Theorem: (Lang's Algebra, p. 264) Let $E$ be any field, $G$ be a finite group of automorphisms of $E$, and $F:=E^G$ be fixed field of $G$: $$ E^G=\{\alpha\in E : \sigma(\alpha)=\alpha \forall \sigma\in G\}. $$ Then $[E:F]=|G|$.
Althugh this theorem has a different proof than the one given by Lang (for ex. in Artin's Galois Theory), the proofs proceed by showing:
If $[E:F]<|G|$ or $[E:F]>|G|$ then we get a contradiction.
Question: Let $V=E$ and $W=\{ \sum_{i} \lambda_i\sigma_i : \lambda_i\in F, \sigma_i\in G\}$.
Then $V$ and $W$ are vector spaces over $F$ of dimension $[E:F]$ and $|G|$ respectively.
Consider the map $$W\times V\rightarrow V, \,\,\,\,\,\,\,\,(\sum_i \lambda_i\sigma_i ,a)\mapsto \sum_i \lambda_i\sigma_i(a).$$ This is a bilinear map. Is it true that this map is non-degenerate?
Question 2. If above map is non-degenerate, is it always the situation in following general situation, such as: If $U,U'$ are finite dimensional vector spaces, $\phi:U\times U'\rightarrow U'$ is a bilinear map, and $\dim U\neq \dim U'$, then this map is degenerate? If yes, can one suggest some lines of proof or some reference?
Question 1. Yes, it is. If $f(a)=0$ for all $a\in V$, then $f=0$ (definition of functions). If $f(a)=0$ for all $f\in W$, take $f=Id_E\in G\subset W$. This shows that $a=0$.
Question 2. No, take $U=F,U'=F^2$, and $\phi:(\lambda,v)\in U\times U'\mapsto \lambda v\in U'$.
This map is clearly $F$-bilinear, and non degenerate: if $\lambda v=0$ for all $v\in U'$, picking $v\neq 0$ shows $\lambda=0$. If $\lambda v=0$ for all $\lambda\in U,$ setting $\lambda=1$ shows $v=0$.