$e_{i}Ae_{j} \neq 0$ implies that $e_{i}rad(A)e_{j} \neq 0$. Let $A$ be a finite dimensional algebra over an algebraically closed field with $rad^2(A)=0$. Let $\{e_{1},\dots,e_{n}\}$ be a complete set of primitive orthogonal idempotents. Suppose that $e_{i}Ae_{j} \neq 0$ for $i \neq j$. Why does this implies that $e_{i}rad(A)e_{j} \neq 0$? 
Would this follow from Malcev's theorem? or why exactly?

By Wedderburn theorem $B:=A/\mathrm{rad}(A)$ is just direct sum of matrix algebras.  Since $e_i$'s are orthogonal to each other, we can assume $B$ is direct sum of the underlying field, where $e_i$ is the projection onto the $i$-th copy.  Schur Lemma says that $e_i B e_j \cong \mathrm{Hom}_B(Be_i,Be_j)=0$.
If we have $\mathrm{Hom}_A(Ae_i,Ae_j)\cong e_iAe_j \neq 0$, but the subspace $\mathrm{Hom}_A(Ae_i,\mathrm{rad}(A)e_j)\cong e_i\mathrm{rad}(A)e_j=0$, then we have $e_iBe_j\neq 0$, a contradiction.
