 Reverse-order law for the Moore--Penrose inverse

For arbitrary real or complex matrices, the Moore-Penrose inverse need not reverse products:For example, $A=(11)$ and $B=(10{)}^T$ give $AB=1$, while $A^{†}=(1/21/2{)}^T$ and $B^{†}=(10)$. Thus $(AB{)}^{†}=1$ but $B^{†}{A}^{†}=1/2$.

A useful sufficient condition is that $A$ have full column rank and $B$ have full row rank, with their shared inner dimension $r$. Then $A^{†}A=I_r$ and $B{B}^{†}=I_r$. Put $C=AB$ and $X=B^{†}{A}^{†}$. Direct multiplication givesThe last two products are Hermitian orthogonal projection matrices. All four Penrose equations therefore hold, proving $X=C^{†}$. These rank conditions are sufficient, rather than necessary; checking the Penrose equations explains why the rule works in a particular case.