rightarrow W_2$ are two linear transformations of vector spaces, we can define the direct sum $$T_1\oplus T_2 : V_1\oplus V_2 \rightarrow W_1\oplus W_2 $$ in a canonical way, when the direct sum is understood to be $external$, without reference to a basis. I guess I am confused about the $internal$ direct sum version of it. 
 A: To make the distinction between internal and extenal direct sum more clear, let me write external ones as $S\times T$ (in fact the underlying set of the external direct sum is the cartesian product set). Then elements of $S\times T$ are pairs $(s,t)$ with $s\in S, t\in T$, and
$$
  T_1\times T_2 : V_1\times V_2 \rightarrow W_1\times W_2
\qquad\text{maps}\quad
  (v_1,v_2)\mapsto (T_1(v_1),T_2(v_2))
$$
for $(v_1,v_2)\in  V_1\times V_2$.
Saying $V=U_1\oplus U_2$ is an internal direct sum of subspaces $U_1,U_2\subseteq V$ just means that the map $\alpha_{U_1,U_2}:U_1\times U_2\to V: (u_1,u_2)\mapsto u_1+u_2$ (which is defined because $u_1,u_2\in V$) is an isomorphism of vector spaces (surjectivity means that $U_1+U_2=V$, injectivity that $U_1\cap U_2=\{0\}$).
Now suppose you have a representation of $G$ on $V$, and $V$ can be written as the direct sum of two invariant subspaces $U_1,U_2$. Then by restriction one has representations $\rho_1,\rho_2$ of $G$ on $U_1$ respectively on $U_2$, called subrepresentations of $G$. The original representation can be recovered from just the subrepresentations by using the composite isomorphism
$$
  V \overset{\alpha_{U,U'}^{-1}}\longrightarrow U\times U'
  \overset{\rho_1\times\rho_2}\longrightarrow U\times U'
  \overset{\alpha_{U,U'}}\longrightarrow V.
$$
Explicitly, to find the image by $\rho(g)$ of $v\in V$, write $v=u_1+u_2$ in the unique possible way with $u_i\in U_i$ for $i=1,2$ (this is applying $\alpha_{U,U'}^{-1}$), then form $\rho_i(g)(u_i)\in U_i$ for $i=1,2$ (this is applying $\rho_1(g)\times\rho_2(g)$) and finally add the results as elements of $V$ (this is applying $\alpha_{U,U'}$). All in all
$$
  \rho(g): v=u_1+u_2 \mapsto \rho_1(g)(u_1)+\rho_2(g)(u_2)
\qquad\text{where $u_1\in U_2,u_2\in U_2$}.
$$
A: If $V = W_{1} \oplus W_{2}$ then for every $v \in V$, $v = u + w$ where $u \in W_{1}$ and $w \in W_{2}$. Hence
$$ \rho_{g}(v) = \rho_{g}(u + w) = \rho_{g}(u) + \rho_{g}(w) = \rho_{g}|_{W_{1}}(u) + \rho_{g}|_{W_{2}}(w) $$
Does it clarify your doubt?
