When I was writing a literature survey on Moore-Penrose pseudoinverse (literatures like this one, and this one), I encountered with the following equality which was named as reverse order law: $$(AB)^+=B^*(A^*ABB^*)^+A^*.$$ They claimed this equality, by noticing $$AB=((A^*)^+A^*A)(BB^*(B^*)^+)=(A^*)^+(A^*ABB^*)(B^*)^+,$$ and then applying a 'general reverse order law' $(PNQ)^+=Q^+N^+P^+$ here. In these papers, this law $(PNQ)^+=Q^+N^+P^+$ was widely used to derive equalities on $(AB)^+$.
My question here is that $(PNQ)^+=Q^+N^+P^+$ cannot be true for arbitrary matrix $P,N,Q$. Are there any extra requirements for it to hold, how is it proved, and how does it fit in the above case? Are there any references?
References on the first discovery of equality $(AB)^+=B^*(A^*ABB^*)^+A^*$ are also appreciated.
Thanks in advance.
The formula $$ (AB)^+ =B^*(A^*ABB^*)^+ A^* $$ doesn't hold in general. If you set $$ F =B^*(A^*ABB^*)^+ A^* $$ then $FABF=F$ and $ABFAB=AB$ (direct computation), but the two other conditions for the pseudo-inverse are only true under certain conditions (unknown for me).
Take for example $$ A=\left(\begin{array}{cc} 1&0\\ 0&0 \end{array}\right)\quad\text{and}\quad B=\left(\begin{array}{cc} 2&1\\ 1&2 \end{array}\right). $$ Then $$ AB=\left(\begin{array}{cc} 2&1\\ 0&0 \end{array}\right)\quad\text{and}\quad (AB)^+=\left(\begin{array}{cc} 2/5&0\\ 1/5&0 \end{array}\right). $$ However $$ F =B^*(A^*ABB^*)^+ A^*=\left(\begin{array}{cc} 14/41&0\\ 13/41&0 \end{array}\right), $$ since $$ (A^*ABB^*)=\left(\begin{array}{cc} 5&4\\ 0&0 \end{array}\right)\quad\text{and}\quad (A^*ABB^*)^+=\left(\begin{array}{cc} 5/41&0\\ 4/41&0 \end{array}\right), $$ and we also compute $$ (FAB)^*=\left(\begin{array}{cc} 28/41 & 26/41 \\ 14/41 & 13/41 \end{array}\right) \ne \left(\begin{array}{cc} 28/41 & 14/41\\ 26/41 & 13/41 \end{array}\right)=FAB. $$
A generally valid formula, which must be known (I made no literature research), is $$ (AB)^+ =(B^*A^*AB)^+ B^*A^*= B^*A^*(AB B^*A^*)^+. $$