BA(e_k)=B(A(e_k))=B(\Sigma_{i=1}^m \alpha_{ik} f_i) = \Sigma_{i=1}^m \alpha_{ik}B(f_i) = \Sigma_{i=1}^m \alpha_{ik} \Sigma_{j=1}^l \beta_{ji}g_j = \Sigma_{j=1}^l (\Sigma_{i=1}^m \beta_{ji}\alpha_{ik})g_j