Endomorphism algebra of irreducible module over a group algebra

Solution 1:

It's actually not "less or equal", its "equal". There are no shenanigans with dimension over skew fields, a subspace generated by one non-zero vector is a line, and is isomorphic to $E$ (as an $E$-module, or $F$-vector space).

So no, for any $v_0\neq 0$, if $Ev_0\subseteq Ev$, then $Ev_0=Ev$.

Solution 2:

Captain Lama has already given an answer. Here I would like to explain detailly.

Theorem Let $E$ be an unital ring. Then $E$ is a division ring (or skew field) if and only if every left $E$-module is free. (see https://planetmath.org/ringswhoseeverymoduleisfree)

Take $0\neq v\in V$, by this theorem, $f:E\rightarrow Ev$ where $e\mapsto ev$ for $e\in E$ is an isomorphism of $E$-modules. So $$ \sum_i k_i(e_iv)=0 \Leftrightarrow \sum_i (k_ie_i)v=0 \Leftrightarrow \sum_i k_ie_i=0 $$ where the last iff holds as $E$ being a division ring. Thus $\dim_FE=\dim_FEv$ for all $0\neq v\in V$, and the answer for the question is no since the assumption $\dim_FE=\dim_FV$.