= Solution
Assume first that $S\ne0$, as required for the irreducible-module conclusion. If $B\to\operatorname{End}_{\mathbb C}(S)$ is surjective, any nonzero invariant subspace contains every image of one of its nonzero vectors under all endomorphisms, hence is all of $S$. Thus $S$ is irreducible.
Conversely an irreducible $B$-module is finite dimensional: for $s\ne0$, $S=Bs$ is a quotient of the finite-dimensional <vector space> $B$. Let $D$ be the image of $B$ in $\operatorname{End}_{\mathbb C}(S)$. It acts faithfully and irreducibly. The subspace $J(D)S$ is a <submodule>, so is zero or all of $S$. The latter alternative would imply $J(D)^kS=S$ for every $k$, contradicting nilpotence of the <Jacobson radical>. Therefore $J(D)S=0$, and faithfulness gives $J(D)=0$.
The <Artin–Wedderburn theorem> makes $D$ a product of <matrix algebras>. A faithful <simple module> forces there to be just one factor, because all other factors would annihilate that <module>. Thus $D\cong M_d(\mathbb C)$ and $S\cong\mathbb C^d$, so the action is the full endomorphism algebra. This is the <Burnside matrix-algebra theorem>.
\b[For nonzero $S$, surjectivity is equivalent to irreducibility.] The zero <module> is a literal exception if it is admitted: its endomorphism algebra is zero, so the action map is surjective, whereas the zero <module> is not irreducible.
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