If the field is $\mathbb C$, There are many ways of seeing this. For $i=1,2$ we may replace $G_i$ by its Zariski closure in $GL(V_i)$ without changing the hypotheses or the conclusion. But if an algebraic subgroup $G\subset GL(V)$ is irreducible, then it is reductive (the unipotent radical will have a fixed space which is G invariant and hence zero). 

We now use the fact that the  reductive group $G_i$  has a Zariski dense compact group $K_i$; we may thus replace $G_i$ by the compact $K_i$ where you have accepted the proof.