Let $m \ge 2$ be an integer, $R = M_{m}(K)$ be the algebra of all $m\times m$ matrices over $K$ an infinite field, if $L = [R,R]$ and $a,p \in R$ are non-zero elements. Then, if $[au, pu] = 0$ for all $u \in L$, then $a= \alpha p$ for some $\alpha \in K$.
I am reading a proof of above lemma and what I don't get is the following:
If $a = \Sigma_{ij}\ a_{ij}e_{ij}$ and $p =\Sigma_{ij} p_{ij}e_{ij}$ for some $a_{ij},p_{ij} \in K\backslash \{0\}$ with $1 \le i,j \le m$. Then, for every $1 \le i \neq j \le m$ we have $0 = e_{kk}[ae_{ij},pe_{ij}] = (a_{ki}p_{ji} - p_{ki}a_{ji})e_{kj}$
I can prove the first equality but have trouble with the second one. Is there some trick to see this or is this just plain calculation? I have tried to calculate the bracket explicitly by writing terms, but it gets a more complicated cause of matrix unit elements ($e_{ij}$). Maybe I missed some trivial detail, any suggestions would be much appreciated. Thanks.
It can be obtained by direct calculation: \begin{aligned} e_{kk}[ae_{ij},pe_{ij}] &=e_{kk}ae_{ij}pe_{ij}-e_{kk}pe_{ij}ae_{ij}\\ &=e_k(e_k^Tae_i)(e_j^Tpe_i)e_j^T-e_k(e_k^Tpe_i)(e_j^Tae_i)e_j^T\\ &=e_k(a_{ki}p_{ji}-p_{ki}a_{ji})e_j^T\\ &=(a_{ki}p_{ji}-p_{ki}a_{ji})e_{kj}. \end{aligned}