Serge Lang Algebra, Sec. XVII, Exercise 9:
Let $E$ be a finite-dimensional vector space over a field $k$. Let $R$ be a semisimple sub-algebra of $\operatorname{End}_k(E)$. Let $a, b \in R$. Assume that $$\ker b_E \supset \ker a_E$$ where $b_E$ is multiplication by $b$ on $E$ and similarly for $a_E$. Show that there exists an element $s \in R$ such that $sa = b$. [Hint: Reduce to $R$ simple. Then $R = \operatorname{End}_D(E_0)$ and $E = E_0^{(n)}$). Let $v_1,\dots, v_r$ be a $D$-basis for $aE$. Define $s$ by $s(av_j)= bv_j$, and extend $s$ by $D$-linearity. Then $sa_E=b_E$, so $sa=b$].
(Simple rings are assumed to be semisimple, and $E^{(n)}$ means the direct sum of $n$ copies of $E$.)
I'm having troubles understanding the hint here. Specifically in the $E=E_0^{(n)}$ part. I think what they meant is that by Artin-Wedderburn's theorem, we may assume $R=M_m(D)$ where $D$ is a division algebra over $k$. Then $M_m(D) \cong \operatorname{End}_D(V^{(m)})$ where $V$ is a simple left $D$-module. But I can't see how $E$ relates to $V^{(n)}$. Perhaps my interpretation is completely off?