Prove that $A^HB^H=(BA)^H$ I want to prove this statement where A and B are matrices and H the Hermitian. Ok, Here is a proof from MathWorld: $$\overline{((ab)^T)_{ij}}=\overline{(b^Ta^T)_{ij}}=\overline{{b^T_{ik}}a^T_{kj}}=(b^Ha^H)_{ij}$$ With one or two steps disregarded. I understand the proof but I fail to see how the coordinate k pops up and disappears. Could you make a brief explanation or if there are otherways to prove this, could you explain them (seems this is the only straightforward way, though) : $A^HB^H=(BA)^H$

The line is written in Einstein sum convention, meaning that there is an implicit sum over the index $k$ that appears twice. With the regular matrix multiplication (row times column) we have for $A = a_{ij}$ and $B=b_{ij}$ $$ (A\cdot B)_{ij} = \sum_k a_{ik} b_{kj} $$