Let $A$ and $B$ are two matrices where $A \in \mathbb{R}^{m\times p}$ and $B \in \mathbb{R}^{p\times n}$ and both $A$ and $B$ are full rank matrices Now I really want to know in what cases
$(AB)^+ = B^+A^+$ ,where $A^+$ is Moore-Penrose Pseudo-inverse of $A$
Here $m,p$ and $n$ are in any order like $m<p<d$ or $m>p<d$ etc.
$$ \mathbf{A} = \left[ \begin{array}{cc} 0 & 1 \\ 3 & 2 \\ 0 & 2 \\ \end{array} \right] , \quad % \mathbf{B} = \left[ \begin{array}{ccc} 0 & 3 & 0 \\ 1 & 2 & 2 \\ \end{array} \right] , \quad % \mathbf{C} = \mathbf{A} \mathbf{B} = \left[ \begin{array}{rrr} 1 & 2 & 2 \\ 2 & 13 & 4 \\ 2 & 4 & 4 \\ \end{array} \right] $$ $$ \mathbf{A}^{\dagger} = \frac{1}{15} \left[ \begin{array}{rrr} -2 & 5 & -4 \\ 3 & 0 & 6 \\ \end{array} \right] , \quad % \mathbf{B}^{\dagger} = \frac{1}{15} \left[ \begin{array}{rr} -2 & 3 \\ 5 & 0 \\ -4 & 6 \\ \end{array} \right] , \quad % \mathbf{C}^{\dagger} = \frac{1}{225} \left[ \begin{array}{rrr} 13 & -10 & 26 \\ -10 & 25 & -20 \\ 26 & -20 & 52 \\ \end{array} \right] $$
Does $\mathbf{C}^{\dagger} = \mathbf{B}^{\dagger}\mathbf{A}^{\dagger}$? $$ \begin{align} \mathbf{B}^{\dagger}\mathbf{A}^{\dagger} &= \frac{1}{15} \left[ \begin{array}{rr} -2 & 3 \\ 5 & 0 \\ -4 & 6 \\ \end{array} \right] \frac{1}{15} \left[ \begin{array}{rrr} -2 & 5 & -4 \\ 3 & 0 & 6 \\ \end{array} \right] \\[3pt] %% & = %% \frac{1}{225} \left[ \begin{array}{rrr} 13 & -10 & 26 \\ -10 & 25 & -20 \\ 26 & -20 & 52 \\ \end{array} \right] \\ &= \mathbf{C}^{\dagger} \end{align} $$
$$\therefore \qquad \left(\mathbf{A}\mathbf{B}\right)^{\dagger} = \mathbf{B}^{\dagger}\mathbf{A}^{\dagger}$$
