Etingof proposition 3.1.4 I am reading Etingof et al's Introduction to Representation Theory. 

First they prove that any irreducible subrepresentation is isomorphic to one of the $V_i$s by Schur's lemma. Further the inclusion of this irreducible after identification with $V_i$ will look like:
 
I do not quite understand what they do after this: 
$G_i$ acts on $n_iV_i$ and therefore on all of $V$. This I understand. But why does it act on the set of subrepresentations of $V$? In the proof what are the matrices $X_i$? If someone could elaborate on these first then I will be able to give another go at the proof. 
I can see that what we want to say is subrepresentation of semisimple is semisimple. I can prove it in other ways. And the matrices $X_i$ can be found by using Schur's lemma and 'Hom and direct sum commutes' property. But I can't follow the argument of Etingof. 
 A: 
The group $G_i=GL_{n_i}(k)$ acts on the set of subrepresentations of $V$.

This implies that for a subrepresentation $W$ of $V$ then $Wg_i$ is also
a subrepresentation of $V$. 
Proof. The only nontrivial part is to check that each element $g_i$ in $G_i$ sends a subrepresentation $W$ of $V$ to another subrepresentation $Wg_i$ of $V$. 
Indeed, fix $a\in A$ then $aw=w'\in W$ for any $w\in W$. We show
$a(wg_i)=w'g_i$. 
We first seperate $w,w'$ into $w=(w_1,w_2), w'=(w_1',w_2')$
where $w_1,w_1'\in \bigoplus_{j\ne i} n_jV_j$ and 
$w_2,w_2'\in n_iV_i$. Note $wg_i=(w_1,w_2g_i)$ so 
$a(wg_i)=w'g_i$ is equivalent to $a(w_1,w_2g_i)=(w_1',w_2'g_i)$.
Since $\bigoplus_{j\ne i} n_jV_j$ and $n_iV_i$ are subrepresentations of 
$V$ so their direct sum is a representation with action defined as
$a(w_1,w_2g_i)=(aw_1,a(w_2g_i))$. From this, we find $a(wg_i)=w'g_i$
is equivalent to $aw_1=w_1',a(w_2)g_i=w_2'g_i$. This holds since $aw=w'$
or $(aw_1,aw_2)=(w_1',w_2')$.
The check that this is indeed a group action is not hard. 

Etingof's proof. 


*

*First show $W$ has an irreducible subrepresentation $P$ isomorphic 
to $V_i$ via identification $v\mapsto (vq_1,\ldots, vq_{n_i})$
for $v\in V_i,q_j\in k$.

*Choose $g_i\in G_i$ so $(q_1,\ldots, q_{n_i})g_i=(1,0,\ldots,0)$
then  $Wg_i$ has irreducible subrepresentation $Pg_i$, which is 
the first summand $V_i$ of $nV_i$. Take the projection $Wg_i\to V_i$
then with $W'=\ker(Wg_i\to V_i)$, we have $Wg_i=V_i\oplus W'$. 
Furthermore, $W'\subset n_1V_1 \oplus \cdots \oplus (n_i-1)V_i
\oplus n_mV_m$ so by inductive hypothesis, $W'$ is semisimple
with inclusion map as defined. From this, we find 
$Wg_i$ is semisimple and has the inclusion map $Wg_i\to V$ 
as described.

*Coming back to the action of $G_i$ on set of subrepresentations of 
$V$, this action preserves the property that if subrepresentation 
$W$ of $V$ is semisimple has inclusion map as described then $Wg_i$
is also semisimple and the inclusion
map $Wg_i$ also has that property (but under different choice
of matrix $X_i$ of course). 

*So now we know $Wg_i$ is semisimple and has inclusion map as described.
From previous argument, $W=(Wg_i)g_i^{-1}$ is semisimple and has inclusion 
map as described. 


So in Etingof's proof, he didn't specifically describe the matrix
$X_i$. He only mentioned that the inclusion $W\to V$ is described 
by some matrices' $X_i$'s. 
