= Reverse-order law for the Moore--Penrose inverse
{c}
For arbitrary real or complex <matrices>, the <Moore-Penrose inverse> need not reverse products:
$$
(AB)^\dagger\ne B^\dagger A^\dagger.
$$
For example, $A=(1\ \ 1)$ and $B=(1\ \ 0)^T$ give $AB=1$, while $A^\dagger=(1/2\ \ 1/2)^T$ and $B^\dagger=(1\ \ 0)$. Thus $(AB)^\dagger=1$ but $B^\dagger A^\dagger=1/2$.
A useful sufficient condition is that $A$ have <full column rank> and $B$ have <full row rank>, with their shared inner dimension $r$. Then $A^\dagger A=I_r$ and $BB^\dagger=I_r$. Put $C=AB$ and $X=B^\dagger A^\dagger$. Direct multiplication gives
$$
CXC=C,\qquad XCX=X,\qquad CX=AA^\dagger,\qquad XC=B^\dagger B.
$$
The last two products are <Hermitian> <orthogonal projection matrices>. All four <Penrose equations> therefore hold, proving $X=C^\dagger$. These rank conditions are sufficient, rather than necessary; checking the <Penrose equations> explains why the rule works in a particular case.
Back to article page
