The <Reverse-order law for the Moore--Penrose inverse> is false in general. Take
$$
A=\begin{pmatrix}1&0\\0&2\\0&0\end{pmatrix},
\qquad
B=\begin{pmatrix}1\\1\end{pmatrix}.
$$
Then
$$
(AB)^\dagger=\begin{pmatrix}1/5&2/5&0\end{pmatrix},
\qquad
B^\dagger A^\dagger=\begin{pmatrix}1/2&1/4&0\end{pmatrix}.
$$
A sufficient condition is
$$
\operatorname{rank}(A)=n,
\qquad
\operatorname{rank}(B)=n,
$$
so that $A$ has <full column rank> and $B$ has <full row rank>. Indeed $A^\dagger A=I_n$ and $BB^\dagger=I_n$; these identities make $B^\dagger A^\dagger$ satisfy all four <Penrose equations> for $AB$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 326 / 1 / b / iv / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/iii/paper-326/1/b/iv/solution