Help understanding this statement in Serre's Schur's Lemma corollary Let $h$ be a linear mapping of $V_1 \rightarrow V_2$
\begin{gather*}
h^0 = \frac{1}{g} \sum_{t \in G}(\rho_t^2)^{-1}h\rho_t^1
\end{gather*}
(1) if $\rho^1$ and $\rho^2$ are not isomorphic, then $h^0 = 0$
(2) if $V_1 = V_2$ and $\rho^1 = \rho^2, h^0$ is a homothety of ratio $(1/n)$Tr$(h)$, $n = \dim(V_1)$
Looking at the matrix forms, we let $\rho_t^1 = (r_{i_1j_1}(t)), \rho_t^2 = (r_{i_2j_2}(t))$, and define $h = (x_{i_2i_1}), h^0 = (x_{i_2i_1}^0)$, so
\begin{gather*}
x_{i_2i_1}^0 = \frac{1}{g}\sum_{t,j_1,j_2}r_{i_2j_2}(t^{-1})x_{j_2j_1}r_{j_1i_1}(t)
\end{gather*}
Serre says "The right hand side is a linear form with respect to $x_{j_2j_1}$; in case (1) this form vanishes for all systems of values of the $x_{j_2j_1}$; thus it's coefficients are zero."
Okay, cool - I can follow you, Serre. Here is where I get confused. He then says, directly by case (1),
\begin{gather*}
\frac{1}{g}\sum_{t,j_1,j_2}r_{i_2j_2}(t^{-1})r_{j_1i_1}(t) = 0
\end{gather*}
Here, I don't see why we lose $x_{j_2j_1}$, especially if it's supposed to have coefficients that are zero.
Later on he says "equating the coefficients of $x_{j_2j_1}$ gives us
\begin{gather*}
\frac{1}{g}\sum_{t,j_1,j_2}r_{i_2j_2}(t^{-1})r_{j_1i_1}(t) ...
\end{gather*}
and, again, I don't see why we lost $x_{j_2j_1}$ - or specifically why it takes on value 1?
 A: Since case (1) says $\rho_1$ and $\rho_2$ are not isomorphic, the equality
$$ 0 =x^0_{i_2i_1} = \frac{1}{g} \sum_{g, j_1, j_2} r_{i_2j_2}(t^{-1})x_{j_2j_1}r_{j_1i_1}(t) $$ holds for any map $h$, i.e., it holds for any system of values $x_{j_2j_1}$. So let's pick a map $h$ for which $x_{j_2j_1} = 1$ for each pair $(j_2,j_1)$: the matrix with $1$ in each component.  
For the second question (now in the case where $h^0 = \lambda$), and he says "equating coefficients of the $x_{j_2j_1}$...", the main question, I think, is "why can we just equate coefficients?" Again we use that the equality 
$$ \frac{1}{g} \sum_{t, j_1, j_2} r_{i_2j_2}(t^{-1}) x_{j_2j_1}r_{j_1i_1}(t) = x^0_{i_2i_1} = \lambda\delta_{i_2i_1} = \frac{1}{n} \sum_{j_1,j_2} \delta_{i_2i_1}\delta_{j_2 j_1}x_{j_2j_1} \tag{$\ast$}$$
holds for any system of values of $x_{j_2j_1}$, i.e., for any map $h$. Fix $x_{j_{2_0}j_{1_0}}$. First let $h$ be the map that has $x_{j_{2_0}j_{1_0}}= 1$ and $0$ elsewhere. Then the equality ($\ast$)
reduces to 
$$ \frac{1}{g}\sum_{t\in G} r_{i_1j_{2_0}}(t^{-1}) r_{j_{1_0}i_1}(t)  = \frac{1}{n} \delta_{i_2i_1}\delta_{j_{2_0} j_{1_0}}.$$ 
Hence for all $(j_1,j_2)$
$$ \frac{1}{g}\sum_{t\in G} r_{i_1j_2}(t^{-1}) r_{j_1i_1}(t)  = \frac{1}{n} \delta_{i_2i_1}\delta_{j_2 j_1}$$
A: You have written that you agree that, in Case (1), the form vanishes for all systems of values of.. What he has done here is just explicitly writing down the coefficients and equating them to zero.
In the second part as well, you can take everything to the LHS, and use the same reasoning to claim that all the coefficients are zero, which is Corollary 3.
